This is a modern-English version of Treatise on light : In which are explained the causes of that which occurs in reflexion, & in refraction and particularly in the strange refraction of Iceland crystal, originally written by Huygens, Christiaan. It has been thoroughly updated, including changes to sentence structure, words, spelling, and grammar—to ensure clarity for contemporary readers, while preserving the original spirit and nuance. If you click on a paragraph, you will see the original text that we modified, and you can toggle between the two versions.

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TREATISE ON LIGHT

In which are explained
The causes of that which occurs
In REFLEXION, & in REFRACTION

In which are explained
The causes of what happens
In REFLECTION, & in REFRACTION

And particularly
In the strange REFRACTION
OF ICELAND CRYSTAL

And especially
In the unusual REFRACTION
OF ICELAND CRYSTAL

By

CHRISTIAAN HUYGENS

Rendered into English

Translated into English

By

By

SILVANUS P. THOMPSON

Silvanus P. Thompson

 

 

University of Chicago Press

 

 

 

[Pg v]

PREFACE

I

wrote this Treatise during my sojourn in France twelve years ago, and I communicated it in the year 1678 to the learned persons who then composed the Royal Academy of Science, to the membership of which the King had done me the honour of calling, me. Several of that body who are still alive will remember having been present when I read it, and above the rest those amongst them who applied themselves particularly to the study of Mathematics; of whom I cannot cite more than the celebrated gentlemen Cassini, Römer, and De la Hire. And, although I have since corrected and changed some parts, the copies which I had made of it at that time may serve for proof that I have yet added nothing to it save some conjectures touching the formation of Iceland Crystal, and a novel observation on the refraction of Rock Crystal. I have desired to relate these particulars to make known how long I have meditated the things which now I publish, and not for the purpose of detracting from the merit of those who, without having seen anything that I have written, may be found to have treated [Pg vi]of like matters: as has in fact occurred to two eminent Geometricians, Messieurs Newton and Leibnitz, with respect to the Problem of the figure of glasses for collecting rays when one of the surfaces is given.

I wrote this Treatise during my time in France twelve years ago, and I shared it in 1678 with the scholars who made up the Royal Academy of Science, to which the King honored me by extending an invitation. Several members who are still around will remember being present when I read it, especially those among them who focused on Mathematics, including the renowned gentlemen Cassini, Römer, and De la Hire. Although I have since revised some parts, the copies I made back then can show that I’ve only added some speculations about the formation of Iceland Crystal and a new observation on the refraction of Rock Crystal. I wanted to share these details to show how long I’ve thought about the topics I’m now publishing, not to diminish the contributions of others who, without having seen my work, may have addressed similar matters: as has indeed happened with two prominent mathematicians, Messieurs Newton and Leibnitz, regarding the problem of designing lenses for collecting rays when one of the surfaces is specified.

One may ask why I have so long delayed to bring this work to the light. The reason is that I wrote it rather carelessly in the Language in which it appears, with the intention of translating it into Latin, so doing in order to obtain greater attention to the thing. After which I proposed to myself to give it out along with another Treatise on Dioptrics, in which I explain the effects of Telescopes and those things which belong more to that Science. But the pleasure of novelty being past, I have put off from time to time the execution of this design, and I know not when I shall ever come to an end if it, being often turned aside either by business or by some new study. Considering which I have finally judged that it was better worth while to publish this writing, such as it is, than to let it run the risk, by waiting longer, of remaining lost.

One might wonder why it took me so long to publish this work. The reason is that I wrote it rather carelessly in the language you see here, intending to translate it into Latin later to gain more attention for the subject. I also planned to release it alongside another treatise on Dioptrics, where I discuss the effects of telescopes and related topics. However, as the excitement of newness faded, I kept postponing this project, and I don't know when I will ever finish it, often distracted by other responsibilities or new studies. Considering all this, I've decided that it’s better to publish this writing as it is than to risk it being lost if I wait any longer.

There will be seen in it demonstrations of those kinds which do not produce as great a certitude as those of Geometry, and which even differ much therefrom, since whereas the Geometers prove their Propositions by fixed and incontestable Principles, here the Principles are verified by the conclusions to be drawn from them; the nature of these things not allowing of this being done otherwise.

There will be demonstrations shown that don't provide as much certainty as those in Geometry, and they are quite different from them. While Geometers prove their statements using established and undeniable principles, here the principles are confirmed by the conclusions drawn from them; the nature of these matters prevents this from being done in any other way.

It is always possible to attain thereby to a degree of probability which very often is scarcely less than complete proof. To wit, when things which have been demonstrated by the Principles that have been assumed correspond perfectly to the phenomena which experiment has brought under observation; especially when there are a great number of [Pg vii]them, and further, principally, when one can imagine and foresee new phenomena which ought to follow from the hypotheses which one employs, and when one finds that therein the fact corresponds to our prevision. But if all these proofs of probability are met with in that which I propose to discuss, as it seems to me they are, this ought to be a very strong confirmation of the success of my inquiry; and it must be ill if the facts are not pretty much as I represent them. I would believe then that those who love to know the Causes of things and who are able to admire the marvels of Light, will find some satisfaction in these various speculations regarding it, and in the new explanation of its famous property which is the main foundation of the construction of our eyes and of those great inventions which extend so vastly the use of them.

It is always possible to achieve a level of probability that is often almost as good as complete proof. For example, when the things demonstrated by the principles we've assumed perfectly match the phenomena observed through experiments, especially when there are many such instances, and furthermore, primarily, when we can imagine and anticipate new phenomena that should follow from the hypotheses we use, and then find that the reality aligns with our expectations. If all these proofs of probability appear in what I intend to discuss, as I believe they do, this should strongly confirm the success of my inquiry; and it would be difficult to argue against the facts being largely as I present them. I believe that those who are curious about the causes of things and who appreciate the wonders of light will find some satisfaction in these various speculations about it, and in the new explanation of its famous properties, which is the main basis for how our eyes are constructed and for those great inventions that greatly expand their usage.

I hope also that there will be some who by following these beginnings will penetrate much further into this question than I have been able to do, since the subject must be far from being exhausted. This appears from the passages which I have indicated where I leave certain difficulties without having resolved them, and still more from matters which I have not touched at all, such as Luminous Bodies of several sorts, and all that concerns Colours; in which no one until now can boast of having succeeded. Finally, there remains much more to be investigated touching the nature of Light which I do not pretend to have disclosed, and I shall owe much in return to him who shall be able to supplement that which is here lacking to me in knowledge. The Hague. The 8 January 1690.[Pg viii]

I also hope that some people, by exploring these initial ideas, will go much deeper into this topic than I have been able to, since there’s still so much to uncover. This is evident in the sections I've pointed out where I leave some challenges unresolved, and even more so in the areas I haven't even touched on, like various types of Luminous Bodies and everything related to Colors; no one has really nailed that down yet. Lastly, there’s so much more to explore regarding the nature of Light, which I don't claim to have fully explained, and I will be very grateful to anyone who can fill in the gaps in my knowledge. The Hague. January 8, 1690.[Pg viii]

[Pg ix]
.

NOTE BY THE TRANSLATOR

C

onsidering the great influence which this Treatise has exercised in the development of the Science of Optics, it seems strange that two centuries should have passed before an English edition of the work appeared. Perhaps the circumstance is due to the mistaken zeal with which formerly everything that conflicted with the cherished ideas of Newton was denounced by his followers. The Treatise on Light of Huygens has, however, withstood the test of time: and even now the exquisite skill with which he applied his conception of the propagation of waves of light to unravel the intricacies of the phenomena of the double refraction of crystals, and of the refraction of the atmosphere, will excite the admiration of the student of Optics. It is true that his wave theory was far from the complete doctrine as subsequently developed by Thomas Young and Augustin Fresnel, and belonged rather to geometrical than to physical Optics. If Huygens had no conception of transverse vibrations, of the principle of interference, or of the existence of the ordered sequence of waves in trains, he nevertheless attained to a remarkably clear understanding of the prin[Pg x]ciples of wave-propagation; and his exposition of the subject marks an epoch in the treatment of Optical problems. It has been needful in preparing this translation to exercise care lest one should import into the author's text ideas of subsequent date, by using words that have come to imply modern conceptions. Hence the adoption of as literal a rendering as possible. A few of the author's terms need explanation. He uses the word "refraction," for example, both for the phenomenon or process usually so denoted, and for the result of that process: thus the refracted ray he habitually terms "the refraction" of the incident ray. When a wave-front, or, as he terms it, a "wave," has passed from some initial position to a subsequent one, he terms the wave-front in its subsequent position "the continuation" of the wave. He also speaks of the envelope of a set of elementary waves, formed by coalescence of those elementary wave-fronts, as "the termination" of the wave; and the elementary wave-fronts he terms "particular" waves. Owing to the circumstance that the French word rayon possesses the double signification of ray of light and radius of a circle, he avoids its use in the latter sense and speaks always of the semi-diameter, not of the radius. His speculations as to the ether, his suggestive views of the structure of crystalline bodies, and his explanation of opacity, slight as they are, will possibly surprise the reader by their seeming modernness. And none can read his investigation of the phenomena found in Iceland spar without marvelling at his insight and sagacity.

Considering the significant impact this Treatise has had on the development of the science of optics, it's surprising that two centuries passed before an English edition of the work was published. This delay might be due to the misguided passion with which Newton's followers denounced anything that contradicted their cherished beliefs. However, Huygens' Treatise on Light has stood the test of time: even today, the exceptional skill he applied in using his wave theory to explain the complexities of double refraction in crystals and atmospheric refraction continues to inspire admiration among optics students. While it's true that his wave theory was not as comprehensive as later developments by Thomas Young and Augustin Fresnel, and was more aligned with geometrical optics than physical optics, Huygens achieved a remarkably clear understanding of wave propagation principles. His explanation of the topic marks a significant milestone in tackling optical problems. In preparing this translation, care has been taken to avoid importing later ideas into the author’s text by using terms that have evolved to imply modern concepts. Hence, a very literal translation was chosen. A few of the author's terms require clarification. For instance, he uses "refraction" both to refer to the process and the result of that process: he habitually calls the refracted ray "the refraction" of the incident ray. When a wave-front, or what he calls a "wave," moves from one position to another, he refers to the wave-front in its new position as "the continuation" of the wave. He also describes the envelope created by the merging of elementary wave-fronts as "the termination" of the wave, and he calls the elementary wave-fronts "particular" waves. Because the French word rayon means both a ray of light and the radius of a circle, he avoids using it in the latter sense and always refers to the semi-diameter instead of the radius. His thoughts on the ether, his insightful ideas about the structure of crystals, and his explanation of opacity, although minimal, might surprise readers with how modern they seem. No one can read his investigation into the phenomena of Iceland spar without marveling at his insight and keen understanding.

S.P.T.

S.P.T.

June, 1912.

June 1912.


[Pg xi]

TABLE OF MATTERS

Contained in this Treatise

CHAP. I. On Rays Propagated in Straight Lines.
That Light is produced by a certain movement.p. 3
That no substance passes from the luminous object to the eyes.p. 3
That Light spreads spherically, almost as Sound does.p. 4
Whether Light takes time to spread.p. 4
Experience seeming to prove that it passes instantaneously.p. 5
Experience proving that it takes time.p. 8
How much its speed is greater than that of Sound.p. 10
In what the emission of Light differs from that of Sound.p. 10
That it is not the same medium which serves for Light and Sound.p. 11
How Sound is propagated.p. 12
How Light is propagated.p. 14
Detailed Remarks on the propagation of Light.p. 15
Why Rays are propagated only in straight lines.p. 20
How Light coming in different directions can cross itself.p. 22
CHAP. II. On Reflexion.
Demonstration of equality of angles of incidence and reflexion.p. 23
Why the incident and reflected rays are in the same plane perpendicular to the reflecting surface.p. 25
That it is not needful for the reflecting surface to be perfectly flat to attain equality of the angles of incidence and reflexion.p. 27
CHAP. III. On Refraction.
That bodies may be transparent without any substance passing through them.p. 29
Proof that the ethereal matter passes through transparent bodies.p. 30
How this matter passing through can render them transparent.p. 31
That the most solid bodies in appearance are of a very loose texture.p. 31
That Light spreads more slowly in water and in glass than in air.p. 32
Third hypothesis to explain transparency, and the retardation which Light suffers.p. 32
On that which makes bodies opaque.p. 34
Demonstration why Refraction obeys the known proportion of Sines.p. 35
Why the incident and refracted Rays produce one another reciprocally.p. 39
Why Reflexion within a triangular glass prism is suddenly augmented when the Light can no longer penetrate.p. 40
That bodies which cause greater Refraction also cause stronger Reflexion.p. 42
Demonstration of the Theorem of Mr. Fermat.p. 43
CHAP. IV. On the Refraction of the Air.
That the emanations of Light in the air are not spherical.p. 45
How consequently some objects appear higher than they are.p. 47
How the Sun may appear on the Horizon before he has risen.p. 49
That the rays of light become curved in the Air of the Atmosphere, and what effects this produces.p. 50
CHAP. V. On the Strange Refraction of Iceland Crystal.
That this Crystal grows also in other countries.p. 52
Who first-wrote about it.p. 53
Description of Iceland Crystal; its substance, shape, and properties.p. 53
That it has two different Refractions.p. 54
That the ray perpendicular to the surface suffers refraction, and that some rays inclined to the surface pass without suffering refraction.p. 55
Observation of the refractions in this Crystal.p. 56
That there is a Regular and an Irregular Refraction.p. 57
The way of measuring the two Refractions of Iceland Crystal.p. 57
Remarkable properties of the Irregular Refraction.p. 60
Hypothesis to explain the double Refraction.p. 61
That Rock Crystal has also a double Refraction.p. 62
Hypothesis of emanations of Light, within Iceland Crystal, of spheroidal form, for the Irregular Refraction.p. 63
How a perpendicular ray can suffer Refraction.p. 64
How the position and form of the spheroidal emanations in this Crystal can be defined.p. 65
Explanation of the Irregular Refraction by these spheroidal emanations.p. 67
Easy way to find the Irregular Refraction of each incident ray.p. 70
Demonstration of the oblique ray which traverses the Crystal without being refracted.p. 73
Other irregularities of Refraction explained.p. 76
That an object placed beneath the Crystal appears double, in two images of different heights.p. 81
Why the apparent heights of one of the images change on changing the position of the eyes above the Crystal.p. 85
Of the different sections of this Crystal which produce yet other refractions, and confirm all this Theory.p. 88
Particular way of polishing the surfaces after it has been cut.p. 91
Surprising phenomenon touching the rays which pass through two separated pieces; the cause of which is not explained.p. 92
Probable conjecture on the internal composition of Iceland Crystal, and of what figure its particles are.p. 95
Tests to confirm this conjecture.p. 97
Calculations which have been supposed in this Chapter.p. 99
CHAP. VI. On the Figures of transparent bodies which serve for Refraction and for Reflexion.
General and easy rule to find these Figures.p. 106
Invention of the Ovals of Mr. Des Cartes for Dioptrics.p. 109
How he was able to find these Lines.p. 114
Way of finding the surface of a glass for perfect refraction, when the other surface is given.p. 116
Remark on what happens to rays refracted at a spherical surface.p. 123
Remark on the curved line which is formed by reflexion in a spherical concave mirror.p. 126

[Pg 1]

TREATISE ON LIGHT

CHAPTER I

ON RAYS PROPAGATED IN STRAIGHT LINES

A

s happens in all the sciences in which Geometry is applied to matter, the demonstrations concerning Optics are founded on truths drawn from experience. Such are that the rays of light are propagated in straight lines; that the angles of reflexion and of incidence are equal; and that in refraction the ray is bent according to the law of sines, now so well known, and which is no less certain than the preceding laws.

As is the case in all sciences where Geometry is applied to physical matter, the demonstrations related to Optics are based on truths from experience. These include the facts that light rays travel in straight lines, the angles of reflection and incidence are equal, and that in refraction, the ray bends according to the well-known law of sines, which is as certain as the previous laws.

The majority of those who have written touching the various parts of Optics have contented themselves with presuming these truths. But some, more inquiring, have desired to investigate the origin and the causes, considering these to be in themselves wonderful effects of Nature. In which they advanced some ingenious things, but not however such that the most intelligent folk do not wish for better and more satisfactory explanations. Wherefore I here desire to propound what I have meditated on the sub[Pg 2]ject, so as to contribute as much as I can to the explanation of this department of Natural Science, which, not without reason, is reputed to be one of its most difficult parts. I recognize myself to be much indebted to those who were the first to begin to dissipate the strange obscurity in which these things were enveloped, and to give us hope that they might be explained by intelligible reasoning. But, on the other hand I am astonished also that even here these have often been willing to offer, as assured and demonstrative, reasonings which were far from conclusive. For I do not find that any one has yet given a probable explanation of the first and most notable phenomena of light, namely why it is not propagated except in straight lines, and how visible rays, coming from an infinitude of diverse places, cross one another without hindering one another in any way.

Most of those who have written about different aspects of Optics have simply accepted these truths without question. However, some, who are more curious, have sought to explore the origins and causes, viewing them as remarkable effects of Nature. They proposed some clever ideas, but they still leave the most intelligent people wanting better and more satisfactory explanations. Therefore, I would like to share my thoughts on the subject, in order to contribute as much as I can to the understanding of this area of Natural Science, which is rightly regarded as one of the most challenging parts. I acknowledge that I owe a lot to those who first began to unravel the strange confusion surrounding these topics and gave us hope that they could be clarified with clear reasoning. Yet, I’m also surprised that even in this field, some have often presented their arguments as if they were conclusive, when they are far from it. I have yet to find anyone who has provided a likely explanation for the first and most significant phenomena of light, specifically why it only travels in straight lines, and how visible rays, coming from countless different sources, can cross each other without interfering in any way.

I shall therefore essay in this book, to give, in accordance with the principles accepted in the Philosophy of the present day, some clearer and more probable reasons, firstly of these properties of light propagated rectilinearly; secondly of light which is reflected on meeting other bodies. Then I shall explain the phenomena of those rays which are said to suffer refraction on passing through transparent bodies of different sorts; and in this part I shall also explain the effects of the refraction of the air by the different densities of the Atmosphere.

I will attempt in this book to provide, in line with the principles accepted in today's philosophy, some clearer and more likely explanations. First, I'll discuss the properties of light that travels in straight lines; second, I'll cover light that gets reflected when it encounters other objects. Then, I'll explain the phenomena of rays that are said to bend when passing through different types of transparent materials; in this section, I'll also discuss how the varying densities of the atmosphere affect the refraction of light.

Thereafter I shall examine the causes of the strange refraction of a certain kind of Crystal which is brought from Iceland. And finally I shall treat of the various shapes of transparent and reflecting bodies by which rays are collected at a point or are turned aside in various ways. From this it will be seen with what facility, following our new Theory, we find not only the Ellipses, Hyperbolas, and [Pg 3]other curves which Mr. Des Cartes has ingeniously invented for this purpose; but also those which the surface of a glass lens ought to possess when its other surface is given as spherical or plane, or of any other figure that may be.

After that, I’ll look into why a certain type of crystal from Iceland bends light in such a strange way. Finally, I’ll discuss the different shapes of transparent and reflective surfaces that gather light at a point or redirect it in various directions. This will show how easily, according to our new theory, we can identify not just the ellipses, hyperbolas, and [Pg 3]other curves that Mr. Des Cartes cleverly created for this purpose, but also the shapes that the surface of a glass lens should have when its other surface is either spherical, flat, or any other shape.

It is inconceivable to doubt that light consists in the motion of some sort of matter. For whether one considers its production, one sees that here upon the Earth it is chiefly engendered by fire and flame which contain without doubt bodies that are in rapid motion, since they dissolve and melt many other bodies, even the most solid; or whether one considers its effects, one sees that when light is collected, as by concave mirrors, it has the property of burning as a fire does, that is to say it disunites the particles of bodies. This is assuredly the mark of motion, at least in the true Philosophy, in which one conceives the causes of all natural effects in terms of mechanical motions. This, in my opinion, we must necessarily do, or else renounce all hopes of ever comprehending anything in Physics.

It’s hard to doubt that light is essentially the movement of some kind of matter. When you think about how it’s created, especially on Earth, it mainly comes from fire and flames, which undeniably involve rapidly moving particles, since they can dissolve and melt even the toughest materials. Or if you consider its effects, you notice that when light is focused, like with concave mirrors, it can burn just like fire does; in other words, it breaks apart the particles of objects. This is definitely a sign of motion, at least according to true philosophy, which explains all natural effects through mechanical movements. In my view, we have to approach it this way, or we’ll have to give up any hope of truly understanding anything in physics.

And as, according to this Philosophy, one holds as certain that the sensation of sight is excited only by the impression of some movement of a kind of matter which acts on the nerves at the back of our eyes, there is here yet one reason more for believing that light consists in a movement of the matter which exists between us and the luminous body.

And according to this philosophy, it’s accepted that the sensation of sight is triggered only by the impression of some movement of a type of matter that affects the nerves at the back of our eyes. This provides yet another reason to believe that light is a movement of the matter that exists between us and the light source.

Further, when one considers the extreme speed with which light spreads on every side, and how, when it comes from different regions, even from those directly opposite, the rays traverse one another without hindrance, one may well understand that when we see a luminous object, it cannot be by any transport of matter coming to us from this object, [Pg 4]in the way in which a shot or an arrow traverses the air; for assuredly that would too greatly impugn these two properties of light, especially the second of them. It is then in some other way that light spreads; and that which can lead us to comprehend it is the knowledge which we have of the spreading of Sound in the air.

Furthermore, when you think about how incredibly fast light travels in all directions, and how, even when it comes from opposite sides, the rays cross each other without any obstruction, it's easy to see that when we perceive a light source, it can't be because any physical matter is coming from that object to us, [Pg 4]like how a bullet or an arrow moves through the air; because that would seriously contradict these two properties of light, especially the second one. So, light spreads in a different way; and what can help us understand it is our knowledge of how sound moves through the air.

We know that by means of the air, which is an invisible and impalpable body, Sound spreads around the spot where it has been produced, by a movement which is passed on successively from one part of the air to another; and that the spreading of this movement, taking place equally rapidly on all sides, ought to form spherical surfaces ever enlarging and which strike our ears. Now there is no doubt at all that light also comes from the luminous body to our eyes by some movement impressed on the matter which is between the two; since, as we have already seen, it cannot be by the transport of a body which passes from one to the other. If, in addition, light takes time for its passage—which we are now going to examine—it will follow that this movement, impressed on the intervening matter, is successive; and consequently it spreads, as Sound does, by spherical surfaces and waves: for I call them waves from their resemblance to those which are seen to be formed in water when a stone is thrown into it, and which present a successive spreading as circles, though these arise from another cause, and are only in a flat surface.

We know that sound travels through air, which is an invisible and formless substance, by a movement that is passed on from one part of the air to another. This movement spreads out equally fast in all directions and creates expanding spherical surfaces that reach our ears. It’s also clear that light comes from a light source to our eyes through some kind of movement acting on the matter in between; since, as we've already discussed, it can't just be the transfer of a physical object from one to the other. If light indeed takes time to travel—which we are about to explore—then this movement on the intervening matter happens in a sequence, and it spreads out like sound does, in spherical surfaces and waves. I refer to them as waves because they resemble the ripples formed in water when a stone is thrown in, which expand outwards in circles, even though those come from a different cause and only occur on a flat surface.

To see then whether the spreading of light takes time, let us consider first whether there are any facts of experience which can convince us to the contrary. As to those which can be made here on the Earth, by striking lights at great distances, although they prove that light takes no sensible time to pass over these distances, one may say with good [Pg 5]reason that they are too small, and that the only conclusion to be drawn from them is that the passage of light is extremely rapid. Mr. Des Cartes, who was of opinion that it is instantaneous, founded his views, not without reason, upon a better basis of experience, drawn from the Eclipses of the Moon; which, nevertheless, as I shall show, is not at all convincing. I will set it forth, in a way a little different from his, in order to make the conclusion more comprehensible.

To determine if the spread of light takes time, let’s first look at any experiences that might suggest otherwise. For those that can be tested here on Earth, like lighting something at great distances, while they demonstrate that light takes no noticeable time to travel these distances, one might reasonably argue that these distances are too small, leading us to conclude that the passage of light is incredibly fast. Mr. Des Cartes believed it was instantaneous, basing his views—reasonably enough—on better experiences drawn from lunar eclipses; however, as I will explain, this is not particularly convincing. I'll present this in a slightly different way to make the conclusion clearer.

Let A be the place of the sun, BD a part of the orbit or annual path of the Earth: ABC a straight line which I suppose to meet the orbit of the Moon, which is represented by the circle CD, at C.

Let A be the position of the sun, BD a section of the Earth's orbit or yearly path: ABC a straight line that I assume intersects the Moon's orbit, represented by the circle CD, at point C.

Now if light requires time, for example one hour, to traverse the space which is between the Earth and the Moon, it will follow that the Earth having arrived at B, the shadow which it casts, or the interruption of the light, will not yet have arrived at the point C, but will only arrive there an hour after. It will then be one hour after, reckoning from the moment when the Earth was at B, [Pg 6]that the Moon, arriving at C, will be obscured: but this obscuration or interruption of the light will not reach the Earth till after another hour. Let us suppose that the Earth in these two hours will have arrived at E. The Earth then, being at E, will see the Eclipsed Moon at C, which it left an hour before, and at the same time will see the sun at A. For it being immovable, as I suppose with Copernicus, and the light moving always in straight lines, it must always appear where it is. But one has always observed, we are told, that the eclipsed Moon appears at the point of the Ecliptic opposite to the Sun; and yet here it would appear in arrear of that point by an amount equal to the angle GEC, the supplement of AEC. This, however, is contrary to experience, since the angle GEC would be very sensible, and about 33 degrees. Now according to our computation, which is given in the Treatise on the causes of the phenomena of Saturn, the distance BA between the Earth and the Sun is about twelve thousand diameters of the Earth, and hence four hundred times greater than BC the distance of the Moon, which is 30 diameters. Then the angle ECB will be nearly four hundred times greater than BAE, which is five minutes; namely, the path which the earth travels in two hours along its orbit; and thus the angle BCE will be nearly 33 degrees; and likewise the angle CEG, which is greater by five minutes.

If light takes about an hour to travel the distance between the Earth and the Moon, then when the Earth reaches point B, the shadow it casts—or the interruption of light—won't have reached point C yet; it will only arrive there an hour later. So, an hour after the Earth was at B, when the Moon reaches C, it will be in shadow; however, this interruption of light won't reach the Earth until another hour passes. If we assume that during these two hours the Earth moves to point E, then at that time, the Earth at E will see the eclipsed Moon at C, which it passed one hour earlier, while simultaneously seeing the Sun at A. Since the Sun is stationary, as I assume with Copernicus, and light always travels in straight lines, it must always be seen where it is. Nevertheless, it has been observed that the eclipsed Moon appears opposite the Sun on the ecliptic; yet here it would seem to lag behind that point by an amount equal to angle GEC, which is the supplement of angle AEC. This contradicts our observations because angle GEC would be quite noticeable, around 33 degrees. According to our calculations in the Treatise on the causes of the phenomena of Saturn, the distance BA between the Earth and the Sun is about twelve thousand Earth diameters, making it four hundred times greater than the distance BC to the Moon, which is 30 diameters. Hence, angle ECB would be nearly four hundred times larger than angle BAE, which is five minutes—the distance the Earth travels in two hours along its orbit. Consequently, angle BCE would be nearly 33 degrees, along with angle CEG, which exceeds it by five minutes.

But it must be noted that the speed of light in this argument has been assumed such that it takes a time of one hour to make the passage from here to the Moon. If one supposes that for this it requires only one minute of time, then it is manifest that the angle CEG will only be 33 minutes; and if it requires only ten seconds of time, [Pg 7]the angle will be less than six minutes. And then it will not be easy to perceive anything of it in observations of the Eclipse; nor, consequently, will it be permissible to deduce from it that the movement of light is instantaneous.

But it should be noted that in this discussion, the speed of light is assumed to take one hour to travel from here to the Moon. If we assume that it only takes one minute, then it’s clear that angle CEG will only be 33 minutes; and if it takes just ten seconds, [Pg 7]the angle will be less than six minutes. At that point, it won’t be easy to see anything in the observations of the eclipse; therefore, it wouldn’t be reasonable to conclude that the movement of light is instantaneous.

It is true that we are here supposing a strange velocity that would be a hundred thousand times greater than that of Sound. For Sound, according to what I have observed, travels about 180 Toises in the time of one Second, or in about one beat of the pulse. But this supposition ought not to seem to be an impossibility; since it is not a question of the transport of a body with so great a speed, but of a successive movement which is passed on from some bodies to others. I have then made no difficulty, in meditating on these things, in supposing that the emanation of light is accomplished with time, seeing that in this way all its phenomena can be explained, and that in following the contrary opinion everything is incomprehensible. For it has always seemed tome that even Mr. Des Cartes, whose aim has been to treat all the subjects of Physics intelligibly, and who assuredly has succeeded in this better than any one before him, has said nothing that is not full of difficulties, or even inconceivable, in dealing with Light and its properties.

It’s true that we’re talking about a strange speed that would be a hundred thousand times faster than sound. Based on my observations, sound travels about 180 toises in one second, or about the time it takes for a pulse to beat. However, this assumption shouldn't seem impossible because it's not about moving a body at such great speed, but rather a successive movement passed from one body to another. So, I have found no problem in pondering these ideas, imagining that the emission of light happens over time. This way, we can explain all its phenomena, while the opposing view leaves everything unclear. It has always seemed to me that even Mr. Des Cartes, who aimed to make all topics in physics understandable and undoubtedly succeeded better than anyone before him, has said nothing that isn’t riddled with difficulties or even impossible when discussing light and its properties.

But that which I employed only as a hypothesis, has recently received great seemingness as an established truth by the ingenious proof of Mr. Römer which I am going here to relate, expecting him himself to give all that is needed for its confirmation. It is founded as is the preceding argument upon celestial observations, and proves not only that Light takes time for its passage, but also demonstrates how much time it takes, and that its velocity is even at least six times greater than that which I have just stated.

But what I initially used as a hypothesis has recently been bolstered into something that looks like an established truth thanks to the clever proof from Mr. Römer, which I'm about to explain, with the expectation that he himself will provide everything necessary for its confirmation. Just like the earlier argument, it's based on celestial observations and not only proves that light takes time to travel, but also shows exactly how much time it takes, demonstrating that its speed is at least six times greater than what I just mentioned.

[Pg 8]For this he makes use of the Eclipses suffered by the little planets which revolve around Jupiter, and which often enter his shadow: and see what is his reasoning. Let A be the Sun, BCDE the annual orbit of the Earth, F Jupiter, GN the orbit of the nearest of his Satellites, for it is this one which is more apt for this investigation than any of the other three, because of the quickness of its revolution. Let G be this Satellite entering into the shadow of Jupiter, H the same Satellite emerging from the shadow.

[Pg 8]To do this, he utilizes the eclipses that occur with the small planets orbiting Jupiter, which often pass into its shadow. Here’s his reasoning: Let A represent the Sun, BCDE the yearly orbit of the Earth, F be Jupiter, and GN the orbit of its closest satellite. This satellite is the most suitable for our analysis due to its rapid orbit. Let G be this satellite as it moves into Jupiter's shadow, and H be the same satellite as it comes out of the shadow.

Let it be then supposed, the Earth being at B some time before the last quadrature, that one has seen the said Satellite emerge from the shadow; it must needs be, if the Earth remains at the same place, that, after 42-1/2 hours, one would again see a similar emergence, because that is the time in which it makes the round of its orbit, and when it would come again into opposition to the Sun. And if the Earth, for instance, were to remain always at B during 30 revolutions of this Satellite, one would see it again emerge from the shadow after 30 times 42-1/2 hours. But the Earth having been carried along during this time to C, increasing thus its distance from Jupiter, it follows that if Light requires time for its passage the illumination of the little planet will be perceived later at [Pg 9]C than it would have been at B, and that there must be added to this time of 30 times 42-1/2 hours that which the Light has required to traverse the space MC, the difference of the spaces CH, BH. Similarly at the other quadrature when the earth has come to E from D while approaching toward Jupiter, the immersions of the Satellite ought to be observed at E earlier than they would have been seen if the Earth had remained at D.

Let’s assume the Earth is at B some time before the last quadrature and that we've seen the Satellite come out of the shadow. It has to be that if the Earth stays in the same spot, after 42.5 hours, we will see a similar emergence again because that’s how long it takes for the Satellite to complete its orbit and return to a position opposite the Sun. If the Earth were to stay at B for 30 cycles of this Satellite, we would see it emerge from the shadow again after 30 times 42.5 hours. However, since the Earth has moved to C during this time, increasing its distance from Jupiter, it follows that because light takes time to travel, we will detect the illumination of the little planet later at C than we would have at B. So, we need to add the travel time of light that took to cross the distance from M to C, which accounts for the difference in distances from C to H and B to H. Similarly, at the other quadrature, when the Earth has moved from D to E while getting closer to Jupiter, we should observe the Satellite entering the shadow at E sooner than we would have seen it if the Earth had remained at D.

Now in quantities of observations of these Eclipses, made during ten consecutive years, these differences have been found to be very considerable, such as ten minutes and more; and from them it has been concluded that in order to traverse the whole diameter of the annual orbit KL, which is double the distance from here to the sun, Light requires about 22 minutes of time.

Now, based on the observations of these eclipses made over ten consecutive years, these differences have proven to be quite significant, amounting to ten minutes or more. From this, it has been concluded that to travel the entire diameter of the annual orbit KL, which is twice the distance from here to the sun, light takes about 22 minutes.

The movement of Jupiter in his orbit while the Earth passed from B to C, or from D to E, is included in this calculation; and this makes it evident that one cannot attribute the retardation of these illuminations or the anticipation of the eclipses, either to any irregularity occurring in the movement of the little planet or to its eccentricity.

The motion of Jupiter in its orbit while the Earth moved from B to C, or from D to E, is factored into this calculation; and this clearly shows that one cannot blame the delay of these illuminations or the early occurrence of the eclipses on any irregularities in the movement of the small planet or its eccentricity.

If one considers the vast size of the diameter KL, which according to me is some 24 thousand diameters of the Earth, one will acknowledge the extreme velocity of Light. For, supposing that KL is no more than 22 thousand of these diameters, it appears that being traversed in 22 minutes this makes the speed a thousand diameters in one minute, that is 16-2/3 diameters in one second or in one beat of the pulse, which makes more than 11 hundred times a hundred thousand toises; since the diameter of the Earth contains 2,865 leagues, reckoned at 25 to the degree, and each [Pg 10]each league is 2,282 Toises, according to the exact measurement which Mr. Picard made by order of the King in 1669. But Sound, as I have said above, only travels 180 toises in the same time of one second: hence the velocity of Light is more than six hundred thousand times greater than that of Sound. This, however, is quite another thing from being instantaneous, since there is all the difference between a finite thing and an infinite. Now the successive movement of Light being confirmed in this way, it follows, as I have said, that it spreads by spherical waves, like the movement of Sound.

If you think about the huge size of the diameter KL, which I estimate to be around 24,000 times the diameter of the Earth, you'll realize just how fast Light travels. If we assume that KL is only 22,000 of these diameters, it turns out that crossing this distance in 22 minutes means it moves a thousand diameters in one minute, which is 16 and 2/3 diameters in one second, or one heartbeat. This translates to over 1,100 times a hundred thousand toises; since the diameter of the Earth has 2,865 leagues, counted at 25 to the degree, and each league is 2,282 Toises, based on the precise measurement that Mr. Picard conducted for the King in 1669. However, Sound, as I mentioned earlier, only travels 180 toises in the same one-second duration: this means Light is over six hundred thousand times faster than Sound. Still, that doesn’t mean it’s instantaneous; there’s a significant difference between something finite and something infinite. As we've confirmed that Light moves successively, it's clear that it spreads out in spherical waves, similar to how Sound moves.

But if the one resembles the other in this respect, they differ in many other things; to wit, in the first production of the movement which causes them; in the matter in which the movement spreads; and in the manner in which it is propagated. As to that which occurs in the production of Sound, one knows that it is occasioned by the agitation undergone by an entire body, or by a considerable part of one, which shakes all the contiguous air. But the movement of the Light must originate as from each point of the luminous object, else we should not be able to perceive all the different parts of that object, as will be more evident in that which follows. And I do not believe that this movement can be better explained than by supposing that all those of the luminous bodies which are liquid, such as flames, and apparently the sun and the stars, are composed of particles which float in a much more subtle medium which agitates them with great rapidity, and makes them strike against the particles of the ether which surrounds them, and which are much smaller than they. But I hold also that in luminous solids such as charcoal or metal made red hot in the fire, this same movement is caused by the violent [Pg 11]agitation of the particles of the metal or of the wood; those of them which are on the surface striking similarly against the ethereal matter. The agitation, moreover, of the particles which engender the light ought to be much more prompt and more rapid than is that of the bodies which cause sound, since we do not see that the tremors of a body which is giving out a sound are capable of giving rise to Light, even as the movement of the hand in the air is not capable of producing Sound.

But while the two are similar in this way, they differ in many other aspects; specifically, in how the movement that causes them is initiated, in how the movement spreads, and in how it is transmitted. When it comes to the production of sound, we know that it is caused by the vibration of an entire body or a significant part of it, which shakes the surrounding air. However, the movement of light must originate from every point of the luminous object; otherwise, we wouldn't be able to see all the different parts of that object, which will become clearer in what follows. I believe that this movement can best be explained by assuming that all liquid luminous bodies, like flames and seemingly the sun and stars, are made up of particles that float in a much finer medium that vibrates them rapidly, causing them to collide with the smaller particles of the surrounding ether. I also hold that in luminous solids, such as charcoal or metal heated in a fire, this same movement is generated by the intense agitation of the particles in the metal or wood; those on the surface collide similarly with the ethereal matter. Furthermore, the agitation of the particles that produce light must be much quicker and more rapid than that of the bodies that produce sound because we don't see that the vibrations of a body producing sound can generate light, just as the motion of a hand in the air cannot create sound.

Now if one examines what this matter may be in which the movement coming from the luminous body is propagated, which I call Ethereal matter, one will see that it is not the same that serves for the propagation of Sound. For one finds that the latter is really that which we feel and which we breathe, and which being removed from any place still leaves there the other kind of matter that serves to convey Light. This may be proved by shutting up a sounding body in a glass vessel from which the air is withdrawn by the machine which Mr. Boyle has given us, and with which he has performed so many beautiful experiments. But in doing this of which I speak, care must be taken to place the sounding body on cotton or on feathers, in such a way that it cannot communicate its tremors either to the glass vessel which encloses it, or to the machine; a precaution which has hitherto been neglected. For then after having exhausted all the air one hears no Sound from the metal, though it is struck.

Now, if you look at what this matter could be in which the movement from the luminous body travels, which I call Ethereal matter, you’ll see that it’s not the same as what carries Sound. The latter is really what we feel and breathe, and taking it away from a place still leaves behind the other type of matter that helps convey Light. This can be demonstrated by sealing a sounding object in a glass container from which the air has been removed using the machine that Mr. Boyle provided, with which he has conducted many impressive experiments. However, in this process, it's important to place the sounding object on cotton or feathers in such a way that it can’t transfer its vibrations to the glass container that holds it or to the machine; a precaution that has been overlooked until now. Because, after removing all the air, you won't hear any Sound from the metal, even when it gets struck.

One sees here not only that our air, which does not penetrate through glass, is the matter by which Sound spreads; but also that it is not the same air but another kind of matter in which Light spreads; since if the air is [Pg 12]removed from the vessel the Light does not cease to traverse it as before.

One can see here not only that our air, which doesn't pass through glass, is the medium through which sound travels; but also that it isn’t the same air but a different type of matter that allows light to spread; since if the air is [Pg 12]removed from the container, light still moves through it as it did before.

And this last point is demonstrated even more clearly by the celebrated experiment of Torricelli, in which the tube of glass from which the quicksilver has withdrawn itself, remaining void of air, transmits Light just the same as when air is in it. For this proves that a matter different from air exists in this tube, and that this matter must have penetrated the glass or the quicksilver, either one or the other, though they are both impenetrable to the air. And when, in the same experiment, one makes the vacuum after putting a little water above the quicksilver, one concludes equally that the said matter passes through glass or water, or through both.

And this final point is even more clearly shown by the famous experiment of Torricelli, where the glass tube, now void of air because the mercury has been removed, still lets light pass through just like when air is present. This proves that a substance different from air exists in this tube, and that this substance must have passed through the glass or the mercury, either one since both are impenetrable to air. In the same experiment, if a vacuum is created after adding a little water above the mercury, it similarly concludes that this substance can pass through glass or water, or both.

As regards the different modes in which I have said the movements of Sound and of Light are communicated, one may sufficiently comprehend how this occurs in the case of Sound if one considers that the air is of such a nature that it can be compressed and reduced to a much smaller space than that which it ordinarily occupies. And in proportion as it is compressed the more does it exert an effort to regain its volume; for this property along with its penetrability, which remains notwithstanding its compression, seems to prove that it is made up of small bodies which float about and which are agitated very rapidly in the ethereal matter composed of much smaller parts. So that the cause of the spreading of Sound is the effort which these little bodies make in collisions with one another, to regain freedom when they are a little more squeezed together in the circuit of these waves than elsewhere.

Regarding the different ways I’ve mentioned that Sound and Light are transmitted, you can understand how this happens with Sound by recognizing that air can be compressed and reduced to a much smaller space than it usually occupies. The more it’s compressed, the harder it tries to return to its original volume; this property, along with its ability to penetrate even when compressed, suggests it consists of small particles that move around and vibrate rapidly in a medium made up of even tinier components. Thus, the reason Sound spreads is the effort these tiny particles make when colliding with each other to regain their freedom when they are squeezed closer together in the pattern of these waves than in other areas.

But the extreme velocity of Light, and other properties which it has, cannot admit of such a propagation of motion, [Pg 13]and I am about to show here the way in which I conceive it must occur. For this, it is needful to explain the property which hard bodies must possess to transmit movement from one to another.

But the extreme speed of light and its other properties don't allow for such a transmission of motion, [Pg 13]and I'm going to explain how I think it must happen. To do this, I need to clarify the property that solid objects must have to transfer movement from one to another.

When one takes a number of spheres of equal size, made of some very hard substance, and arranges them in a straight line, so that they touch one another, one finds, on striking with a similar sphere against the first of these spheres, that the motion passes as in an instant to the last of them, which separates itself from the row, without one's being able to perceive that the others have been stirred. And even that one which was used to strike remains motionless with them. Whence one sees that the movement passes with an extreme velocity which is the greater, the greater the hardness of the substance of the spheres.

When you take a series of identical spheres made of a very hard material and line them up so that they touch each other, you’ll notice that when you hit the first sphere with another sphere, the motion instantly transfers to the last one in the line, causing it to pop out without any noticeable movement in the other spheres. Even the sphere you used to strike the first one stays still with the others. This shows that the movement travels with incredible speed, and that speed increases with the hardness of the material of the spheres.

But it is still certain that this progression of motion is not instantaneous, but successive, and therefore must take time. For if the movement, or the disposition to movement, if you will have it so, did not pass successively through all these spheres, they would all acquire the movement at the same time, and hence would all advance together; which does not happen. For the last one leaves the whole row and acquires the speed of the one which was pushed. Moreover there are experiments which demonstrate that all the bodies which we reckon of the hardest kind, such as quenched steel, glass, and agate, act as springs and bend somehow, not only when extended as rods but also when they are in the form of spheres or of other shapes. That is to say they yield a little in themselves at the place where they are struck, and immediately regain their former figure. For I have found that on striking with a ball of glass or of agate against a large and quite thick [Pg 14]thick piece of the same substance which had a flat surface, slightly soiled with breath or in some other way, there remained round marks, of smaller or larger size according as the blow had been weak or strong. This makes it evident that these substances yield where they meet, and spring back: and for this time must be required.

But it’s clear that this progression of motion isn’t instantaneous, but rather successive, so it takes time. If the movement—or the readiness for movement, if you like—didn’t pass through each sphere in order, they would all start moving at the same time and advance together, which doesn’t happen. The last one leaves the entire line and picks up the speed of the one that was pushed. Additionally, experiments show that even the hardest materials we know, like quenched steel, glass, and agate, act like springs and bend in some way, not just when stretched as rods, but also when they’re shaped like spheres or other forms. In other words, they yield a bit at the point of impact and then immediately return to their original shape. For instance, I found that when I struck a glass or agate ball against a large, thick piece of the same material with a flat surface, slightly soiled with breath or otherwise, round marks were left behind, varying in size depending on whether the hit was weak or strong. This clearly shows that these substances give way at the point of contact and then bounce back, which requires some time.

Now in applying this kind of movement to that which produces Light there is nothing to hinder us from estimating the particles of the ether to be of a substance as nearly approaching to perfect hardness and possessing a springiness as prompt as we choose. It is not necessary to examine here the causes of this hardness, or of that springiness, the consideration of which would lead us too far from our subject. I will say, however, in passing that we may conceive that the particles of the ether, notwithstanding their smallness, are in turn composed of other parts and that their springiness consists in the very rapid movement of a subtle matter which penetrates them from every side and constrains their structure to assume such a disposition as to give to this fluid matter the most overt and easy passage possible. This accords with the explanation which Mr. Des Cartes gives for the spring, though I do not, like him, suppose the pores to be in the form of round hollow canals. And it must not be thought that in this there is anything absurd or impossible, it being on the contrary quite credible that it is this infinite series of different sizes of corpuscles, having different degrees of velocity, of which Nature makes use to produce so many marvellous effects.

Now, when applying this kind of movement to the things that produce light, nothing stops us from considering the particles of the ether to be made of a material that's nearly perfectly hard and has a springiness as quick as we desire. We don’t need to dive into the reasons for this hardness or springiness, as that would take us too far off-topic. However, I will briefly mention that we might imagine the particles of the ether, despite their small size, are made up of even smaller parts, and their springiness comes from the very fast movement of a subtle matter that fills them from all sides, forcing their structure to form in a way that allows this fluid matter to pass through as easily as possible. This aligns with Mr. Des Cartes' explanation for springs, although I don’t, like him, believe that the pores are shaped like round hollow tubes. And we shouldn't think there’s anything absurd or impossible about this; on the contrary, it’s quite plausible that Nature uses this infinite series of differently sized particles, each with varying speeds, to create so many amazing effects.

But though we shall ignore the true cause of springiness we still see that there are many bodies which possess this property; and thus there is nothing strange in supposing [Pg 15]that it exists also in little invisible bodies like the particles of the Ether. Also if one wishes to seek for any other way in which the movement of Light is successively communicated, one will find none which agrees better, with uniform progression, as seems to be necessary, than the property of springiness; because if this movement should grow slower in proportion as it is shared over a greater quantity of matter, in moving away from the source of the light, it could not conserve this great velocity over great distances. But by supposing springiness in the ethereal matter, its particles will have the property of equally rapid restitution whether they are pushed strongly or feebly; and thus the propagation of Light will always go on with an equal velocity.

But even though we’ll disregard the true reason for springiness, we can still see that many objects have this quality; so it's not surprising to assume that it also exists in tiny, invisible particles like those of the Ether. If someone wants to find another way in which the movement of Light is transmitted in succession, they won’t find one that fits as well, with a steady progression, as the property of springiness does. If this movement were to slow down as it spreads over a larger amount of matter, moving away from the light source, it wouldn’t be able to maintain its high speed over long distances. However, by assuming springiness in the ethereal material, its particles will have the ability to return to their original state quickly, whether they are pushed hard or softly; thus, the propagation of Light will always occur at a constant speed.

And it must be known that although the particles of the ether are not ranged thus in straight lines, as in our row of spheres, but confusedly, so that one of them touches several others, this does not hinder them from transmitting their movement and from spreading it always forward. As to this it is to be remarked that there is a law of motion serving for this propagation, and verifiable by experiment. It is that when a sphere, such as A here, touches several other similar spheres CCC, if it is struck by another sphere B in such a way as to exert an impulse against all the spheres CCC which touch it, it transmits to them the whole of its movement, and remains after that motionless like the sphere B. And without supposing that the ethereal particles are of spherical form (for I see indeed no need to suppose them so) one may well understand that this property of communicating an impulse [Pg 16]does not fail to contribute to the aforesaid propagation of movement.

And it should be noted that even though the particles of the ether aren’t arranged in straight lines like our row of spheres, but rather in a chaotic manner, with one touching several others, this doesn’t prevent them from transmitting their movement and pushing it forward. It’s important to point out that there is a law of motion that governs this propagation, which can be tested through experiments. This law states that when a sphere, like A here, touches several other similar spheres CCC, and if it gets hit by another sphere B in such a way that it impacts all the spheres CCC that are in contact with it, it passes on its entire movement to them and then stays still like sphere B. And although we don’t have to assume that the ethereal particles are shaped like spheres (since I see no need to do so), it’s clear that this ability to transfer an impulse [Pg 16]plays a role in the previously mentioned propagation of movement.

Equality of size seems to be more necessary, because otherwise there ought to be some reflexion of movement backwards when it passes from a smaller particle to a larger one, according to the Laws of Percussion which I published some years ago.

Equality of size seems more essential because, otherwise, there should be some kind of backward motion when it moves from a smaller particle to a larger one, according to the Laws of Percussion that I published a few years ago.

However, one will see hereafter that we have to suppose such an equality not so much as a necessity for the propagation of light as for rendering that propagation easier and more powerful; for it is not beyond the limits of probability that the particles of the ether have been made equal for a purpose so important as that of light, at least in that vast space which is beyond the region of atmosphere and which seems to serve only to transmit the light of the Sun and the Stars.

However, one will see later that we need to assume such an equality not so much as a requirement for the spread of light but to make that spread easier and stronger; because it’s not out of the question that the particles of the ether have been made equal for a purpose as significant as that of light, at least in the immense space beyond the atmosphere, which seems to exist solely to transmit the light of the Sun and the Stars.

I have then shown in what manner one may conceive Light to spread successively, by spherical waves, and how it is possible that this spreading is accomplished with as great a velocity as that which experiments and celestial observations demand. Whence it may be further remarked that although the particles are supposed to be in continual movement (for there are many reasons for this) the successive propagation of the waves cannot be hindered by this; because the propagation consists nowise in the transport of those particles but merely in a small agitation which they cannot help communicating to those surrounding, notwithstanding any movement which may act on them causing them to be changing positions amongst themselves.

I have shown how light can be thought of as spreading out in spherical waves and how this spreading can happen as quickly as experiments and observations of celestial bodies require. It's also worth noting that even though the particles are thought to be constantly moving (and there are many reasons for this), the gradual movement of the waves isn’t disrupted by this. This is because the propagation doesn’t rely on the movement of those particles, but rather on a slight disturbance that they can’t help but pass on to the nearby particles, despite any movement that might cause them to shift positions among themselves.

But we must consider still more particularly the origin of these waves, and the manner in which they spread. And, first, it follows from what has been said on the production [Pg 17]of Light, that each little region of a luminous body, such as the Sun, a candle, or a burning coal, generates its own waves of which that region is the centre. Thus in the flame of a candle, having distinguished the points A, B, C, concentric circles described about each of these points represent the waves which come from them. And one must imagine the same about every point of the surface and of the part within the flame.

But we need to pay closer attention to the source of these waves and how they propagate. First, as we discussed regarding the production [Pg 17]of Light, each small area of a light source, like the Sun, a candle, or a burning coal, creates its own waves that radiate from that area. For example, in the flame of a candle, if we identify points A, B, and C, the concentric circles drawn around each of these points represent the waves emanating from them. We should think of this similarly for every point on the surface and within the flame.

But as the percussions at the centres of these waves possess no regular succession, it must not be supposed that the waves themselves follow one another at equal distances: and if the distances marked in the figure appear to be such, it is rather to mark the progression of one and the same wave at equal intervals of time than to represent several of them issuing from one and the same centre.

But since the shocks at the centers of these waves don't occur in a consistent pattern, we shouldn’t assume that the waves themselves come at equal intervals. If the distances shown in the figure seem to be equal, that’s more about illustrating the movement of a single wave over equal time intervals rather than showing several waves coming from the same center.

After all, this prodigious quantity of waves which traverse one another without confusion and without effacing one another must not be deemed inconceivable; it being certain that one and the same particle of matter can serve for many waves coming from different sides or even from contrary directions, not only if it is struck by blows which follow one another closely but even for those which act on it at the same instant. It can do so because the spreading of the movement is successive. This may be proved by the row of equal spheres of hard matter, spoken of above. If against this row there are pushed from two opposite sides at the same time two similar spheres A and [Pg 18]D, one will see each of them rebound with the same velocity which it had in striking, yet the whole row will remain in its place, although the movement has passed along its whole length twice over. And if these contrary movements happen to meet one another at the middle sphere, B, or at some other such as C, that sphere will yield and act as a spring at both sides, and so will serve at the same instant to transmit these two movements.

After all, this huge amount of waves that pass through each other without getting mixed up or canceling each other out shouldn't be considered impossible. It's clear that a single particle of matter can be affected by multiple waves coming from different angles or even from opposite directions. This is true not just when they're hit in quick succession but also when they impact at the same time. It can handle this because the spread of motion happens gradually. This can be demonstrated by the row of equal spheres made of solid material mentioned earlier. If, at the same time, two similar spheres A and [Pg 18]D are pushed toward this row from opposite sides, you'll see each bounce back with the same speed they had when they hit, yet the entire row will stay in place, even though the motion has traveled through its entire length twice. And if these opposing movements happen to meet at the middle sphere, B, or at another one like C, that sphere will give way and act like a spring, transmitting both movements at the same time.

But what may at first appear full strange and even incredible is that the undulations produced by such small movements and corpuscles, should spread to such immense distances; as for example from the Sun or from the Stars to us. For the force of these waves must grow feeble in proportion as they move away from their origin, so that the action of each one in particular will without doubt become incapable of making itself felt to our sight. But one will cease to be astonished by considering how at a great distance from the luminous body an infinitude of waves, though they have issued from different points of this body, unite together in such a way that they sensibly compose one single wave only, which, consequently, ought to have enough force to make itself felt. Thus this infinite number of waves which originate at the same instant from all points of a fixed star, big it may be as the Sun, make practically only one single wave which may well have force enough to produce an impression on our eyes. Moreover from each luminous point there may come many thousands of waves in the smallest imaginable time, by the frequent percussion of the corpuscles which strike the [Pg 19]Ether at these points: which further contributes to rendering their action more sensible.

But what might initially seem really strange and even unbelievable is that the tiny movements and particles create waves that can travel such vast distances, like from the Sun or stars to us. The strength of these waves must weaken as they move away from their source, so the effect of each individual wave will likely become too faint for us to notice. However, it’s less surprising when we consider that at a great distance from the light source, numerous waves, even though they come from different points of that body, merge in such a way that they effectively create one single wave, which should have enough power to be noticeable. Thus, the countless waves originating at the same moment from all points of a star, even one as large as the Sun, collectively form almost one single wave that might indeed be strong enough to make an impression on our eyes. Additionally, from each light point, many thousands of waves can emerge in the shortest possible time, due to the constant impact of the particles hitting the [Pg 19]Ether at these locations, which further enhances their perceptibility.

There is the further consideration in the emanation of these waves, that each particle of matter in which a wave spreads, ought not to communicate its motion only to the next particle which is in the straight line drawn from the luminous point, but that it also imparts some of it necessarily to all the others which touch it and which oppose themselves to its movement. So it arises that around each particle there is made a wave of which that particle is the centre. Thus if DCF is a wave emanating from the luminous point A, which is its centre, the particle B, one of those comprised within the sphere DCF, will have made its particular or partial wave KCL, which will touch the wave DCF at C at the same moment that the principal wave emanating from the point A has arrived at DCF; and it is clear that it will be only the region C of the wave KCL which will touch the wave DCF, to wit, that which is in the straight line drawn through AB. Similarly the other particles of the sphere DCF, such as bb, dd, etc., will each make its own wave. But each of these waves can be infinitely feeble only as compared with the wave DCF, to the composition of which all the others contribute by the part of their surface which is most distant from the centre A.

There’s also the point that when these waves spread, each particle of matter involved shouldn’t just pass its motion to the next particle directly in line from the light source, but should also share some of it with all the other particles that touch it and resist its movement. This leads to each particle creating a wave with itself as the center. For example, if DCF is a wave coming from the light source A, which is its center, particle B, which is part of the DCF sphere, will create its own partial wave KCL. This wave will touch the wave DCF at point C just as the main wave from point A reaches DCF. It’s clear that only point C of the wave KCL will touch wave DCF, specifically the part that lies on the straight line between A and B. In the same way, other particles within the DCF sphere, like bb, dd, and so on, will generate their own waves. However, each of these waves will be incredibly weak compared to wave DCF, as they all contribute to the formation of DCF from the parts of their surfaces that are furthest away from the center A.

[Pg 20]One sees, in addition, that the wave DCF is determined by the distance attained in a certain space of time by the movement which started from the point A; there being no movement beyond this wave, though there will be in the space which it encloses, namely in parts of the particular waves, those parts which do not touch the sphere DCF. And all this ought not to seem fraught with too much minuteness or subtlety, since we shall see in the sequel that all the properties of Light, and everything pertaining to its reflexion and its refraction, can be explained in principle by this means. This is a matter which has been quite unknown to those who hitherto have begun to consider the waves of light, amongst whom are Mr. Hooke in his Micrographia, and Father Pardies, who, in a treatise of which he let me see a portion, and which he was unable to complete as he died shortly afterward, had undertaken to prove by these waves the effects of reflexion and refraction. But the chief foundation, which consists in the remark I have just made, was lacking in his demonstrations; and for the rest he had opinions very different from mine, as may be will appear some day if his writing has been preserved.

[Pg 20]Additionally, the wave DCF is defined by the distance covered in a specific amount of time by the movement that started from point A; there is no movement beyond this wave, although there will be within the area it encloses, specifically in segments of the individual waves that do not touch the DCF sphere. This should not seem overly detailed or complex, since we will later see that all the properties of Light, including its reflection and refraction, can fundamentally be explained this way. This aspect has been largely overlooked by those who have previously examined light waves, including Mr. Hooke in his Micrographia, and Father Pardies, who had started a treatise that he shared a part of with me before he passed away shortly after and could not complete. His work aimed to demonstrate the effects of reflection and refraction through these waves, but he lacked the foundational observation I just mentioned, and he held views that were quite different from my own, which may become evident someday if his writings have been preserved.

To come to the properties of Light. We remark first that each portion of a wave ought to spread in such a way that its extremities lie always between the same straight lines drawn from the luminous point. Thus the portion BG of the wave, having the luminous point A as its centre, will spread into the arc CE bounded by the straight lines ABC, AGE. For although the particular waves produced by the particles comprised within the space CAE spread also outside this space, they yet do not concur at the same instant to compose a wave which terminates the [Pg 21]movement, as they do precisely at the circumference CE, which is their common tangent.

To discuss the properties of Light, we first note that each part of a wave should spread in such a way that its edges always fall between the same straight lines drawn from the light source. For example, the segment BG of the wave, centered on the light point A, will spread into the arc CE, which is bounded by the straight lines ABC and AGE. Even though the individual waves created by the particles within the space CAE also spread beyond this area, they do not come together at the same moment to form a wave that ends the [Pg 21]movement, as they do precisely at the circumference CE, which acts as their common tangent.

And hence one sees the reason why light, at least if its rays are not reflected or broken, spreads only by straight lines, so that it illuminates no object except when the path from its source to that object is open along such lines.

And that's why light, as long as its rays aren't reflected or bent, travels only in straight lines. It only shines on an object when there's a clear path from its source to that object along those lines.

For if, for example, there were an opening BG, limited by opaque bodies BH, GI, the wave of light which issues from the point A will always be terminated by the straight lines AC, AE, as has just been shown; the parts of the partial waves which spread outside the space ACE being too feeble to produce light there.

For instance, if there were an opening BG, surrounded by opaque objects BH and GI, the light wave coming from point A will always end at the straight lines AC and AE, as has been demonstrated; the parts of the partial waves that extend beyond the area ACE are too weak to create light there.

Now, however small we make the opening BG, there is always the same reason causing the light there to pass between straight lines; since this opening is always large enough to contain a great number of particles of the ethereal matter, which are of an inconceivable smallness; so that it appears that each little portion of the wave necessarily advances following the straight line which comes from the luminous point. Thus then we may take the rays of light as if they were straight lines.

Now, no matter how small we make the opening BG, there's always the same reason for the light to travel in straight lines; this opening is always big enough to include a large number of tiny particles of ethereal matter, which are incredibly small; so it seems that each tiny section of the wave must advance along the straight line coming from the light source. Therefore, we can treat light rays as if they were straight lines.

It appears, moreover, by what has been remarked touching the feebleness of the particular waves, that it is not needful that all the particles of the Ether should be equal amongst themselves, though equality is more apt for the propagation of the movement. For it is true that inequality will cause a particle by pushing against another larger one to strive to recoil with a part of its movement; but it will thereby merely generate backwards towards the luminous point some partial waves incapable of causing light, and not a wave compounded of many as CE was.

It seems, additionally, based on observations about the weakness of the specific waves, that it's not necessary for all the particles of the Ether to be equal to each other, even though equality is better for spreading the movement. It's true that inequality will make a smaller particle push against a larger one and try to pull back with some of its movement; however, this will just create some smaller waves that can't produce light, rather than a wave made up of many like CE was.

Another property of waves of light, and one of the most [Pg 22]marvellous, is that when some of them come from different or even from opposing sides, they produce their effect across one another without any hindrance. Whence also it comes about that a number of spectators may view different objects at the same time through the same opening, and that two persons can at the same time see one another's eyes. Now according to the explanation which has been given of the action of light, how the waves do not destroy nor interrupt one another when they cross one another, these effects which I have just mentioned are easily conceived. But in my judgement they are not at all easy to explain according to the views of Mr. Des Cartes, who makes Light to consist in a continuous pressure merely tending to movement. For this pressure not being able to act from two opposite sides at the same time, against bodies which have no inclination to approach one another, it is impossible so to understand what I have been saying about two persons mutually seeing one another's eyes, or how two torches can illuminate one another.

Another property of light waves, and one of the most amazing, is that when some of them come from different or even opposing sides, they can affect each other without any interference. This also allows multiple spectators to view different objects at the same time through the same opening, and two people can see each other’s eyes simultaneously. Based on the explanation of how light operates, where waves do not destroy or interrupt each other when they cross, these effects are easy to understand. However, in my opinion, they are not easy to explain according to Mr. Des Cartes’ view, which states that light is just a continuous pressure aimed at movement. This pressure cannot act from two opposite sides at the same time on bodies that have no inclination to move toward each other, making it impossible to comprehend what I've said about two people seeing each other’s eyes or how two torches can illuminate each other.


CHAPTER II

ON REFLEXION

H

aving explained the effects of waves of light which spread in a homogeneous matter, we will examine next that which happens to them on encountering other bodies. We will first make evident how the Reflexion of light is explained by these same waves, and why it preserves equality of angles.

Having explained how waves of light travel through a uniform medium, we will next look at what happens when they hit other objects. We will first clarify how the reflection of light can be understood through these same waves and why it maintains equal angles.

[Pg 23]Let there be a surface AB; plane and polished, of some metal, glass, or other body, which at first I will consider as perfectly uniform (reserving to myself to deal at the end of this demonstration with the inequalities from which it cannot be exempt), and let a line AC, inclined to AD, represent a portion of a wave of light, the centre of which is so distant that this portion AC may be considered as a straight line; for I consider all this as in one plane, imagining to myself that the plane in which this figure is, cuts the sphere of the wave through its centre and intersects the plane AB at right angles. This explanation will suffice once for all.

[Pg 23]Imagine a surface AB; flat and smooth, made of some metal, glass, or another material, which I will initially treat as completely uniform (I’ll address any imperfections at the end of this explanation). Now, let a line AC, slanted to AD, represent a segment of a light wave, so distant that we can consider AC as a straight line. I envision everything happening in one plane, picturing that this plane containing the figure cuts through the center of the wave sphere and intersects the plane AB at a right angle. This explanation will be sufficient for now.

The piece C of the wave AC, will in a certain space of time advance as far as the plane AB at B, following the straight line CB, which may be supposed to come from the luminous centre, and which in consequence is perpendicular to AC. Now in this same space of time the portion A of the same wave, which has been hindered from communicating its movement beyond the plane AB, or at least partly so, ought to have continued its movement in the matter which is above this plane, and this along a distance equal to CB, making its [Pg 24]own partial spherical wave, according to what has been said above. Which wave is here represented by the circumference SNR, the centre of which is A, and its semi-diameter AN equal to CB.

The part C of the wave AC will, in a certain amount of time, move as far as the plane AB at point B, following the straight line CB, which can be thought of as originating from the light source and is therefore perpendicular to AC. During this same time, the portion A of the same wave, which has been prevented from transmitting its movement beyond the plane AB, or at least partially so, should have continued its movement in the matter above this plane, covering a distance equal to CB, creating its own partial spherical wave, as mentioned earlier. This wave is illustrated here by the circumference SNR, with A as the center and its radius AN equal to CB.

If one considers further the other pieces H of the wave AC, it appears that they will not only have reached the surface AB by straight lines HK parallel to CB, but that in addition they will have generated in the transparent air, from the centres K, K, K, particular spherical waves, represented here by circumferences the semi-diameters of which are equal to KM, that is to say to the continuations of HK as far as the line BG parallel to AC. But all these circumferences have as a common tangent the straight line BN, namely the same which is drawn from B as a tangent to the first of the circles, of which A is the centre, and AN the semi-diameter equal to BC, as is easy to see.

If we look more closely at the other parts H of the wave AC, we see that they will not only have reached the surface AB along straight lines HK that are parallel to CB, but they will also have created specific spherical waves in the transparent air from the centers K, K, K, represented here by circles whose radii are equal to KM, which means extending HK all the way to the line BG that is parallel to AC. However, all these circles share a common tangent, which is the straight line BN. This is the same line drawn from B as a tangent to the first of the circles, where A is the center and AN is the radius equal to BC, which is easy to understand.

It is then the line BN (comprised between B and the point N where the perpendicular from the point A falls) which is as it were formed by all these circumferences, and which terminates the movement which is made by the reflexion of the wave AC; and it is also the place where the movement occurs in much greater quantity than anywhere else. Wherefore, according to that which has been explained, BN is the propagation of the wave AC at the moment when the piece C of it has arrived at B. For there is no other line which like BN is a common tangent to all the aforesaid circles, except BG below the plane AB; which line BG would be the propagation of the wave if the movement could have spread in a medium homogeneous with that which is above the plane. And if one wishes to see how the wave AC has come successively to BN, one has only to draw in the same figure the straight lines KO [Pg 25]parallel to BN, and the straight lines KL parallel to AC. Thus one will see that the straight wave AC has become broken up into all the OKL parts successively, and that it has become straight again at NB.

It’s the line BN (between B and point N, where the perpendicular from point A meets) that is formed by all these circles, and it marks the end of the movement created by the reflection of the wave AC; it's also where the movement happens in much greater amounts than anywhere else. Therefore, as explained, BN is the propagation of wave AC at the moment when piece C of it reaches B. There isn’t another line like BN that is a common tangent to all the mentioned circles, except for BG below the plane AB; that line BG would be the propagation of the wave if the movement could spread in a medium that was the same as the one above the plane. If you want to see how wave AC gradually reached BN, you just need to draw straight lines KO [Pg 25]parallel to BN, and straight lines KL parallel to AC. This way, you’ll notice that the straight wave AC has broken into all the OKL parts successively, and that it straightens out again at NB.

Now it is apparent here that the angle of reflexion is made equal to the angle of incidence. For the triangles ACB, BNA being rectangular and having the side AB common, and the side CB equal to NA, it follows that the angles opposite to these sides will be equal, and therefore also the angles CBA, NAB. But as CB, perpendicular to CA, marks the direction of the incident ray, so AN, perpendicular to the wave BN, marks the direction of the reflected ray; hence these rays are equally inclined to the plane AB.

Now it's clear that the angle of reflection is equal to the angle of incidence. For triangles ACB and BNA, which are right-angled and share the side AB, and with side CB equal to NA, it follows that the angles opposite these sides will be equal, which means the angles CBA and NAB are also equal. Just as CB, which is perpendicular to CA, indicates the direction of the incoming ray, AN, which is perpendicular to wave BN, indicates the direction of the reflected ray; therefore, these rays are equally inclined to the plane AB.

But in considering the preceding demonstration, one might aver that it is indeed true that BN is the common tangent of the circular waves in the plane of this figure, but that these waves, being in truth spherical, have still an infinitude of similar tangents, namely all the straight lines which are drawn from the point B in the surface generated by the straight line BN about the axis BA. It remains, therefore, to demonstrate that there is no difficulty herein: and by the same argument one will see why the incident ray and the reflected ray are always in one and the same plane perpendicular to the reflecting plane. I say then that the wave AC, being regarded only as a line, produces no light. For a visible ray of light, however narrow it may be, has always some width, and consequently it is necessary, in representing the wave whose progression constitutes the ray, to put instead of a line AC some plane figure such as the circle HC in the following figure, by supposing, as we have done, the luminous point to be infinitely distant. [Pg 26]Now it is easy to see, following the preceding demonstration, that each small piece of this wave HC having arrived at the plane AB, and there generating each one its particular wave, these will all have, when C arrives at B, a common plane which will touch them, namely a circle BN similar to CH; and this will be intersected at its middle and at right angles by the same plane which likewise intersects the circle CH and the ellipse AB.

But when examining the previous demonstration, one might argue that it’s true that BN is the common tangent of the circular waves in this figure. However, since these waves are actually spherical, there are countless similar tangents—specifically, all the straight lines drawn from point B on the surface formed by the straight line BN around the axis BA. Therefore, it remains to show that there is no issue with this: and using the same reasoning, you will understand why the incident ray and the reflected ray are always in the same plane that is perpendicular to the reflecting plane. I say that the wave AC, when considered only as a line, produces no light. For any visible ray of light, no matter how narrow, always has some width, so in representing the wave that constitutes the ray, instead of a line AC, we need to use a plane figure like the circle HC in the following figure, assuming, as we have done, that the luminous point is infinitely distant. [Pg 26]Now it’s easy to see, following the previous demonstration, that each small segment of this wave HC that reaches the plane AB generates its own wave, and when C arrives at B, they will all share a common plane that will touch them, namely a circle BN similar to CH; this will be intersected at its midpoint and at right angles by the same plane that also intersects the circle CH and the ellipse AB.

One sees also that the said spheres of the partial waves cannot have any common tangent plane other than the circle BN; so that it will be this plane where there will be more reflected movement than anywhere else, and which will therefore carry on the light in continuance from the wave CH.

One can also see that the mentioned spheres of the partial waves cannot share any common tangent plane other than the circle BN; therefore, this will be the plane where there is more reflected movement than anywhere else, and it will consequently carry the light continuously from the wave CH.

I have also stated in the preceding demonstration that the movement of the piece A of the incident wave is not able to communicate itself beyond the plane AB, or at least not wholly. Whence it is to be remarked that though the movement of the ethereal matter might communicate itself partly to that of the reflecting body, this could in nothing alter the velocity of progression of the waves, on which [Pg 27]the angle of reflexion depends. For a slight percussion ought to generate waves as rapid as strong percussion in the same matter. This comes about from the property of bodies which act as springs, of which we have spoken above; namely that whether compressed little or much they recoil in equal times. Equally so in every reflexion of the light, against whatever body it may be, the angles of reflexion and incidence ought to be equal notwithstanding that the body might be of such a nature that it takes away a portion of the movement made by the incident light. And experiment shows that in fact there is no polished body the reflexion of which does not follow this rule.

I've also mentioned in the previous explanation that the movement of piece A of the incident wave cannot fully transfer beyond the plane AB. Thus, it’s important to note that while the movement of the ether might partly influence that of the reflecting body, it doesn't change the speed of the waves, which is what the angle of reflection depends on. A slight impact should generate waves just as fast as a strong impact in the same medium. This happens because bodies behave like springs, as we discussed earlier; whether they are compressed a little or a lot, they bounce back in the same amount of time. Likewise, in every reflection of light, regardless of the object, the angles of reflection and incidence should be equal, even if the object absorbs some of the movement from the incident light. Experiments demonstrate that there is no polished surface whose reflection does not adhere to this principle.

But the thing to be above all remarked in our demonstration is that it does not require that the reflecting surface should be considered as a uniform plane, as has been supposed by all those who have tried to explain the effects of reflexion; but only an evenness such as may be attained by the particles of the matter of the reflecting body being set near to one another; which particles are larger than those of the ethereal matter, as will appear by what we shall say in treating of the transparency and opacity of bodies. For the surface consisting thus of particles put together, and the ethereal particles being above, and smaller, it is evident that one could not demonstrate the equality of the angles of incidence and reflexion by similitude to that which happens to a ball thrown against a wall, of which writers have always made use. In our way, on the other hand, the thing is explained without difficulty. For the smallness of the particles of quicksilver, for example, being such that one must conceive millions of them, in the smallest visible surface proposed, arranged like a heap of grains of sand which has been flattened as much as it is capable of being, [Pg 28]this surface then becomes for our purpose as even as a polished glass is: and, although it always remains rough with respect to the particles of the Ether it is evident that the centres of all the particular spheres of reflexion, of which we have spoken, are almost in one uniform plane, and that thus the common tangent can fit to them as perfectly as is requisite for the production of light. And this alone is requisite, in our method of demonstration, to cause equality of the said angles without the remainder of the movement reflected from all parts being able to produce any contrary effect.

But the key point to note in our demonstration is that it does not require the reflecting surface to be considered a uniform plane, as everyone who has tried to explain the effects of reflection has assumed. Instead, it only needs to be smooth enough that the particles of the reflecting material are closely packed together; these particles are larger than those of the ethereal matter, as we will discuss when we talk about the transparency and opacity of materials. Since the surface is made up of closely arranged particles, and the ethereal particles are smaller and above, it's clear that we can't show the equality of the angles of incidence and reflection in the same way as with a ball thrown against a wall, which is what many writers have used as an analogy. In our approach, however, this can be explained easily. For instance, the tiny particles of mercury are so small that you can imagine millions of them on the smallest visible surface, arranged like a flattened pile of sand. [Pg 28]This surface, then, becomes as smooth as polished glass for our purposes; and even though it always remains rough in comparison to the particles of Ether, it's evident that the centers of all the individual reflection spheres we have discussed are nearly aligned in a uniform plane. Thus, the common tangent can align with them as perfectly as necessary to generate light. This alone is sufficient in our method of demonstration to ensure the equality of these angles, without the rest of the reflected movement from all parts causing any contrary effect.


CHAPTER III

ON REFRACTION

I

n the same way as the effects of Reflexion have been explained by waves of light reflected at the surface of polished bodies, we will explain transparency and the phenomena of refraction by waves which spread within and across diaphanous bodies, both solids, such as glass, and liquids, such as water, oils, etc. But in order that it may not seem strange to suppose this passage of waves in the interior of these bodies, I will first show that one may conceive it possible in more than one mode.

In the same way that the effects of reflection have been explained by light waves bouncing off the surface of shiny objects, we will explain transparency and the phenomena of refraction through waves that travel within and across clear materials, both solid, like glass, and liquid, like water, oil, etc. However, to avoid any confusion about the idea of these waves moving inside these materials, I will first demonstrate that it's conceivable in more than one way.

First, then, if the ethereal matter cannot penetrate transparent bodies at all, their own particles would be able to communicate successively the movement of the waves, the same as do those of the Ether, supposing that, like those, they are of a nature to act as a spring. And this is [Pg 29]easy to conceive as regards water and other transparent liquids, they being composed of detached particles. But it may seem more difficult as regards glass and other transparent and hard bodies, because their solidity does not seem to permit them to receive movement except in their whole mass at the same time. This, however, is not necessary because this solidity is not such as it appears to us, it being probable rather that these bodies are composed of particles merely placed close to one another and held together by some pressure from without of some other matter, and by the irregularity of their shapes. For primarily their rarity is shown by the facility with which there passes through them the matter of the vortices of the magnet, and that which causes gravity. Further, one cannot say that these bodies are of a texture similar to that of a sponge or of light bread, because the heat of the fire makes them flow and thereby changes the situation of the particles amongst themselves. It remains then that they are, as has been said, assemblages of particles which touch one another without constituting a continuous solid. This being so, the movement which these particles receive to carry on the waves of light, being merely communicated from some of them to others, without their going for that purpose out of their places or without derangement, it may very well produce its effect without prejudicing in any way the apparent solidity of the compound.

First, if the ethereal matter can’t penetrate transparent objects at all, their own particles should be able to transmit the movement of the waves, just like those of the Ether, assuming that they behave like a spring. This is easy to imagine for water and other transparent liquids, since they consist of separate particles. However, it might seem more challenging for glass and other hard, transparent materials because their solidity suggests they can only move as a whole. But this isn’t actually necessary, since their solidity is not as it seems; it’s likely that these materials are made up of particles that are simply close together and held by some external pressure from other matter, as well as by the irregular shapes of the particles. Their porosity is evidenced by how easily the matter from magnetic vortices and gravitating forces passes through them. Moreover, one can’t compare their texture to that of a sponge or light bread because heat from fire causes them to melt, which alters the arrangement of the particles. Thus, they are, as mentioned, collections of particles that touch each other without forming a continuous solid. Therefore, the movement that these particles receive to transmit light waves can be shared from some to others without them shifting out of place or becoming disordered, allowing the waves to propagate without affecting the apparent solidity of the material.

By pressure from without, of which I have spoken, must not be understood that of the air, which would not be sufficient, but that of some other more subtle matter, a pressure which I chanced upon by experiment long ago, namely in the case of water freed from air, which remains suspended in a tube open at its lower end, notwithstanding [Pg 30]that the air has been removed from the vessel in which this tube is enclosed.

By external pressure, which I mentioned earlier, I don't mean the pressure of air, as that wouldn't be enough. I'm talking about some other finer substance, a pressure I discovered through experiments a long time ago—specifically with water that has been freed from air. This water stays suspended in a tube that's open at the bottom, even though the air has been taken out of the container that holds this tube. [Pg 30]

One can then in this way conceive of transparency in a solid without any necessity that the ethereal matter which serves for light should pass through it, or that it should find pores in which to insinuate itself. But the truth is that this matter not only passes through solids, but does so even with great facility; of which the experiment of Torricelli, above cited, is already a proof. Because on the quicksilver and the water quitting the upper part of the glass tube, it appears that it is immediately filled with ethereal matter, since light passes across it. But here is another argument which proves this ready penetrability, not only in transparent bodies but also in all others.

One can think of transparency in a solid without needing the ethereal matter that carries light to pass through it or to find tiny openings to get in. But the reality is that this matter not only passes through solids but does so quite easily, as shown by Torricelli's experiment mentioned earlier. When the mercury and water leave the upper part of the glass tube, it seems to fill immediately with ethereal matter since light can travel through it. But here's another point that shows this easy penetrability happens not just in transparent bodies but in all types.

When light passes across a hollow sphere of glass, closed on all sides, it is certain that it is full of ethereal matter, as much as the spaces outside the sphere. And this ethereal matter, as has been shown above, consists of particles which just touch one another. If then it were enclosed in the sphere in such a way that it could not get out through the pores of the glass, it would be obliged to follow the movement of the sphere when one changes its place: and it would require consequently almost the same force to impress a certain velocity on this sphere, when placed on a horizontal plane, as if it were full of water or perhaps of quicksilver: because every body resists the velocity of the motion which one would give to it, in proportion to the quantity of matter which it contains, and which is obliged to follow this motion. But on the contrary one finds that the sphere resists the impress of movement only in proportion to the quantity of matter of the glass of which it is made. Then it must be that the ethereal matter which [Pg 31]is inside is not shut up, but flows through it with very great freedom. We shall demonstrate hereafter that by this process the same penetrability may be inferred also as relating to opaque bodies.

When light moves through a hollow glass sphere that's closed on all sides, it definitely has some kind of ethereal matter inside, just like the space outside the sphere. This ethereal matter, as previously mentioned, is made up of particles that barely touch each other. If it were trapped inside the sphere so it couldn't escape through the glass, it would have to move along with the sphere when it was moved. Consequently, it would take nearly the same amount of force to give this sphere a certain speed on a flat surface as it would to do so if it were filled with water or even mercury. This is because every object resists the speed that is applied to it based on the amount of matter it contains, which also has to move along. However, we find that the sphere only resists movement based on the amount of glass it's made of. Therefore, the ethereal matter inside it isn't trapped but instead flows through it quite freely. Later, we will show that this same ability to penetrate also applies to opaque objects.

The second mode then of explaining transparency, and one which appears more probably true, is by saying that the waves of light are carried on in the ethereal matter, which continuously occupies the interstices or pores of transparent bodies. For since it passes through them continuously and freely, it follows that they are always full of it. And one may even show that these interstices occupy much more space than the coherent particles which constitute the bodies. For if what we have just said is true: that force is required to impress a certain horizontal velocity on bodies in proportion as they contain coherent matter; and if the proportion of this force follows the law of weights, as is confirmed by experiment, then the quantity of the constituent matter of bodies also follows the proportion of their weights. Now we see that water weighs only one fourteenth part as much as an equal portion of quicksilver: therefore the matter of the water does not occupy the fourteenth part of the space which its mass obtains. It must even occupy much less of it, since quicksilver is less heavy than gold, and the matter of gold is by no means dense, as follows from the fact that the matter of the vortices of the magnet and of that which is the cause of gravity pass very freely through it.

The second way to explain transparency, which seems more likely to be true, is to say that light waves travel through an ethereal substance that continuously fills the spaces or pores of transparent materials. Since light moves through them easily and continuously, it means these materials are always filled with it. Moreover, we can demonstrate that these spaces take up much more volume than the solid particles making up the materials. If what we just mentioned is accurate—that a force is needed to give objects a certain horizontal speed based on the amount of solid matter they contain—and if the amount of this force follows the law of weights, as experiments confirm, then the amount of matter in objects also relates to their weights. For instance, we see that water weighs only one-fourteenth as much as the same volume of mercury, so the matter in water doesn’t occupy one-fourteenth of the space that its mass implies. It likely occupies even less space because mercury is lighter than gold, and gold’s matter is not very dense. This is evident since the matter in the vortices generated by magnets and the matter causing gravity can move freely through it.

But it may be objected here that if water is a body of so great rarity, and if its particles occupy so small a portion of the space of its apparent bulk, it is very strange how it yet resists Compression so strongly without permitting itself to be condensed by any force which one has [Pg 32]hitherto essayed to employ, preserving even its entire liquidity while subjected to this pressure.

But someone might argue that if water is such a rare substance, and if its particles take up such a small part of its apparent volume, it's quite strange how it resists compression so effectively without being condensed by any force that has been tried so far, maintaining its complete liquid form even under this pressure. [Pg 32]

This is no small difficulty. It may, however, be resolved by saying that the very violent and rapid motion of the subtle matter which renders water liquid, by agitating the particles of which it is composed, maintains this liquidity in spite of the pressure which hitherto any one has been minded to apply to it.

This is not a minor issue. It can, however, be explained by saying that the very fast and intense movement of the tiny particles that make up water keeps it liquid, even under the pressure that anyone has thought to apply to it.

The rarity of transparent bodies being then such as we have said, one easily conceives that the waves might be carried on in the ethereal matter which fills the interstices of the particles. And, moreover, one may believe that the progression of these waves ought to be a little slower in the interior of bodies, by reason of the small detours which the same particles cause. In which different velocity of light I shall show the cause of refraction to consist.

The rarity of transparent substances, as we've mentioned, makes it easy to understand that waves could move through the ethereal matter that fills the spaces between particles. Additionally, it's reasonable to think that these waves should travel a bit slower inside objects due to the slight detours caused by the particles themselves. I will explain that this difference in the speed of light is the reason for refraction.

Before doing so, I will indicate the third and last mode in which transparency may be conceived; which is by supposing that the movement of the waves of light is transmitted indifferently both in the particles of the ethereal matter which occupy the interstices of bodies, and in the particles which compose them, so that the movement passes from one to the other. And it will be seen hereafter that this hypothesis serves excellently to explain the double refraction of certain transparent bodies.

Before doing so, I will point out the third and final way that transparency can be understood; which is by suggesting that the movement of light waves travels equally through both the particles of the ethereal matter found in the gaps between objects and the particles that make up those objects, allowing the movement to transfer from one to the other. It will be shown later that this hypothesis works very well to explain the double refraction observed in certain transparent materials.

Should it be objected that if the particles of the ether are smaller than those of transparent bodies (since they pass through their intervals), it would follow that they can communicate to them but little of their movement, it may be replied that the particles of these bodies are in turn composed of still smaller particles, and so it will be [Pg 33]these secondary particles which will receive the movement from those of the ether.

Should it be argued that if the particles of ether are smaller than those of transparent objects (since they can pass through their gaps), then they can only transfer a little of their movement to them, it can be countered that the particles of these objects are further made up of even smaller particles, and so it will be [Pg 33]these smaller particles that will receive the movement from those of the ether.

Furthermore, if the particles of transparent bodies have a recoil a little less prompt than that of the ethereal particles, which nothing hinders us from supposing, it will again follow that the progression of the waves of light will be slower in the interior of such bodies than it is outside in the ethereal matter.

Furthermore, if the particles of clear materials react a bit slower than the particles in the ether, which we can assume, it would mean that the speed of light waves inside those materials will be slower than it is outside in the ether.

All this I have found as most probable for the mode in which the waves of light pass across transparent bodies. To which it must further be added in what respect these bodies differ from those which are opaque; and the more so since it might seem because of the easy penetration of bodies by the ethereal matter, of which mention has been made, that there would not be any body that was not transparent. For by the same reasoning about the hollow sphere which I have employed to prove the smallness of the density of glass and its easy penetrability by the ethereal matter, one might also prove that the same penetrability obtains for metals and for every other sort of body. For this sphere being for example of silver, it is certain that it contains some of the ethereal matter which serves for light, since this was there as well as in the air when the opening of the sphere was closed. Yet, being closed and placed upon a horizontal plane, it resists the movement which one wishes to give to it, merely according to the quantity of silver of which it is made; so that one must conclude, as above, that the ethereal matter which is enclosed does not follow the movement of the sphere; and that therefore silver, as well as glass, is very easily penetrated by this matter. Some of it is therefore present continuously and in quantities between the particles of silver and of all other opaque [Pg 34]bodies: and since it serves for the propagation of light it would seem that these bodies ought also to be transparent, which however is not the case.

All of this seems to be the most likely explanation for how light waves move through transparent objects. Additionally, we should consider how these objects differ from opaque ones; it might seem that because the ethereal matter can easily penetrate objects, there wouldn’t be any that aren’t transparent. Using the same logic I applied to the hollow sphere to demonstrate the low density of glass and its ability to be easily penetrated by ethereal matter, one could argue that this same penetrability applies to metals and all other types of matter. For example, if the sphere were made of silver, it would definitely contain some ethereal matter that is necessary for light, since it was present alongside the air when the sphere was sealed. However, when it is closed and sitting on a flat surface, it resists any movement applied to it purely based on the amount of silver it contains. Thus, we must conclude, as previously mentioned, that the ethereal matter inside does not move with the sphere; therefore, silver, like glass, is easily penetrated by this matter. Some of it is continuously present in quantities between the particles of silver and all other opaque [Pg 34]bodies. Since this matter is vital for the transmission of light, it would suggest that these bodies should also be transparent, which is not the case.

Whence then, one will say, does their opacity come? Is it because the particles which compose them are soft; that is to say, these particles being composed of others that are smaller, are they capable of changing their figure on receiving the pressure of the ethereal particles, the motion of which they thereby damp, and so hinder the continuance of the waves of light? That cannot be: for if the particles of the metals are soft, how is it that polished silver and mercury reflect light so strongly? What I find to be most probable herein, is to say that metallic bodies, which are almost the only really opaque ones, have mixed amongst their hard particles some soft ones; so that some serve to cause reflexion and the others to hinder transparency; while, on the other hand, transparent bodies contain only hard particles which have the faculty of recoil, and serve together with those of the ethereal matter for the propagation of the waves of light, as has been said.

Where, then, one might ask, does their opacity come from? Is it because the particles that make them up are soft? In other words, do these particles consist of smaller particles that can change shape when pressed by ethereal particles, which dampen their motion and prevent the continuation of light waves? That can't be the case: if the particles of the metals are soft, how do polished silver and mercury reflect light so well? What seems most likely to me is that metallic bodies, which are nearly the only truly opaque ones, have a mix of hard and soft particles; some of these particles enable reflection while others prevent transparency. In contrast, transparent bodies are made up only of hard particles that can recoil and work with those of the ethereal matter to propagate light waves, as has been explained.

Let us pass now to the explanation of the effects of Refraction, assuming, as we have done, the passage of waves of light through transparent bodies, and the diminution of velocity which these same waves suffer in them.

Let’s now explain the effects of refraction, assuming, as we have, that light waves travel through transparent materials and that these waves experience a decrease in speed as they do so.

The chief property of Refraction is that a ray of light, such as AB, being in the air, and falling obliquely upon the polished surface of a transparent body, such as FG, is [Pg 35]broken at the point of incidence B, in such a way that with the straight line DBE which cuts the surface perpendicularly it makes an angle CBE less than ABD which it made with the same perpendicular when in the air. And the measure of these angles is found by describing, about the point B, a circle which cuts the radii AB, BC. For the perpendiculars AD, CE, let fall from the points of intersection upon the straight line DE, which are called the Sines of the angles ABD, CBE, have a certain ratio between themselves; which ratio is always the same for all inclinations of the incident ray, at least for a given transparent body. This ratio is, in glass, very nearly as 3 to 2; and in water very nearly as 4 to 3; and is likewise different in other diaphanous bodies.

The main property of Refraction is that when a ray of light, like AB, is in the air and strikes the polished surface of a transparent material, like FG, at an angle, it gets [Pg 35]bent at the point of impact B. This happens in such a way that the straight line DBE, which meets the surface at a right angle, forms an angle CBE that is smaller than angle ABD, which it made with the same perpendicular while in the air. To find these angles, you can draw a circle around point B that intersects the lines AB and BC. The perpendiculars AD and CE, dropped from the intersection points onto the straight line DE, are called the Sines of angles ABD and CBE, and they maintain a specific ratio relative to each other. This ratio stays consistent for all angles of the incoming ray, at least for a given transparent material. For glass, this ratio is roughly 3 to 2; for water, it's about 4 to 3; and it varies in other transparent materials.

Another property, similar to this, is that the refractions are reciprocal between the rays entering into a transparent body and those which are leaving it. That is to say that if the ray AB in entering the transparent body is refracted into BC, then likewise CB being taken as a ray in the interior of this body will be refracted, on passing out, into BA.

Another property, similar to this, is that the refractions are reciprocal between the rays entering a transparent object and those that are exiting it. In other words, if the ray AB enters the transparent object and is refracted into BC, then similarly, if CB is considered a ray inside this object, it will be refracted back out into BA.

To explain then the reasons of these phenomena according to our principles, let AB be the straight line which [Pg 36]represents a plane surface bounding the transparent substances which lie towards C and towards N. When I say plane, that does not signify a perfect evenness, but such as has been understood in treating of reflexion, and for the same reason. Let the line AC represent a portion of a wave of light, the centre of which is supposed so distant that this portion may be considered as a straight line. The piece C, then, of the wave AC, will in a certain space of time have advanced as far as the plane AB following the straight line CB, which may be imagined as coming from the luminous centre, and which consequently will cut AC at right angles. Now in the same time the piece A would have come to G along the straight line AG, equal and parallel to CB; and all the portion of wave AC would be at GB if the matter of the transparent body transmitted the movement of the wave as quickly as the matter of the Ether. But let us suppose that it transmits this movement less quickly, by one-third, for instance. Movement will then be spread from the point A, in the matter of the transparent body through a distance equal to two-thirds of CB, making its own particular spherical wave according to what has been said before. This wave is then represented by the circumference SNR, the centre of which is A, and its semi-diameter equal to two-thirds of CB. Then if one considers in order the other pieces H of the wave AC, it appears that in the same time that the piece C reaches B they will not only have arrived at the surface AB along the straight lines HK parallel to CB, but that, in addition, they will have generated in the diaphanous substance from the centres K, partial waves, represented here by circumferences the semi-diameters of which are equal to two-thirds of the lines KM, that is to say, to [Pg 37]two-thirds of the prolongations of HK down to the straight line BG; for these semi-diameters would have been equal to entire lengths of KM if the two transparent substances had been of the same penetrability.

To explain the reasons for these phenomena based on our principles, let AB be the straight line which [Pg 36]represents a flat surface bounding the transparent materials that lie toward C and toward N. When I say flat, I don’t mean perfectly even, but as interpreted in the discussion of reflection, and for the same reason. Let the line AC represent a part of a light wave, the center of which is assumed to be so far away that this part can be treated as a straight line. The piece C of the wave AC will, in a certain period, have moved as far as the plane AB along the straight line CB, which can be imagined as coming from the light source and which will thus intersect AC at right angles. Now during the same time, piece A would have reached G along the straight line AG, which is equal and parallel to CB; and the entire wave AC would be at GB if the material of the transparent body transmitted the wave movement as quickly as the material of the Ether. But let’s assume it transmits this movement slower, by one-third, for example. The movement will then spread from point A in the material of the transparent body through a distance equal to two-thirds of CB, creating its own particular spherical wave as previously described. This wave is represented by the circumference SNR, with A as its center and its semi-diameter equal to two-thirds of CB. Then, if we consider the other pieces H of the wave AC in order, it becomes clear that at the same time piece C reaches B, they will not only have arrived at surface AB along the straight lines HK parallel to CB, but additionally, they will have generated in the transparent substance from centers K, partial waves represented here by circumferences with semi-diameters equal to two-thirds of lines KM, which is to say, to [Pg 37]two-thirds of the extensions of HK down to the straight line BG; for these semi-diameters would have been equal to the full lengths of KM if the two transparent substances had equal penetrability.

Now all these circumferences have for a common tangent the straight line BN; namely the same line which is drawn as a tangent from the point B to the circumference SNR which we considered first. For it is easy to see that all the other circumferences will touch the same BN, from B up to the point of contact N, which is the same point where AN falls perpendicularly on BN.

Now all these circles share a common tangent, the straight line BN; specifically, the same line drawn as a tangent from point B to the circle SNR that we looked at first. It's clear that all the other circles will touch the same BN, from point B to the contact point N, which is where AN meets BN at a right angle.

It is then BN, which is formed by small arcs of these circumferences, which terminates the movement that the wave AC has communicated within the transparent body, and where this movement occurs in much greater amount than anywhere else. And for that reason this line, in accordance with what has been said more than once, is the propagation of the wave AC at the moment when its piece C has reached B. For there is no other line below the plane AB which is, like BN, a common tangent to all these partial waves. And if one would know how the wave AC has come progressively to BN, it is necessary only to draw in the same figure the straight lines KO parallel to BN, and all the lines KL parallel to AC. Thus one will see that the wave CA, from being a straight line, has become broken in all the positions LKO successively, and that it has again become a straight line at BN. This being evident by what has already been demonstrated, there is no need to explain it further.

It is then BN, which is formed by small arcs of these circles, that concludes the movement that the wave AC has transmitted within the transparent medium, where this movement is significantly greater than anywhere else. For this reason, this line, as mentioned several times before, is the propagation of the wave AC at the moment when its point C has reached B. There is no other line below the plane AB that is, like BN, a common tangent to all these partial waves. If you want to understand how the wave AC has gradually approached BN, you only need to draw the straight lines KO parallel to BN, and all the lines KL parallel to AC in the same figure. In this way, you will see that the wave CA, which starts as a straight line, has become broken at all the points LKO successively, and that it has again become a straight line at BN. This is clear from what has already been demonstrated, so there's no need to explain it further.

Now, in the same figure, if one draws EAF, which cuts the plane AB at right angles at the point A, since AD is perpendicular to the wave AC, it will be DA which will [Pg 38]mark the ray of incident light, and AN which was perpendicular to BN, the refracted ray: since the rays are nothing else than the straight lines along which the portions of the waves advance.

Now, in the same figure, if you draw EAF, which intersects the plane AB at a right angle at point A, and since AD is perpendicular to the wave AC, it will be DA that will [Pg 38]mark the incident light ray, and AN, which is perpendicular to BN, the refracted ray: because the rays are simply the straight lines along which the parts of the waves move forward.

Whence it is easy to recognize this chief property of refraction, namely that the Sine of the angle DAE has always the same ratio to the Sine of the angle NAF, whatever be the inclination of the ray DA: and that this ratio is the same as that of the velocity of the waves in the transparent substance which is towards AE to their velocity in the transparent substance towards AF. For, considering AB as the radius of a circle, the Sine of the angle BAC is BC, and the Sine of the angle ABN is AN. But the angle BAC is equal to DAE, since each of them added to CAE makes a right angle. And the angle ABN is equal to NAF, since each of them with BAN makes a right angle. Then also the Sine of the angle DAE is to the Sine of NAF as BC is to AN. But the ratio of BC to AN was the same as that of the velocities of light in the substance which is towards AE and in that which is towards AF; therefore also the Sine of the angle DAE will be to the Sine of the angle NAF the same as the said velocities of light.

It’s easy to see this main feature of refraction: the sine of angle DAE always has the same ratio to the sine of angle NAF, no matter the angle of the ray DA. This ratio is the same as the ratio of the speed of waves in the transparent medium towards AE to their speed in the transparent medium towards AF. If we consider AB as the radius of a circle, the sine of angle BAC is BC, and the sine of angle ABN is AN. The angle BAC is equal to DAE because each of them added to CAE makes a right angle. Similarly, angle ABN is equal to NAF since each one combined with BAN makes a right angle. Thus, the sine of angle DAE is to the sine of NAF as BC is to AN. The ratio of BC to AN is the same as the ratio of the speeds of light in the medium towards AE and AF; therefore, the sine of angle DAE will be to the sine of angle NAF as those light speeds are.

To see, consequently, what the refraction will be when the waves of light pass into a substance in which the movement travels more quickly than in that from which they emerge (let us again assume the ratio of 3 to 2), it is only necessary to repeat all the same construction and demonstration which we have just used, merely substituting everywhere 3/2 instead of 2/3. And it will be found by the same reasoning, in this other figure, that when the piece C of the wave AC shall have reached the surface AB at B, [Pg 39]all the portions of the wave AC will have advanced as far as BN, so that BC the perpendicular on AC is to AN the perpendicular on BN as 2 to 3. And there will finally be this same ratio of 2 to 3 between the Sine of the angle BAD and the Sine of the angle FAN.

To determine what the refraction will be when light waves enter a substance where the speed is faster than in the medium they are leaving (let's again use the ratio of 3 to 2), all we need to do is repeat the same construction and demonstration we just performed, simply replacing 2/3 with 3/2 throughout. Using the same reasoning in this other figure, when part C of wave AC reaches the surface AB at point B, [Pg 39]all sections of wave AC will have progressed to point BN, meaning that BC, the perpendicular to AC, is to AN, the perpendicular to BN, as 2 is to 3. Ultimately, there will also be the same ratio of 2 to 3 between the Sine of angle BAD and the Sine of angle FAN.

Hence one sees the reciprocal relation of the refractions of the ray on entering and on leaving one and the same transparent body: namely that if NA falling on the external surface AB is refracted into the direction AD, so the ray AD will be refracted on leaving the transparent body into the direction AN.

Hence one sees the mutual relationship of the refractions of the ray when it enters and exits the same transparent body: namely, if the ray NA hitting the external surface AB is bent into the direction AD, then the ray AD will be bent on leaving the transparent body into the direction AN.

One sees also the reason for a noteworthy accident which happens in this refraction: which is this, that after a certain obliquity of the incident ray DA, it begins to be quite unable to penetrate into the other transparent substance. For if the angle DAQ or CBA is such that in the triangle ACB, CB is equal to 2/3 of AB, or is greater, then AN cannot form one side of the triangle ANB, since it becomes equal to or greater than AB: so that the portion of wave BN cannot be found anywhere, neither consequently can AN, which ought to be perpendicular to it. And thus the incident ray DA does not then pierce the surface AB.

One can also see the reason for a notable incident that occurs during this refraction: after a certain angle of the incoming ray DA, it can no longer penetrate the other transparent material. If the angle DAQ or CBA is such that in triangle ACB, CB is equal to 2/3 of AB, or larger, then AN cannot be one side of triangle ANB, since it becomes equal to or larger than AB. As a result, the segment of the wave BN cannot be found anywhere, and consequently, AN, which should be perpendicular to it, cannot either. Therefore, the incoming ray DA does not break through the surface AB.

[Pg 40]When the ratio of the velocities of the waves is as two to three, as in our example, which is that which obtains for glass and air, the angle DAQ must be more than 48 degrees 11 minutes in order that the ray DA may be able to pass by refraction. And when the ratio of the velocities is as 3 to 4, as it is very nearly in water and air, this angle DAQ must exceed 41 degrees 24 minutes. And this accords perfectly with experiment.

[Pg 40]When the ratio of the wave speeds is two to three, like in our example with glass and air, the angle DAQ needs to be more than 48 degrees 11 minutes for ray DA to pass through refraction. When the ratio is three to four, which is close to the situation with water and air, this angle DAQ must be greater than 41 degrees 24 minutes. This matches perfectly with experimental results.

But it might here be asked: since the meeting of the wave AC against the surface AB ought to produce movement in the matter which is on the other side, why does no light pass there? To which the reply is easy if one remembers what has been said before. For although it generates an infinitude of partial waves in the matter which is at the other side of AB, these waves never have a common tangent line (either straight or curved) at the same moment; and so there is no line terminating the propagation of the wave AC beyond the plane AB, nor any place where the movement is gathered together in sufficiently great quantity to produce light. And one will easily see the truth of this, namely that CB being larger than 2/3 of AB, the waves excited beyond the plane AB will have no common tangent if about the centres K one then draws circles having radii equal to 3/2 of the lengths LB to which they correspond. For all these circles will be enclosed in one another and will all pass beyond the point B.

But one might ask: since the wave AC meeting the surface AB should create movement in the matter on the other side, why doesn't any light get through? The answer is straightforward if you recall what has been mentioned before. Even though it creates countless partial waves in the matter beyond AB, these waves never align along a common tangent line (whether straight or curved) at the same time; therefore, there’s no line that allows the wave AC to continue spreading beyond the plane AB, nor is there a spot where the movement builds up enough to generate light. It’s easy to see this truth: since CB is greater than 2/3 of AB, the waves generated beyond the plane AB won't have a common tangent if you draw circles centered around K with radii equal to 3/2 of the lengths LB they correspond to. All these circles will be nested within one another and will all extend past point B.

Now it is to be remarked that from the moment when the angle DAQ is smaller than is requisite to permit the refracted ray DA to pass into the other transparent substance, one finds that the interior reflexion which occurs at the surface AB is much augmented in brightness, as [Pg 41]is easy to realize by experiment with a triangular prism; and for this our theory can afford this reason. When the angle DAQ is still large enough to enable the ray DA to pass, it is evident that the light from the portion AC of the wave is collected in a minimum space when it reaches BN. It appears also that the wave BN becomes so much the smaller as the angle CBA or DAQ is made less; until when the latter is diminished to the limit indicated a little previously, this wave BN is collected together always at one point. That is to say, that when the piece C of the wave AC has then arrived at B, the wave BN which is the propagation of AC is entirely reduced to the same point B. Similarly when the piece H has reached K, the part AH is entirely reduced to the same point K. This makes it evident that in proportion as the wave CA comes to meet the surface AB, there occurs a great quantity of movement along that surface; which movement ought also to spread within the transparent body and ought to have much re-enforced the partial waves which produce the interior reflexion against the surface AB, according to the laws of reflexion previously explained.

Now, it's important to note that once the angle DAQ is smaller than what's needed for the refracted ray DA to enter the other transparent material, the internal reflection that happens at the surface AB becomes much brighter, as [Pg 41] can be easily demonstrated with a triangular prism. Our theory explains this. When the angle DAQ is still large enough for the ray DA to pass through, it's clear that the light from the section AC of the wave is focused into a smaller area by the time it reaches BN. It also seems that wave BN gets smaller as the angle CBA or DAQ decreases; until, when the latter is minimized as previously indicated, wave BN converges at a single point. This means that when the piece C of wave AC reaches B, wave BN, which is the continuation of AC, is completely concentrated at point B. Similarly, when piece H gets to K, part AH is entirely focused at point K. This illustrates that as wave CA approaches surface AB, there is significant movement along that surface, which should also propagate within the transparent body and substantially enhance the partial waves that create the internal reflection against surface AB, following the reflection laws we've discussed earlier.

And because a slight diminution of the angle of incidence DAQ causes the wave BN, however great it was, to be reduced to zero, (for this angle being 49 degrees 11 minutes in the glass, the angle BAN is still 11 degrees 21 minutes, and the same angle being reduced by one degree only the angle BAN is reduced to zero, and so the wave BN reduced to a point) thence it comes about that the interior reflexion from being obscure becomes suddenly bright, so soon as the angle of incidence is such that it no longer gives passage to the refraction.

And because a small decrease in the angle of incidence DAQ causes the wave BN, no matter how strong it was, to drop to zero (since this angle is 49 degrees 11 minutes in glass, the angle BAN remains at 11 degrees 21 minutes, and by reducing the angle by just one degree, the angle BAN goes to zero, making the wave BN diminish to a point), this results in the interior reflection shifting from being dim to suddenly bright as soon as the angle of incidence reaches a point where it no longer allows refraction to occur.

[Pg 42]Now as concerns ordinary external reflexion, that is to say which occurs when the angle of incidence DAQ is still large enough to enable the refracted ray to penetrate beyond the surface AB, this reflexion should occur against the particles of the substance which touches the transparent body on its outside. And it apparently occurs against the particles of the air or others mingled with the ethereal particles and larger than they. So on the other hand the external reflexion of these bodies occurs against the particles which compose them, and which are also larger than those of the ethereal matter, since the latter flows in their interstices. It is true that there remains here some difficulty in those experiments in which this interior reflexion occurs without the particles of air being able to contribute to it, as in vessels or tubes from which the air has been extracted.

[Pg 42]Now regarding ordinary external reflection, which happens when the angle of incidence DAQ is still large enough for the refracted ray to pass beyond the surface AB, this reflection should occur against the particles of the substance that touches the transparent body from the outside. It seems to happen against the particles of air or others mixed with ethereal particles that are larger than those. Conversely, the external reflection of these bodies occurs against the particles that make them up, which are also larger than the ethereal matter, since the latter flows in their spaces. It is true that there remains some difficulty in those experiments where this internal reflection occurs without the air particles being able to contribute, such as in vessels or tubes from which the air has been removed.

Experience, moreover, teaches us that these two reflexions are of nearly equal force, and that in different transparent bodies they are so much the stronger as the refraction of these bodies is the greater. Thus one sees manifestly that the reflexion of glass is stronger than that of water, and that of diamond stronger than that of glass.

Experience also shows us that these two reflections are almost equally strong, and that in different transparent materials, they become even stronger as the refraction of these materials increases. So, it’s clear that the reflection of glass is stronger than that of water, and the reflection of diamond is stronger than that of glass.

I will finish this theory of refraction by demonstrating a remarkable proposition which depends on it; namely, that a ray of light in order to go from one point to another, when these points are in different media, is refracted in such wise at the plane surface which joins these two media that it employs the least possible time: and exactly the same happens in the case of reflexion against a plane surface. Mr. Fermat was the first to propound this property of refraction, holding with us, and directly counter to the opinion of Mr. Des Cartes, that light passes [Pg 43]more slowly through glass and water than through air. But he assumed besides this a constant ratio of Sines, which we have just proved by these different degrees of velocity alone: or rather, what is equivalent, he assumed not only that the velocities were different but that the light took the least time possible for its passage, and thence deduced the constant ratio of the Sines. His demonstration, which may be seen in his printed works, and in the volume of letters of Mr. Des Cartes, is very long; wherefore I give here another which is simpler and easier.

I will conclude this theory of refraction by showing a remarkable idea that relies on it: a ray of light traveling from one point to another, when these points are in different media, is refracted at the surface connecting these two media in such a way that it takes the least possible time. The same principle applies to reflection off a flat surface. Mr. Fermat was the first to suggest this property of refraction, agreeing with us and directly opposing Mr. Des Cartes, who believed that light travels more slowly through glass and water than through air. However, he also assumed a constant ratio of sines, which we've just demonstrated using these different speeds alone. In other words, he assumed not only that the speeds were different but that light took the least time possible to travel, which allowed him to derive the constant ratio of the sines. His proof, found in his published works and in the letters of Mr. Des Cartes, is quite lengthy; therefore, I will present a simpler and easier one here.

Let KF be the plane surface; A the point in the medium which the light traverses more easily, as the air; C the point in the other which is more difficult to penetrate, as water. And suppose that a ray has come from A, by B, to C, having been refracted at B according to the law demonstrated a little before; that is to say that, having drawn PBQ, which cuts the plane at right angles, let the sine of the angle ABP have to the sine of the angle CBQ the same ratio as the velocity of light in the medium where A is to the velocity of light in the medium where C is. It is to be shown that the time of passage of light along AB and BC taken together, is the shortest that can be. Let us assume that it may have come by other lines, and, in the first place, along AF, FC, so [Pg 44]that the point of refraction F may be further from B than the point A; and let AO be a line perpendicular to AB, and FO parallel to AB; BH perpendicular to FO, and FG to BC.

Let KF be the flat surface; A the point in the medium where light travels more easily, like air; C the point in the medium that is harder for light to penetrate, like water. Suppose a ray of light has traveled from A, by B, to C, being refracted at B according to the law demonstrated earlier; that is, if we draw PBQ, which intersects the plane at a right angle, the sine of angle ABP will have the same ratio to the sine of angle CBQ as the speed of light in the medium at A has to the speed of light in the medium at C. We will show that the total time taken for light to travel along AB and BC is the shortest possible. Let's assume that it could take other paths, first along AF, FC, so [Pg 44]that the point of refraction F is farther from B than point A; let AO be a line perpendicular to AB, and FO parallel to AB; BH perpendicular to FO, and FG to BC.

Since then the angle HBF is equal to PBA, and the angle BFG equal to QBC, it follows that the sine of the angle HBF will also have the same ratio to the sine of BFG, as the velocity of light in the medium A is to its velocity in the medium C. But these sines are the straight lines HF, BG, if we take BF as the semi-diameter of a circle. Then these lines HF, BG, will bear to one another the said ratio of the velocities. And, therefore, the time of the light along HF, supposing that the ray had been OF, would be equal to the time along BG in the interior of the medium C. But the time along AB is equal to the time along OH; therefore the time along OF is equal to the time along AB, BG. Again the time along FC is greater than that along GC; then the time along OFC will be longer than that along ABC. But AF is longer than OF, then the time along AFC will by just so much more exceed the time along ABC.

Since then, the angle HBF is equal to PBA, and the angle BFG is equal to QBC. This means that the sine of angle HBF will have the same ratio to the sine of BFG as the speed of light in medium A is to its speed in medium C. These sines correspond to the straight lines HF and BG, assuming BF is the semi-diameter of a circle. Thus, the lines HF and BG will have that same ratio of speeds. Therefore, the time for light traveling along HF, if the ray were OF, would be equal to the time along BG in medium C. Since the time along AB is equal to the time along OH, the time along OF is equal to the time along AB and BG. Additionally, the time along FC is greater than the time along GC, so the time along OFC will be longer than along ABC. Since AF is longer than OF, the time along AFC will exceed the time along ABC by that much more.

Now let us assume that the ray has come from A to C along AK, KC; the point of refraction K being nearer to A than the point B is; and let CN be the perpendicular upon BC, KN parallel to BC: BM perpendicular upon KN, and KL upon BA.

Now let’s assume that the ray travels from A to C along AK and KC; with the point of refraction K being closer to A than point B is; and let CN be the perpendicular to BC, KN parallel to BC: BM perpendicular to KN, and KL to BA.

Here BL and KM are the sines of angles BKL, KBM; that is to say, of the angles PBA, QBC; and therefore they are to one another as the velocity of light in the medium A is to the velocity in the medium C. Then the time along LB is equal to the time along KM; and since the time along BC is equal to the time along MN, the [Pg 45]time along LBC will be equal to the time along KMN. But the time along AK is longer than that along AL: hence the time along AKN is longer than that along ABC. And KC being longer than KN, the time along AKC will exceed, by as much more, the time along ABC. Hence it appears that the time along ABC is the shortest possible; which was to be proven.

Here, BL and KM represent the sines of angles BKL and KBM, which correspond to angles PBA and QBC. Therefore, they relate to each other the same way the speed of light in medium A relates to the speed in medium C. The time taken along LB equals the time taken along KM. Since the time along BC is the same as the time along MN, the time along LBC will equal the time along KMN. However, the time along AK is longer than the time along AL; thus, the time along AKN is longer than the time along ABC. And since KC is longer than KN, the time along AKC will be even longer than the time along ABC. Therefore, it shows that the time along ABC is the shortest possible, which was the point to prove.


CHAPTER IV

ON THE REFRACTION OF THE AIR

W

e have shown how the movement which constitutes light spreads by spherical waves in any homogeneous matter. And it is evident that when the matter is not homogeneous, but of such a constitution that the movement is communicated in it more rapidly toward one side than toward another, these waves cannot be spherical: but that they must acquire their figure according to the different distances over which the successive movement passes in equal times.

We have demonstrated how the movement that makes up light spreads as spherical waves in any uniform material. It's clear that when the material is not uniform, and the movement travels more quickly in one direction than in another, these waves cannot be spherical. Instead, they must take on a shape based on the varying distances the movement covers in equal time intervals.

It is thus that we shall in the first place explain the refractions which occur in the air, which extends from here to the clouds and beyond. The effects of which refractions are very remarkable; for by them we often see objects which the rotundity of the Earth ought otherwise to hide; such as Islands, and the tops of mountains when one is at sea. Because also of them the Sun and the Moon appear as risen before in fact they have, and appear to set [Pg 46]later: so that at times the Moon has been seen eclipsed while the Sun appeared still above the horizon. And so also the heights of the Sun and of the Moon, and those of all the Stars always appear a little greater than they are in reality, because of these same refractions, as Astronomers know. But there is one experiment which renders this refraction very evident; which is that of fixing a telescope on some spot so that it views an object, such as a steeple or a house, at a distance of half a league or more. If then you look through it at different hours of the day, leaving it always fixed in the same way, you will see that the same spots of the object will not always appear at the middle of the aperture of the telescope, but that generally in the morning and in the evening, when there are more vapours near the Earth, these objects seem to rise higher, so that the half or more of them will no longer be visible; and so that they seem lower toward mid-day when these vapours are dissipated.

First, we will explain the refractions that happen in the air, which stretches from here to the clouds and beyond. The effects of these refractions are quite remarkable; they often allow us to see objects that the curvature of the Earth would normally hide, like islands and the tops of mountains when we are at sea. Because of these refractions, the Sun and the Moon seem to rise before they actually do and appear to set later; sometimes the Moon has been seen eclipsed while the Sun still looks like it's above the horizon. Additionally, the heights of the Sun, Moon, and all the stars always seem slightly higher than they really are due to these same refractions, as astronomers are aware. There is one experiment that clearly demonstrates this refraction: if you set up a telescope to view an object, like a steeple or a house, at a distance of half a league or more, and look through it at different times of the day while keeping it fixed, you’ll notice that the same points of the object won’t always appear in the center of the telescope's view. Generally, in the morning and evening, when there are more vapors near the Earth, these objects appear to rise higher, making half or more of them invisible; while around midday, when the vapors clear, they seem lower.

Those who consider refraction to occur only in the surfaces which separate transparent bodies of different nature, would find it difficult to give a reason for all that I have just related; but according to our Theory the thing is quite easy. It is known that the air which surrounds us, besides the particles which are proper to it and which float in the ethereal matter as has been explained, is full also of particles of water which are raised by the action of heat; and it has been ascertained further by some very definite experiments that as one mounts up higher the density of air diminishes in proportion. Now whether the particles of water and those of air take part, by means of the particles of ethereal matter, in the movement which constitutes light, but have a less prompt recoil than these, [Pg 47]or whether the encounter and hindrance which these particles of air and water offer to the propagation of movement of the ethereal progress, retard the progression, it follows that both kinds of particles flying amidst the ethereal particles, must render the air, from a great height down to the Earth, gradually less easy for the spreading of the waves of light.

Those who think that refraction only happens at the surfaces that separate transparent materials of different types would struggle to explain everything I've just mentioned; however, according to our Theory, it's quite simple. It's known that the air around us, in addition to the particles that make it up and that float in the ethereal matter as previously described, is also filled with water particles that are lifted by heat. Moreover, some very clear experiments have shown that as you go higher, the density of the air decreases proportionally. Now, whether the water particles and air particles participate, through the ethereal matter, in the movement that creates light but have a slower response than it, or whether the obstruction and resistance that these air and water particles present to the movement of the ethereal flow slow down its progression, it follows that both types of particles moving among the ethereal particles must make the air, from a great height down to the Earth, progressively less accommodating for the spreading of light waves.

Whence the configuration of the waves ought to become nearly such as this figure represents: namely, if A is a light, or the visible point of a steeple, the waves which start from it ought to spread more widely upwards and less widely downwards, but in other directions more or less as they approximate to these two extremes. This being so, it necessarily follows that every line intersecting one of these waves at right angles will pass above the point A, always excepting the one line which is perpendicular to the horizon.[Pg 48]

The shape of the waves should look almost like what this figure shows: if A represents a light, or the visible point of a steeple, then the waves that start from it should spread out wider upwards and narrower downwards, while in other directions, they spread more or less as they get closer to these two extremes. Given this, it follows that every line that crosses one of these waves at a right angle will always be above point A, except for the one line that is perpendicular to the horizon.[Pg 48]

Let BC be the wave which brings the light to the spectator who is at B, and let BD be the straight line which intersects this wave at right angles. Now because the ray or straight line by which we judge the spot where the object appears to us is nothing else than the perpendicular to the wave that reaches our eye, as will be understood by what was said above, it is manifest that the point A will be perceived as being in the line BD, and therefore higher than in fact it is.

Let BC be the wave that brings light to the viewer at B, and let BD be the straight line that intersects this wave at a right angle. Now, since the ray or straight line we use to determine where the object appears to us is just the perpendicular to the wave that reaches our eye, as explained above, it’s clear that point A will be seen as being on the line BD, and therefore higher than it actually is.

Similarly if the Earth be AB, and the top of the Atmosphere CD, which probably is not a well defined spherical surface (since we know that the air becomes rare in proportion as one ascends, for above there is so much less of it to press down upon it), the waves of light from the sun coming, for instance, in such a way that so long as they have not reached the Atmosphere CD the straight line AE intersects them perpendicularly, they ought, when they enter the Atmosphere, to advance more quickly in elevated regions than in regions nearer to the Earth. So that if [Pg 49]CA is the wave which brings the light to the spectator at A, its region C will be the furthest advanced; and the straight line AF, which intersects this wave at right angles, and which determines the apparent place of the Sun, will pass above the real Sun, which will be seen along the line AE. And so it may occur that when it ought not to be visible in the absence of vapours, because the line AE encounters the rotundity of the Earth, it will be perceived in the line AF by refraction. But this angle EAF is scarcely ever more than half a degree because the attenuation of the vapours alters the waves of light but little. Furthermore these refractions are not altogether constant in all weathers, particularly at small elevations of 2 or 3 degrees; which results from the different quantity of aqueous vapours rising above the Earth.

Similarly, if the Earth is AB, and the top of the Atmosphere is CD, which probably isn’t a perfectly defined spherical surface (since we know that air gets thinner as you go higher, due to having less of it pressing down), the waves of light from the sun arriving, for example, in such a way that as long as they haven’t reached the Atmosphere CD, the straight line AE intersects them perpendicularly, should, when they enter the Atmosphere, move faster in higher regions than in areas closer to the Earth. So, if [Pg 49]CA is the wave that brings light to the viewer at A, its region C will be the furthest along; and the straight line AF, which intersects this wave at right angles and determines the apparent position of the Sun, will pass above the actual Sun, which will be seen along the line AE. Thus, it might happen that when it shouldn’t be visible in the absence of vapors, because line AE hits the curvature of the Earth, it will be seen along line AF due to refraction. However, this angle EAF is rarely more than half a degree because the thinning of the vapors only slightly alters the light waves. Furthermore, these refractions are not entirely constant in all weather conditions, especially at low elevations of 2 or 3 degrees, which is due to the varying amounts of water vapor rising above the Earth.

And this same thing is the cause why at certain times a distant object will be hidden behind another less distant one, and yet may at another time be able to be seen, although the spot from which it is viewed is always the same. But the reason for this effect will be still more evident from what we are going to remark touching the curvature of rays. It appears from the things explained above that the progression or propagation of a small part of a wave of light is properly what one calls a ray. Now these rays, instead of being straight as they are in homogeneous media, ought to be curved in an atmosphere of unequal penetrability. For they necessarily follow from the object to the eye the line which intersects at right angles all the progressions of the waves, as in the first figure the line AEB does, as will be shown hereafter; and it is this line which determines what interposed bodies would or would not hinder us from seeing the object. For [Pg 50]although the point of the steeple A appears raised to D, it would yet not appear to the eye B if the tower H was between the two, because it crosses the curve AEB. But the tower E, which is beneath this curve, does not hinder the point A from being seen. Now according as the air near the Earth exceeds in density that which is higher, the curvature of the ray AEB becomes greater: so that at certain times it passes above the summit E, which allows the point A to be perceived by the eye at B; and at other times it is intercepted by the same tower E which hides A from this same eye.

And this is why sometimes a distant object is blocked by another closer object, but at other times it can be seen, even though the viewing spot remains the same. The reason for this effect will become clearer when we discuss the bending of light rays. From the previous explanations, we understand that the movement of a small section of a light wave is what we refer to as a ray. In homogeneous media, these rays are straight, but in an atmosphere with varying density, they bend. They follow a path that intersects perpendicularly with the wave's progress, as shown in the first figure with line AEB, which we will explain later. This path determines which objects will block our view of the distant object. For example, although the steeple point A seems elevated to point D, it won't be visible to observer B if the tower H is in between, as it obstructs the curve AEB. However, tower E, which is beneath this curve, doesn't block the view of point A. The curvature of ray AEB increases as the density of air near the Earth exceeds that of higher altitudes. This means that at certain times the ray can pass above the summit E, allowing observer B to see point A; at other times, it is blocked by the same tower E, hiding A from view.

But to demonstrate this curvature of the rays conformably to all our preceding Theory, let us imagine that AB is a small portion of a wave of light coming from the side C, which we may consider as a straight line. Let us also suppose that it is perpendicular to the Horizon, the portion B being nearer to the Earth than the portion A; and that because the vapours are less hindering at A than at B, the particular wave which comes from the point A spreads through a certain space AD while the particular wave which starts from the point B spreads through a shorter space BE; AD and BE being parallel to the Horizon. Further, supposing the straight lines FG, HI, etc., to be [Pg 51]drawn from an infinitude of points in the straight line AB and to terminate on the line DE (which is straight or may be considered as such), let the different penetrabilities at the different heights in the air between A and B be represented by all these lines; so that the particular wave, originating from the point F, will spread across the space FG, and that from the point H across the space HI, while that from the point A spreads across the space AD.

But to show this bending of the rays according to our earlier theory, let’s imagine that AB is a small part of a light wave coming from the side C, which we can think of as a straight line. Let’s also assume it is perpendicular to the horizon, with point B being closer to the Earth than point A. Because the vapors are less obstructive at A than at B, the wave coming from point A spreads across a certain distance AD, while the wave starting from point B spreads across a shorter distance BE; AD and BE are both parallel to the horizon. Furthermore, let’s suppose the straight lines FG, HI, etc., are [Pg 51]drawn from countless points along the straight line AB and end on the line DE (which is straight or can be considered straight). The varying degrees of penetration at different heights in the air between A and B can be represented by all these lines; so the wave starting from point F spreads across the distance FG, the wave from point H spreads across the distance HI, and the wave from point A spreads across the distance AD.

Now if about the centres A, B, one describes the circles DK, EL, which represent the spreading of the waves which originate from these two points, and if one draws the straight line KL which touches these two circles, it is easy to see that this same line will be the common tangent to all the other circles drawn about the centres F, H, etc.; and that all the points of contact will fall within that part of this line which is comprised between the perpendiculars AK, BL. Then it will be the line KL which will terminate the movement of the particular waves originating from the points of the wave AB; and this movement will be stronger between the points KL, than anywhere else at the same instant, since an infinitude of circumferences concur to form this straight line; and consequently KL will be the propagation of the portion of wave AB, as has been said in explaining reflexion and ordinary refraction. Now it appears that AK and BL dip down toward the side where the air is less easy to penetrate: for AK being longer than BL, and parallel to it, it follows that the lines AB and KL, being prolonged, would meet at the side L. But the angle K is a right angle: hence KAB is necessarily acute, and consequently less than DAB. If one investigates in the same way the progression of the portion of the wave KL, one will find that after a further time it has [Pg 52]arrived at MN in such a manner that the perpendiculars KM, LN, dip down even more than do AK, BL. And this suffices to show that the ray will continue along the curved line which intersects all the waves at right angles, as has been said.

Now, if we draw circles DK and EL around points A and B, which illustrate the spreading of the waves from these two locations, and we then draw the straight line KL that touches both circles, it's clear that this line will also be the common tangent to all the other circles centered around points F, H, etc. All the contact points will be within the segment of this line between the perpendiculars AK and BL. This means that line KL will mark the end of the movement of the specific waves coming from points along wave AB, and the movement will be stronger between points KL than at any other point at the same time, since countless circles contribute to forming this straight line. Consequently, KL will represent the propagation of segment wave AB, as previously explained regarding reflection and ordinary refraction. It appears that AK and BL incline toward the side where the air is harder to move through: since AK is longer than BL and parallel to it, if we extend lines AB and KL, they would meet at point L. However, angle K is a right angle, which means angle KAB must be acute and thus less than angle DAB. If we similarly examine how the wave portion KL progresses, we will find that after some time it reaches MN in such a way that the perpendiculars KM and LN dip down even more than AK and BL. This is enough to demonstrate that the ray will continue along the curved line that intersects all the waves at right angles, as previously mentioned.


CHAPTER V

ON THE STRANGE REFRACTION OF ICELAND CRYSTAL

1.

Below is a short piece of text (5 words or fewer). Modernize it into contemporary English if there's enough context, but do not add or omit any information. If context is insufficient, return it unchanged. Do not add commentary, and do not modify any placeholders. If you see placeholders of the form __A_TAG_PLACEHOLDER_x__, you must keep them exactly as-is so they can be replaced with links. 1.

T

here is brought from Iceland, which is an Island in the North Sea, in the latitude of 66 degrees, a kind of Crystal or transparent stone, very remarkable for its figure and other qualities, but above all for its strange refractions. The causes of this have seemed to me to be worthy of being carefully investigated, the more so because amongst transparent bodies this one alone does not follow the ordinary rules with respect to rays of light. I have even been under some necessity to make this research, because the refractions of this Crystal seemed to overturn our preceding explanation of regular refraction; which explanation, on the contrary, they strongly confirm, as will be seen after they have been brought under the same principle. In Iceland are found great lumps of this Crystal, some of which I have seen of 4 or 5 pounds. But it occurs also in other countries, for I have had some of the same sort which had been found in France near the town of Troyes in Champagne, and some others which came from the Island of Corsica, though both were [Pg 53]less clear and only in little bits, scarcely capable of letting any effect of refraction be observed.

Here is a type of crystal or transparent stone brought from Iceland, which is an island in the North Sea, at a latitude of 66 degrees. It's notable for its shape and other properties, but especially for its unusual refractions. I think the reasons behind this are worth investigating carefully, especially since this particular transparent stone doesn't follow the usual rules regarding light rays. I've even felt it necessary to conduct this research because the crystal's refractions seemed to challenge our previous understanding of regular refraction; however, they ultimately support that explanation, as will be shown once they are examined under the same principle. Large chunks of this crystal can be found in Iceland, some weighing 4 or 5 pounds. It also occurs in other countries; I've had some of a similar type found in France near the town of Troyes in Champagne, as well as some from the island of Corsica, although both were less clear and in small pieces, barely allowing any observable effects of refraction.

2. The first knowledge which the public has had about it is due to Mr. Erasmus Bartholinus, who has given a description of Iceland Crystal and of its chief phenomena. But here I shall not desist from giving my own, both for the instruction of those who may not have seen his book, and because as respects some of these phenomena there is a slight difference between his observations and those which I have made: for I have applied myself with great exactitude to examine these properties of refraction, in order to be quite sure before undertaking to explain the causes of them.

2. The first information the public received about it comes from Mr. Erasmus Bartholinus, who described Iceland Crystal and its main phenomena. However, I will also share my observations, both for the benefit of those who might not have read his book and because I’ve noticed some slight differences between his findings and mine. I have carefully studied these refractive properties to ensure I'm confident before explaining their causes.

3. As regards the hardness of this stone, and the property which it has of being easily split, it must be considered rather as a species of Talc than of Crystal. For an iron spike effects an entrance into it as easily as into any other Talc or Alabaster, to which it is equal in gravity.

3. When it comes to the hardness of this stone and its ability to be easily split, it should be viewed more as a type of Talc than Crystal. A iron spike can penetrate it just as easily as it can any other Talc or Alabaster, which it is equal to in weight.

4. The pieces of it which are found have the figure of an oblique parallelepiped; each of the six faces being a parallelogram; and it admits of being split in three directions parallel to two of these opposed faces. Even in such wise, if you will, that all the six faces are equal and similar rhombuses. The figure here added represents a piece of this Crystal. The obtuse angles of all the parallelograms, as C, D, here, are angles of 101 degrees 52 minutes, [Pg 54]and consequently the acute angles, such as A and B, are of 78 degrees 8 minutes.

4. The pieces that are found have the shape of an oblique parallelepiped, with each of the six faces being a parallelogram. They can be split in three directions parallel to two of these opposing faces. In fact, all six faces are equal and similar rhombuses. The figure shown here represents a piece of this crystal. The obtuse angles of all the parallelograms, like C and D, are 101 degrees 52 minutes, [Pg 54] and therefore the acute angles, like A and B, are 78 degrees 8 minutes.

5. Of the solid angles there are two opposite to one another, such as C and E, which are each composed of three equal obtuse plane angles. The other six are composed of two acute angles and one obtuse. All that I have just said has been likewise remarked by Mr. Bartholinus in the aforesaid treatise; if we differ it is only slightly about the values of the angles. He recounts moreover some other properties of this Crystal; to wit, that when rubbed against cloth it attracts straws and other light things as do amber, diamond, glass, and Spanish wax. Let a piece be covered with water for a day or more, the surface loses its natural polish. When aquafortis is poured on it it produces ebullition, especially, as I have found, if the Crystal has been pulverized. I have also found by experiment that it may be heated to redness in the fire without being in anywise altered or rendered less transparent; but a very violent fire calcines it nevertheless. Its transparency is scarcely less than that of water or of Rock Crystal, and devoid of colour. But rays of light pass through it in another fashion and produce those marvellous refractions the causes of which I am now going to try to explain; reserving for the end of this Treatise the statement of my conjectures touching the formation and extraordinary configuration of this Crystal.

5. There are two solid angles that are opposite each other, like C and E, and each is made up of three equal obtuse plane angles. The other six are made up of two acute angles and one obtuse angle. Everything I've just mentioned has also been noted by Mr. Bartholinus in the mentioned treatise; if we disagree, it’s only slightly about the values of the angles. He also describes some other properties of this crystal; for example, when it is rubbed against cloth, it attracts straws and other light objects, similar to amber, diamond, glass, and Spanish wax. If a piece is covered with water for a day or longer, the surface loses its natural shine. When aquafortis is poured on it, it causes bubbling, especially, as I've found, if the crystal has been powdered. From my experiments, I've also discovered that it can be heated until it's red in the fire without being changed or losing its transparency; however, an extremely hot fire will calcine it. Its transparency is almost as good as that of water or rock crystal, and it has no color. But rays of light pass through it in a different way and create those amazing refractions that I’m now going to try to explain; I will reserve my thoughts on the formation and unique structure of this crystal for the end of this treatise.

6. In all other transparent bodies that we know there is but one sole and simple refraction; but in this substance there are two different ones. The effect is that objects seen through it, especially such as are placed right against it, appear double; and that a ray of sunlight, falling on one of its surfaces, parts itself into two rays and traverses the Crystal thus.

6. In all other clear materials we know of, there's only one straightforward refraction, but in this substance, there are two different ones. The result is that objects seen through it, particularly those positioned directly against it, look double; and when a ray of sunlight hits one of its surfaces, it splits into two rays and passes through the Crystal like that.

[Pg 55]7. It is again a general law in all other transparent bodies that the ray which falls perpendicularly on their surface passes straight on without suffering refraction, and that an oblique ray is always refracted. But in this Crystal the perpendicular ray suffers refraction, and there are oblique rays which pass through it quite straight.

[Pg 55]7. It's a common rule for all other clear materials that a ray hitting their surface straight on continues moving in a straight line without bending, while an angled ray always bends. However, in this crystal, the straight ray does bend, and there are angled rays that go through it without changing direction at all.

8. But in order to explain these phenomena more particularly, let there be, in the first place, a piece ABFE of the same Crystal, and let the obtuse angle ACB, one of the three which constitute the equilateral solid angle C, be divided into two equal parts by the straight line CG, and let it be conceived that the Crystal is intersected by a plane which passes through this line and through the side CF, which plane will necessarily be perpendicular to [Pg 56]the surface AB; and its section in the Crystal will form a parallelogram GCFH. We will call this section the principal section of the Crystal.

8. To explain these phenomena more clearly, let's consider a piece ABFE of the same crystal. We'll divide the obtuse angle ACB, one of the three angles that make up the equilateral solid angle C, into two equal parts using the straight line CG. Imagine that the crystal is cut by a plane that goes through this line and the side CF; this plane will necessarily be perpendicular to the surface AB. The intersection in the crystal will create a parallelogram GCFH. We will refer to this intersection as the principal section of the crystal.

9. Now if one covers the surface AB, leaving there only a small aperture at the point K, situated in the straight line CG, and if one exposes it to the sun, so that his rays face it perpendicularly above, then the ray IK will divide itself at the point K into two, one of which will continue to go on straight by KL, and the other will separate itself along the straight line KM, which is in the plane GCFH, and which makes with KL an angle of about 6 degrees 40 minutes, tending from the side of the solid angle C; and on emerging from the other side of the Crystal it will turn again parallel to JK, along MZ. And as, in this extraordinary refraction, the point M is seen by the refracted ray MKI, which I consider as going to the eye at I, it necessarily follows that the point L, by virtue of the same refraction, will be seen by the refracted ray LRI, so that LR will be parallel to MK if the distance from the eye KI is supposed very great. The point L appears then as being in the straight line IRS; but the same point appears also, by ordinary refraction, to be in the straight line IK, hence it is necessarily judged to be double. And similarly if L be a small hole in a sheet of paper or other substance which is laid against the Crystal, it will appear when turned towards daylight as if there were two holes, which will seem the wider apart from one another the greater the thickness of the Crystal.

9. Now, if you cover the surface AB, leaving only a small opening at point K, located on the straight line CG, and then expose it to the sun so that the rays hit it directly from above, the ray IK will split at point K into two parts. One will continue straight as KL, and the other will separate along the straight line KM, which is in the plane GCFH and forms an angle of about 6 degrees 40 minutes with KL, moving from the direction of the solid angle C. When it exits the other side of the Crystal, it will turn again and run parallel to JK along MZ. Since point M is visible to the refracted ray MKI, which I consider as heading toward the eye at I, it follows that point L, because of the same refraction, will also be seen by the refracted ray LRI, making LR parallel to MK if we assume the distance from the eye KI is very large. Point L then appears to be on the straight line IRS; however, the same point also appears, due to ordinary refraction, to be on the straight line IK, so it must be judged as double. Similarly, if L is a small hole in a piece of paper or another material placed against the Crystal, it will look like there are two holes when turned toward daylight, and they will seem to be farther apart the thicker the Crystal is.

10. Again, if one turns the Crystal in such wise that an incident ray NO, of sunlight, which I suppose to be in the plane continued from GCFH, makes with GC an [Pg 57]angle of 73 degrees and 20 minutes, and is consequently nearly parallel to the edge CF, which makes with FH an angle of 70 degrees 57 minutes, according to the calculation which I shall put at the end, it will divide itself at the point O into two rays, one of which will continue along OP in a straight line with NO, and will similarly pass out of the other side of the crystal without any refraction; but the other will be refracted and will go along OQ. And it must be noted that it is special to the plane through GCF and to those which are parallel to it, that all incident rays which are in one of these planes continue to be in it after they have entered the Crystal and have become double; for it is quite otherwise for rays in all other planes which intersect the Crystal, as we shall see afterwards.

10. Again, if you turn the crystal in such a way that an incoming ray of sunlight, which I assume is in the extended plane from GCFH, creates an angle of 73 degrees and 20 minutes with GC, and is thus nearly parallel to the edge CF, which makes an angle of 70 degrees 57 minutes with FH, according to the calculations I’ll provide at the end, it will split at point O into two rays. One will continue along OP in a straight line with NO and will exit from the other side of the crystal without any refraction; the other will be refracted and will go along OQ. It’s important to note that this characteristic holds true for the plane through GCF and those parallel to it, where all incident rays that are in these planes remain in them after entering the crystal and becoming double. However, the same is not true for rays in all other planes that intersect the crystal, as we will see later.

11. I recognized at first by these experiments and by some others that of the two refractions which the ray suffers in this Crystal, there is one which follows the ordinary rules; and it is this to which the rays KL and OQ belong. This is why I have distinguished this ordinary refraction from the other; and having measured it by exact observation, I found that its proportion, considered as to the Sines of the angles which the incident and refracted rays make with the perpendicular, was very precisely that of 5 to 3, as was found also by Mr. Bartholinus, and consequently much greater than that of Rock Crystal, or of glass, which is nearly 3 to 2.

11. I initially recognized through these experiments and others that of the two refractions the ray experiences in this crystal, one follows the usual rules; this is the one to which the rays KL and OQ belong. That’s why I’ve differentiated this ordinary refraction from the other. After measuring it with precise observations, I found that its ratio, based on the sines of the angles formed by the incident and refracted rays with the perpendicular, was very specifically 5 to 3, as Mr. Bartholinus also discovered, and therefore much greater than that of rock crystal or glass, which is about 3 to 2.

12. The mode of making these observations exactly is as follows. Upon a leaf of paper fixed on a thoroughly flat table there is traced a black line AB, and two others, CED and KML, which cut it at right angles and are more or less distant from one another according [Pg 58]as it is desired to examine a ray that is more or less oblique. Then place the Crystal upon the intersection E so that the line AB concurs with that which bisects the obtuse angle of the lower surface, or with some line parallel to it. Then by placing the eye directly above the line AB it will appear single only; and one will see that the portion viewed through the Crystal and the portions which appear outside it, meet together in a straight line: but the line CD will appear double, and one can distinguish the image which is due to regular refraction by the circumstance that when one views it with both eyes it seems raised up more than the other, or again by the circumstance that, when the Crystal is turned around on the paper, this image remains stationary, whereas the other image shifts and moves entirely around. Afterwards let the eye be placed at I (remaining [Pg 59]always in the plane perpendicular through AB) so that it views the image which is formed by regular refraction of the line CD making a straight line with the remainder of that line which is outside the Crystal. And then, marking on the surface of the Crystal the point H where the intersection E appears, this point will be directly above E. Then draw back the eye towards O, keeping always in the plane perpendicular through AB, so that the image of the line CD, which is formed by ordinary refraction, may appear in a straight line with the line KL viewed without refraction; and then mark on the Crystal the point N where the point of intersection E appears.

12. The way to accurately make these observations is as follows. On a flat table, place a sheet of paper and draw a black line AB, along with two others, CED and KML, that intersect it at right angles and are spaced out based on how inclined you want to examine a ray. Then, position the Crystal at the intersection point E so that line AB aligns with the line bisecting the obtuse angle of the lower surface, or with any parallel line. By looking directly above line AB, it will appear as a single line; you’ll notice that the area viewed through the Crystal and the parts visible outside of it align in a straight line. However, line CD will appear double, and you can identify the image produced by regular refraction because when viewed with both eyes, it seems higher than the other, or if you rotate the Crystal on the paper, this image stays in place while the other image shifts around. Next, position your eye at I (always remaining in the perpendicular plane through AB) so that you can see the image created by regular refraction of line CD in a straight line with the part of that line outside the Crystal. Then, mark on the surface of the Crystal at point H where intersection E is visible; this point will be directly above E. After that, move your eye back towards O, while staying in the perpendicular plane through AB, so that the image of line CD formed by regular refraction appears in a straight line with line KL viewed without refraction; then, mark on the Crystal at point N where point of intersection E is visible.

13. Then one will know the length and position of the lines NH, EM, and of HE, which is the thickness of the Crystal: which lines being traced separately upon a plan, and then joining NE and NM which cuts HE at P, the proportion of the refraction will be that of EN to NP, because these lines are to one another as the sines of the angles NPH, NEP, which are equal to those which the incident ray ON and its refraction NE make with the perpendicular to the surface. This proportion, as I have said, is sufficiently precisely as 5 to 3, and is always the same for all inclinations of the incident ray.

13. Then one can determine the length and position of the lines NH, EM, and HE, which represent the thickness of the Crystal. When these lines are drawn separately on a diagram, and NE and NM are connected to intersect HE at point P, the ratio of the refraction will correspond to that of EN to NP. This is because these lines relate to each other like the sines of the angles NPH and NEP, which are equal to those formed by the incident ray ON and its refraction NE with the perpendicular line to the surface. As I mentioned, this ratio is roughly 5 to 3, and it remains constant for all angles of the incident ray.

14. The same mode of observation has also served me for examining the extraordinary or irregular refraction of this Crystal. For, the point H having been found and marked, as aforesaid, directly above the point E, I observed the appearance of the line CD, which is made by the extraordinary refraction; and having placed the eye at Q, so that this appearance made a straight line with the line KL viewed without refraction, I ascertained the triangles REH, RES, and consequently the angles RSH, [Pg 60]RES, which the incident and the refracted ray make with the perpendicular.

14. I used the same method of observation to study the unusual or irregular refraction of this crystal. Once I found and marked point H directly above point E, I noticed the line CD, which is created by the extraordinary refraction. By positioning my eye at Q so that this appearance aligned in a straight line with the line KL seen without refraction, I calculated the triangles REH, RES, and consequently the angles RSH, [Pg 60]RES, which the incident and refracted rays form with the perpendicular.

15. But I found in this refraction that the ratio of FR to RS was not constant, like the ordinary refraction, but that it varied with the varying obliquity of the incident ray.

15. But I found in this refraction that the ratio of FR to RS was not constant, like regular refraction, but that it changed with the angle of the incoming ray.

16. I found also that when QRE made a straight line, that is, when the incident ray entered the Crystal without being refracted (as I ascertained by the circumstance that then the point E viewed by the extraordinary refraction appeared in the line CD, as seen without refraction) I found, I say, then that the angle QRG was 73 degrees 20 minutes, as has been already remarked; and so it is not the ray parallel to the edge of the Crystal, which crosses it in a straight line without being refracted, as Mr. Bartholinus believed, since that inclination is only 70 degrees 57 minutes, as was stated above. And this is to be noted, in order that no one may search in vain for the cause of the singular property of this ray in its parallelism to the edges mentioned.

16. I also discovered that when QRE forms a straight line, meaning when the incident ray enters the Crystal without being refracted (I verified this because point E, viewed through the extraordinary refraction, appeared in line CD, just as it does without refraction), I found that the angle QRG was 73 degrees 20 minutes, as previously mentioned. Therefore, it is not the ray parallel to the edge of the Crystal that travels through it in a straight line without refraction, as Mr. Bartholinus thought, since that angle is only 70 degrees 57 minutes, as stated before. This is important to note so that no one searches in vain for the reason behind the unique property of this ray being parallel to the mentioned edges.

17. Finally, continuing my observations to discover the [Pg 61]nature of this refraction, I learned that it obeyed the following remarkable rule. Let the parallelogram GCFH, made by the principal section of the Crystal, as previously determined, be traced separately. I found then that always, when the inclinations of two rays which come from opposite sides, as VK, SK here, are equal, their refractions KX and KT meet the bottom line HF in such wise that points X and T are equally distant from the point M, where the refraction of the perpendicular ray IK falls; and this occurs also for refractions in other sections of this Crystal. But before speaking of those, which have also other particular properties, we will investigate the causes of the phenomena which I have already reported.

17. Finally, as I continued my observations to uncover the [Pg 61]nature of this refraction, I discovered that it follows a remarkable rule. If we separately trace the parallelogram GCFH, created by the principal section of the Crystal, as mentioned earlier, I found that whenever the angles of two rays coming from opposite sides, like VK and SK here, are equal, their refractions KX and KT hit the bottom line HF in such a way that points X and T are equally distant from point M, where the refraction of the perpendicular ray IK lands; and this also occurs for refractions in other sections of this Crystal. But before discussing those, which have their own specific properties, we will explore the causes of the phenomena I have already described.

It was after having explained the refraction of ordinary transparent bodies by means of the spherical emanations of light, as above, that I resumed my examination of the nature of this Crystal, wherein I had previously been unable to discover anything.

It was after explaining how light bends through regular clear materials using the spherical waves of light, as mentioned above, that I continued my investigation into the nature of this Crystal, which I had previously found nothing in.

18. As there were two different refractions, I conceived that there were also two different emanations of waves of light, and that one could occur in the ethereal matter extending through the body of the Crystal. Which matter, being present in much larger quantity than is that of the particles which compose it, was alone capable of causing transparency, according to what has been explained heretofore. I attributed to this emanation of waves the regular refraction which is observed in this stone, by supposing these waves to be ordinarily of spherical form, and having a slower progression within the Crystal than they have outside it; whence proceeds refraction as I have demonstrated.

18. Since there were two different refractions, I thought there were also two different sources of light waves, and that one could happen in the ethereal matter that runs through the body of the Crystal. This matter, being present in much larger amounts than the particles that make it up, was solely responsible for causing transparency, as explained earlier. I attributed the regular refraction observed in this stone to this source of waves, assuming these waves are usually spherical in shape and move slower inside the Crystal than outside of it; this is how refraction occurs, as I have demonstrated.

19. As to the other emanation which should produce [Pg 62]the irregular refraction, I wished to try what Elliptical waves, or rather spheroidal waves, would do; and these I supposed would spread indifferently both in the ethereal matter diffused throughout the crystal and in the particles of which it is composed, according to the last mode in which I have explained transparency. It seemed to me that the disposition or regular arrangement of these particles could contribute to form spheroidal waves (nothing more being required for this than that the successive movement of light should spread a little more quickly in one direction than in the other) and I scarcely doubted that there were in this crystal such an arrangement of equal and similar particles, because of its figure and of its angles with their determinate and invariable measure. Touching which particles, and their form and disposition, I shall, at the end of this Treatise, propound my conjectures and some experiments which confirm them.

19. Regarding the other phenomenon that should create [Pg 62]the irregular refraction, I wanted to see what Elliptical waves, or more accurately, spheroidal waves would do; I thought these would move evenly in both the ethereal matter spread throughout the crystal and in the particles that make it up, based on the last explanation I gave about transparency. It seemed to me that the arrangement or orderly setup of these particles could help create spheroidal waves (the only requirement being that the light's successive movement spreads slightly faster in one direction than another), and I had little doubt that there was such an arrangement of equal and similar particles in this crystal, due to its shape and the angles which have a specific and consistent measurement. As for those particles, and their shape and arrangement, I will present my theories and some experiments that support them at the end of this Treatise.

20. The double emission of waves of light, which I had imagined, became more probable to me after I had observed a certain phenomenon in the ordinary [Rock] Crystal, which occurs in hexagonal form, and which, because of this regularity, seems also to be composed of particles, of definite figure, and ranged in order. This was, that this crystal, as well as that from Iceland, has a double refraction, though less evident. For having had cut from it some well polished Prisms of different sections, I remarked in all, in viewing through them the flame of a candle or the lead of window panes, that everything appeared double, though with images not very distant from one another. Whence I understood the reason why this substance, though so transparent, is useless for Telescopes, when they have ever so little length.

20. The idea of double waves of light that I had imagined seemed more likely to me after I noticed a specific phenomenon in regular [Rock] Crystal, which has a hexagonal shape and, due to its symmetry, appears to be made up of orderly arranged particles with distinct shapes. This crystal, like the one from Iceland, also shows double refraction, although it’s less obvious. When I had some well-polished prisms cut from it in various shapes, I observed that when looking through them at a candle flame or window panes, everything appeared doubled, but the images weren't very far apart. From this, I understood why this material, despite being so transparent, isn’t useful for telescopes if they are even slightly long.

[Pg 63]21. Now this double refraction, according to my Theory hereinbefore established, seemed to demand a double emission of waves of light, both of them spherical (for both the refractions are regular) and those of one series a little slower only than the others. For thus the phenomenon is quite naturally explained, by postulating substances which serve as vehicle for these waves, as I have done in the case of Iceland Crystal. I had then less trouble after that in admitting two emissions of waves in one and the same body. And since it might have been objected that in composing these two kinds of crystal of equal particles of a certain figure, regularly piled, the interstices which these particles leave and which contain the ethereal matter would scarcely suffice to transmit the waves of light which I have localized there, I removed this difficulty by regarding these particles as being of a very rare texture, or rather as composed of other much smaller particles, between which the ethereal matter passes quite freely. This, moreover, necessarily follows from that which has been already demonstrated touching the small quantity of matter of which the bodies are built up.

[Pg 63]21. Now, according to my earlier established theory, this double refraction seems to require a double emission of light waves, both of which are spherical (since both refractions are regular), and one series moves slightly slower than the other. This phenomenon can be naturally explained by assuming the existence of substances that serve as carriers for these waves, as I previously described with Iceland Crystal. After that, I found it easier to accept the idea of two emissions of waves in a single body. And since it could be argued that when creating these two types of crystal from equal particles of a certain shape, stacked neatly, the gaps left by these particles would hardly allow for the transmission of the light waves I identified, I addressed this issue by considering these particles to be made of very fine material, or rather, composed of much smaller particles that allow the ethereal matter to pass through quite freely. This conclusion also follows from what has already been demonstrated about the small amount of matter that makes up these bodies.

22. Supposing then these spheroidal waves besides the spherical ones, I began to examine whether they could serve to explain the phenomena of the irregular refraction, and how by these same phenomena I could determine the figure and position of the spheroids: as to which I obtained at last the desired success, by proceeding as follows.

22. So, assuming these spheroidal waves in addition to the spherical ones, I started to look into whether they could help explain the irregular refraction phenomena, and how I could use these same phenomena to figure out the shape and position of the spheroids: I ultimately achieved the desired success by following these steps.

23. I considered first the effect of waves so formed, as respects the ray which falls perpendicularly on the flat surface of a transparent body in which they should spread in this manner. I took AB for the exposed region of the surface. And, since a ray perpendicular to a plane, and [Pg 64]coming from a very distant source of light, is nothing else, according to the precedent Theory, than the incidence of a portion of the wave parallel to that plane, I supposed the straight line RC, parallel and equal to AB, to be a portion of a wave of light, in which an infinitude of points such as RHhC come to meet the surface AB at the points AKkB. Then instead of the hemispherical partial waves which in a body of ordinary refraction would spread from each of these last points, as we have above explained in treating of refraction, these must here be hemi-spheroids. The axes (or rather the major diameters) of these I supposed to be oblique to the plane AB, as is AV the semi-axis or semi-major diameter of the spheroid SVT, which represents the partial wave coming from the point A, after the wave RC has reached AB. I say axis or major diameter, because the same ellipse SVT may be considered as the section of a spheroid of which the axis is AZ perpendicular to AV. But, for the present, without yet deciding one or other, we will consider these spheroids only in those sections of them which make ellipses in the plane of this figure. Now taking a certain space of time during which the wave SVT has spread from A, it would needs be that from all the other points KkB there should proceed, in the same time, waves similar to SVT and similarly situated. And the common tangent NQ of all these semi-ellipses would be the propagation of the wave RC which fell on AB, and [Pg 65]would be the place where this movement occurs in much greater amount than anywhere else, being made up of arcs of an infinity of ellipses, the centres of which are along the line AB.

23. First, I thought about how waves would behave when they spread in this way, particularly with regard to the ray that hits the flat surface of a transparent object directly. I labeled AB as the exposed area of the surface. Since a ray coming straight down onto a plane from a far-off light source is essentially just part of a wave that runs parallel to that plane, I imagined a straight line RC, which is parallel and equal to AB, as a segment of a light wave. This wave has countless points such as RHhC that meet the surface AB at the points AKkB. Instead of the hemispherical partial waves that would normally spread from each of these points in a material where light refracts, which we've previously explained, here they should be hemi-spheroids. I assumed that the axes (or major diameters) of these would be angled to the plane AB, with AV as the semi-axis or semi-major diameter of the spheroid SVT, representing the partial wave coming from point A after the wave RC hits AB. I refer to it as the axis or major diameter because the same ellipse SVT could be seen as a section of a spheroid where the axis is AZ, standing perpendicularly to AV. For now, without making a choice between these options, we'll only consider the spheroids and their sections that form ellipses in the plane of this figure. Taking a specific time period during which the wave SVT has spread from A, it’s necessary that waves similar to SVT and similarly positioned also emanate from all the other points KkB during the same time. The common tangent NQ of all these semi-ellipses would represent the propagation of the wave RC that fell on AB, and [Pg 65]this would be the location where this movement is happening more intensively than anywhere else, consisting of arcs from an infinite number of ellipses, the centers of which are along the line AB.

24. Now it appeared that this common tangent NQ was parallel to AB, and of the same length, but that it was not directly opposite to it, since it was comprised between the lines AN, BQ, which are diameters of ellipses having A and B for centres, conjugate with respect to diameters which are not in the straight line AB. And in this way I comprehended, a matter which had seemed to me very difficult, how a ray perpendicular to a surface could suffer refraction on entering a transparent body; seeing that the wave RC, having come to the aperture AB, went on forward thence, spreading between the parallel lines AN, BQ, yet itself remaining always parallel to AB, so that here the light does not spread along lines perpendicular to its waves, as in ordinary refraction, but along lines cutting the waves obliquely.

24. It seemed that this common tangent NQ was parallel to AB and the same length, but it wasn't directly opposite to it. Instead, it was located between the lines AN and BQ, which are the diameters of ellipses centered at A and B, conjugate to diameters that aren't in a straight line with AB. In this way, I understood something that had seemed really difficult: how a ray perpendicular to a surface could bend when entering a transparent object. The wave RC, after reaching the opening AB, continued forward, spreading between the parallel lines AN and BQ, while always remaining parallel to AB. Here, the light doesn't spread along lines that are perpendicular to its waves, like in regular refraction, but along lines that cut across the waves at an angle.

25. Inquiring subsequently what might be the position and form of these spheroids in the crystal, I considered that all the six faces produced precisely the same refractions. Taking, then, the parallelopiped AFB, of which the obtuse solid angle C is contained between the three equal plane angles, and imagining in it the three principal sections, one of which is perpendicular to the face DC and passes through the edge CF, another perpendicular to the face BF passing through the edge [Pg 66]CA, and the third perpendicular to the face AF passing through the edge BC; I knew that the refractions of the incident rays belonging to these three planes were all similar. But there could be no position of the spheroid which would have the same relation to these three sections except that in which the axis was also the axis of the solid angle C. Consequently I saw that the axis of this angle, that is to say the straight line which traversed the crystal from the point C with equal inclination to the edges CF, CA, CB was the line which determined the position of the axis of all the spheroidal waves which one imagined to originate from some point, taken within or on the surface of the crystal, since all these spheroids ought to be alike, and have their axes parallel to one another.

25. Later, I wondered about the position and shape of these spheroids in the crystal. I realized that all six faces caused exactly the same refractions. So, considering the parallelepiped AFB, where the obtuse solid angle C is formed by three equal plane angles, I visualized the three main sections: one perpendicular to face DC, passing through edge CF; another perpendicular to face BF, passing through edge [Pg 66]CA; and the third perpendicular to face AF, passing through edge BC. I understood that the refractions of incoming rays related to these three planes were all the same. However, there could be no arrangement of the spheroid that would maintain the same relationship to these three sections, except for the one where the axis was also the axis of solid angle C. Therefore, I concluded that the axis of this angle, which is the straight line running through the crystal from point C with equal inclination to edges CF, CA, and CB, defined the position of the axis of all the spheroidal waves imagined to originate from some point inside or on the surface of the crystal, since all these spheroids should be identical and have parallel axes.

26. Considering after this the plane of one of these three sections, namely that through GCF, the angle of which is 109 degrees 3 minutes, since the angle F was shown above to be 70 degrees 57 minutes; and, imagining a spheroidal wave about the centre C, I knew, because I have just explained it, that its axis must be in the same plane, the half of which axis I have marked CS in the next figure: and seeking by calculation (which will be given with others at the end of this discourse) the value of the angle CGS, I found it 45 degrees 20 minutes.

26. Considering the plane of one of these three sections, specifically the one through GCF, which has an angle of 109 degrees 3 minutes, and knowing that angle F was shown earlier to be 70 degrees 57 minutes; and, picturing a spheroidal wave around center C, I realized, as I just explained, that its axis must be in the same plane, half of which I marked as CS in the next figure. By calculating (which will be provided along with others at the end of this discussion), I found the value of angle CGS to be 45 degrees 20 minutes.

27. To know from this the form of this spheroid, that is to say the proportion of the semi-diameters CS, CP, of its elliptical section, which are perpendicular to one another, I considered that the point M where the ellipse is touched by the straight line FH, parallel to CG, ought to be so situated that CM makes with the perpendicular CL an angle of 6 degrees 40 minutes; since, this being so, this ellipse satisfies what has been said about the refraction of [Pg 67]the ray perpendicular to the surface CG, which is inclined to the perpendicular CL by the same angle. This, then, being thus disposed, and taking CM at 100,000 parts, I found by the calculation which will be given at the end, the semi-major diameter CP to be 105,032, and the semi-axis CS to be 93,410, the ratio of which numbers is very nearly 9 to 8; so that the spheroid was of the kind which resembles a compressed sphere, being generated by the revolution of an ellipse about its smaller diameter. I found also the value of CG the semi-diameter parallel to the tangent ML to be 98,779.

27. To determine the shape of this spheroid, meaning the ratio of the semi-diameters CS and CP of its elliptical section, which are perpendicular to each other, I figured that the point M where the ellipse touches the straight line FH, which is parallel to CG, should be positioned so that CM forms an angle of 6 degrees 40 minutes with the perpendicular CL. If this is the case, then this ellipse meets the criteria discussed regarding the refraction of [Pg 67]the ray that is perpendicular to the surface CG, which is tilted at the same angle as the perpendicular CL. With this arrangement and setting CM at 100,000 parts, I calculated, as will be detailed later, that the semi-major diameter CP is 105,032, and the semi-axis CS is 93,410, giving a ratio of very close to 9 to 8. Thus, the spheroid resembles a compressed sphere, formed by the revolution of an ellipse around its shorter diameter. I also calculated that the value of CG, the semi-diameter parallel to the tangent ML, is 98,779.

28. Now passing to the investigation of the refractions which obliquely incident rays must undergo, according to our hypothesis of spheroidal waves, I saw that these refractions depended on the ratio between the velocity of movement of the light outside the crystal in the ether, and that within the crystal. For supposing, for example, this proportion to be such that while the light in the crystal forms the spheroid GSP, as I have just said, it forms outside a sphere the semi-diameter of which is equal to the line N which will be determined hereafter, the following is the way of finding the refraction of the incident rays. Let there be such a ray RC falling upon the [Pg 68]surface CK. Make CO perpendicular to RC, and across the angle KCO adjust OK, equal to N and perpendicular to CO; then draw KI, which touches the Ellipse GSP, and from the point of contact I join IC, which will be the required refraction of the ray RC. The demonstration of this is, it will be seen, entirely similar to that of which we made use in explaining ordinary refraction. For the refraction of the ray RC is nothing else than the progression of the portion C of the wave CO, continued in the crystal. Now the portions H of this wave, during the time that O came to K, will have arrived at the surface CK along the straight lines Hx, and will moreover have produced in the crystal around the centres x some hemi-spheroidal partial waves similar to the hemi-spheroidal GSPg, and similarly disposed, and of which the major [Pg 69]and minor diameters will bear the same proportions to the lines xv (the continuations of the lines Hx up to KB parallel to CO) that the diameters of the spheroid GSPg bear to the line CB, or N. And it is quite easy to see that the common tangent of all these spheroids, which are here represented by Ellipses, will be the straight line IK, which consequently will be the propagation of the wave CO; and the point I will be that of the point C, conformably with that which has been demonstrated in ordinary refraction.

28. Now, moving on to examine the refractions that obliquely incident rays must go through, based on our idea of spheroidal waves, I observed that these refractions depend on the ratio between the speed of light moving outside the crystal in the ether and the speed inside the crystal. For instance, if this ratio is such that while the light in the crystal creates the spheroid GSP, as I've just mentioned, it forms a sphere outside with a semi-diameter equal to the line N, which will be determined later, here's how we find the refraction of the incident rays. Let's say there's a ray RC striking the surface CK. Make CO perpendicular to RC and adjust OK across the angle KCO, making it equal to N and also perpendicular to CO; then draw KI, which touches the Ellipse GSP, and from the contact point I draw IC, which will give us the needed refraction of the ray RC. You'll see that the proof for this is quite similar to what we used when explaining ordinary refraction. The refraction of ray RC is simply the progression of the portion C of the wave CO, continuing in the crystal. Now the portions H of this wave, while O moves to K, will have reached the surface CK along the straight lines Hx, and will have also created in the crystal around the centers x some hemi-spheroidal partial waves similar to the hemi-spheroidal GSPg, similarly arranged, with the major [Pg 69]and minor diameters maintaining the same proportions to the lines xv (the continuations of the lines Hx up to KB parallel to CO) as the diameters of the spheroid GSPg maintain to the line CB, or N. It’s easy to see that the common tangent of all these spheroids, represented here by Ellipses, will be the straight line IK, which therefore represents the propagation of the wave CO; and point I will correspond to point C, in line with what we've established in ordinary refraction.

Now as to finding the point of contact I, it is known that one must find CD a third proportional to the lines CK, CG, and draw DI parallel to CM, previously determined, which is the conjugate diameter to CG; for then, by drawing KI it touches the Ellipse at I.

Now, to find the point of contact I, it's known that you need to find CD as a third proportional to the lines CK and CG, and draw DI parallel to CM, which has already been established as the conjugate diameter to CG; because then, by drawing KI, it touches the ellipse at I.

29. Now as we have found CI the refraction of the ray RC, similarly one will find Ci the refraction of the ray rC, which comes from the opposite side, by making Co perpendicular to rC and following out the rest of the construction as before. Whence one sees that if the ray rC is inclined equally with RC, the line Cd will necessarily be equal to CD, because Ck is equal to CK, and Cg to CG. And in consequence Ii will be cut at E into equal parts by the line CM, to which DI and di are parallel. And because CM is the conjugate diameter to CG, it follows that iI will be parallel to gG. Therefore if one prolongs the refracted rays CI, Ci, until they meet the tangent ML at T and t, the distances MT, Mt, will also be equal. And so, by our hypothesis, we explain perfectly the phenomenon mentioned above; to wit, that when there are two rays equally inclined, but coming from opposite sides, as here the rays RC, rc, their refractions diverge equally from the line [Pg 70]followed by the refraction of the ray perpendicular to the surface, by considering these divergences in the direction parallel to the surface of the crystal.

29. Now that we've found CI, the refraction of the ray RC, we can also determine Ci, the refraction of the ray rC, which comes from the opposite side, by making Co perpendicular to rC and completing the rest of the setup as before. From this, we can see that if the ray rC is angled the same as RC, the line Cd will be equal to CD since Ck equals CK and Cg equals CG. Consequently, Ii will be divided into equal parts at E by the line CM, to which DI and di are parallel. Since CM is the conjugate diameter to CG, it follows that iI will be parallel to gG. Therefore, if we extend the refracted rays CI and Ci until they intersect the tangent ML at T and t, the distances MT and Mt will also be equal. Thus, according to our hypothesis, we can fully explain the phenomenon mentioned earlier; specifically, when there are two rays equally inclined but coming from opposite sides, as in the case of the rays RC and rc, their refractions diverge equally from the line [Pg 70] followed by the refraction of the ray perpendicular to the surface, considering these divergences in the direction parallel to the surface of the crystal.

30. To find the length of the line N, in proportion to CP, CS, CG, it must be determined by observations of the irregular refraction which occurs in this section of the crystal; and I find thus that the ratio of N to GC is just a little less than 8 to 5. And having regard to some other observations and phenomena of which I shall speak afterwards, I put N at 156,962 parts, of which the semi-diameter CG is found to contain 98,779, making this ratio 8 to 5-1/29. Now this proportion, which there is between the line N and CG, may be called the Proportion of the Refraction; similarly as in glass that of 3 to 2, as will be manifest when I shall have explained a short process in the preceding way to find the irregular refractions.

30. To determine the length of line N in relation to CP, CS, and CG, we need to look at the irregular refraction happening in this part of the crystal. I’ve found that the ratio of N to GC is just a bit less than 8 to 5. Considering some other observations and phenomena I’ll discuss later, I set N at 156,962 parts, with the semi-diameter CG measuring 98,779, making this ratio 8 to 5-1/29. This proportion between line N and CG can be referred to as the Proportion of the Refraction; just like in glass where it’s 3 to 2, as will become clear when I explain a brief method for finding the irregular refractions.

31. Supposing then, in the next figure, as previously, the surface of the crystal gG, the Ellipse GPg, and the line N; and CM the refraction of the perpendicular ray FC, from which it diverges by 6 degrees 40 minutes. Now let there be some other ray RC, the refraction of which must be found.

31. Assuming then, in the next diagram, as before, the surface of the crystal gG, the Ellipse GPg, and the line N; and CM the refraction of the perpendicular ray FC, from which it diverges by 6 degrees and 40 minutes. Now let there be another ray RC, the refraction of which needs to be determined.

About the centre C, with semi-diameter CG, let the circumference gRG be described, cutting the ray RC at R; and let RV be the perpendicular on CG. Then as the line N is to CG let CV be to CD, and let DI be drawn parallel to CM, cutting the Ellipse gMG at I; then joining CI, this will be the required refraction of the ray RC. Which is demonstrated thus.

About the center C, with a radius CG, let the circle gRG be drawn, intersecting the ray RC at R; and let RV be the perpendicular to CG. Then, as the line N is to CG, let CV be to CD, and let DI be drawn parallel to CM, intersecting the ellipse gMG at I; then by connecting CI, this will be the required refraction of the ray RC. This is demonstrated as follows.

Let CO be perpendicular to CR, and across the angle OCG let OK be adjusted, equal to N and perpendicular to CO, and let there be drawn the straight line KI, which if it [Pg 71]is demonstrated to be a tangent to the Ellipse at I, it will be evident by the things heretofore explained that CI is the refraction of the ray RC. Now since the angle RCO is a right angle, it is easy to see that the right-angled triangles RCV, KCO, are similar. As then, CK is to KO, so also is RC to CV. But KO is equal to N, and RC to CG: then as CK is to N so will CG be to CV. But as N is to CG, so, by construction, is CV to CD. Then as CK is to CG so is CG to CD. And because DI is parallel to CM, the conjugate diameter to CG, it follows that KI touches the Ellipse at I; which remained to be shown.

Let CO be perpendicular to CR, and across the angle OCG let OK be set up, equal to N and perpendicular to CO. Next, draw the straight line KI, which, if it [Pg 71]is proven to be a tangent to the Ellipse at I, it becomes clear from what has been explained earlier that CI is the refraction of the ray RC. Since the angle RCO is a right angle, it's easy to see that the right-angled triangles RCV and KCO are similar. Just as CK is to KO, RC is to CV. Since KO equals N, and RC equals CG, then as CK is to N, CG will be to CV. And just as N is to CG, by construction, CV will be to CD. Therefore, as CK is to CG, CG is to CD. Because DI is parallel to CM, the conjugate diameter to CG, it follows that KI touches the Ellipse at I; which needed to be demonstrated.

32. One sees then that as there is in the refraction of [Pg 72]ordinary media a certain constant proportion between the sines of the angles which the incident ray and the refracted ray make with the perpendicular, so here there is such a proportion between CV and CD or IE; that is to say between the Sine of the angle which the incident ray makes with the perpendicular, and the horizontal intercept, in the Ellipse, between the refraction of this ray and the diameter CM. For the ratio of CV to CD is, as has been said, the same as that of N to the semi-diameter CG.

32. It can be seen that just like in the refraction of ordinary materials, there's a consistent ratio between the sines of the angles formed by the incident ray and the refracted ray with the perpendicular. Similarly, there's a ratio between CV and CD or IE; specifically, this refers to the sine of the angle that the incident ray makes with the perpendicular, and the horizontal distance in the ellipse between the refraction of this ray and the diameter CM. The ratio of CV to CD is, as mentioned, the same as that of N to the semi-diameter CG.

33. I will add here, before passing away, that in comparing together the regular and irregular refraction of this crystal, there is this remarkable fact, that if ABPS be the spheroid by which light spreads in the Crystal in a certain space of time (which spreading, as has been said, serves for the irregular refraction), then the inscribed sphere BVST is the extension in the same space of time of the light which serves for the regular refraction.

33. Before I pass on, I want to note that when comparing the regular and irregular refraction of this crystal, there's an interesting observation: if ABPS represents the spheroid through which light disperses in the crystal in a specific amount of time (this dispersion is what accounts for the irregular refraction), then the inscribed sphere BVST shows the extension of light in the same amount of time, which is responsible for the regular refraction.

For we have stated before this, that the line N being the radius of a spherical wave of light in air, while in the crystal it spread through the spheroid ABPS, the ratio of N to CS will be 156,962 to 93,410. But it has also been stated that the proportion of the regular refraction was 5 to 3; that is to say, that N being the radius of a spherical wave of light in air, its extension in the crystal would, in the same space of time, form a sphere the radius of which would be to N as 3 to 5. Now 156,962 is to 93,410 as 5 to 3 less 1/41. So that it is sufficiently nearly, and may [Pg 73]be exactly, the sphere BVST, which the light describes for the regular refraction in the crystal, while it describes the spheroid BPSA for the irregular refraction, and while it describes the sphere of radius N in air outside the crystal.

For we have mentioned earlier that the line N represents the radius of a spherical light wave in air, while in the crystal it spreads through the spheroid ABPS. The ratio of N to CS will be 156,962 to 93,410. It was also noted that the ratio of regular refraction is 5 to 3; that is to say, that N, as the radius of a spherical light wave in air, would, in the same amount of time, form a sphere in the crystal with a radius that is to N as 3 is to 5. Now, 156,962 is to 93,410 as 5 to 3 minus 1/41. So, it is close enough and may [Pg 73]be exactly the sphere BVST, which the light describes for the regular refraction in the crystal, while it describes the spheroid BPSA for the irregular refraction, and while it describes the sphere with radius N in air outside the crystal.

Although then there are, according to what we have supposed, two different propagations of light within the crystal, it appears that it is only in directions perpendicular to the axis BS of the spheroid that one of these propagations occurs more rapidly than the other; but that they have an equal velocity in the other direction, namely, in that parallel to the same axis BS, which is also the axis of the obtuse angle of the crystal.

Although there are, based on our assumptions, two different ways light propagates within the crystal, it seems that only in directions perpendicular to the axis BS of the spheroid does one of these propagations happen faster than the other. In the other direction, which is parallel to the same axis BS, both propagations have the same speed.

34. The proportion of the refraction being what we have just seen, I will now show that there necessarily follows thence that notable property of the ray which falling obliquely on the surface of the crystal enters it without suffering refraction. For supposing the same things as before, and that the ray makes with the same surface gG the angle RCG of [Pg 74]73 degrees 20 minutes, inclining to the same side as the crystal (of which ray mention has been made above); if one investigates, by the process above explained, the refraction CI, one will find that it makes exactly a straight line with RC, and that thus this ray is not deviated at all, conformably with experiment. This is proved as follows by calculation.

34. Considering the proportion of refraction we've just discussed, I will now demonstrate that a significant property of a ray striking the surface of the crystal at an angle results in it entering without undergoing refraction. Assuming the same conditions as before, and that the ray forms an angle RCG of 73 degrees 20 minutes with the surface gG, tilting toward the same side as the crystal (which we've referenced before); if we analyze the refraction CI using the previously explained method, we'll see that it aligns perfectly in a straight line with RC, indicating that this ray is not deviated at all, which is consistent with experimental evidence. This is proven through calculation.

CG or CR being, as precedently, 98,779; CM being 100,000; and the angle RCV 73 degrees 20 minutes, CV will be 28,330. But because CI is the refraction of the ray RC, the proportion of CV to CD is 156,962 to 98,779, namely, that of N to CG; then CD is 17,828.

CG or CR is still 98,779; CM is 100,000; and the angle RCV is 73 degrees 20 minutes, so CV will be 28,330. However, since CI is the refraction of the ray RC, the ratio of CV to CD is 156,962 to 98,779, which is the same as that of N to CG; therefore, CD is 17,828.

Now the rectangle gDC is to the square of DI as the square of CG is to the square of CM; hence DI or CE will be 98,353. But as CE is to EI, so will CM be to MT, which will then be 18,127. And being added to ML, which is 11,609 (namely the sine of the angle LCM, which is 6 degrees 40 minutes, taking CM 100,000 as radius) we get LT 27,936; and this is to LC 99,324 as CV to VR, that is to say, as 29,938, the tangent of the complement of the angle RCV, which is 73 degrees 20 minutes, is to the radius of the Tables. Whence it appears that RCIT is a straight line; which was to be proved.

Now the rectangle gDC is to the square of DI as the square of CG is to the square of CM; therefore, DI or CE will be 98,353. Just as CE is to EI, CM will be to MT, which will then be 18,127. When we add this to ML, which is 11,609 (the sine of angle LCM, measuring 6 degrees 40 minutes, with CM taken as 100,000 for the radius), we get LT as 27,936; and this is to LC 99,324 as CV is to VR, meaning as 29,938, the tangent of the complement of angle RCV, which is 73 degrees 20 minutes, is to the radius of the Tables. Thus, it shows that RCIT is a straight line; which was to be proved.

35. Further it will be seen that the ray CI in emerging through the opposite surface of the crystal, ought to pass out quite straight, according to the following demonstration, which proves that the reciprocal relation of refraction obtains in this crystal the same as in other transparent bodies; that is to say, that if a ray RC in meeting the surface of the crystal CG is refracted as CI, the ray CI emerging through the opposite parallel surface of the [Pg 75]crystal, which I suppose to be IB, will have its refraction IA parallel to the ray RC.

35. It will also be evident that the ray CI, when it comes out through the opposite surface of the crystal, should pass straight out, based on the following demonstration. This shows that the reciprocal relationship of refraction applies in this crystal just like in other transparent materials. In other words, if ray RC is refracted as CI when it hits the surface of the crystal CG, then the ray CI emerging through the opposite parallel surface of the [Pg 75]crystal, which I assume to be IB, will have its refraction IA parallel to the ray RC.

Let the same things be supposed as before; that is to say, let CO, perpendicular to CR, represent a portion of a wave the continuation of which in the crystal is IK, so that the piece C will be continued on along the straight line CI, while O comes to K. Now if one takes a second period of time equal to the first, the piece K of the wave IK will, in this second period, have advanced along the straight line KB, equal and parallel to CI, because every piece of the wave CO, on arriving at the surface CK, ought to go on in the crystal the same as the piece C; and in this same time there will be formed in the air from the point I a partial spherical wave having a semi-diameter IA equal to KO, since KO has been traversed in an equal time. Similarly, if one considers some other point of the wave IK, such as h, it will go along hm, parallel to CI, to meet the surface IB, while the point K traverses Kl equal to hm; and while this accomplishes the remainder lB, there will start from the point m a partial wave the semi-diameter of which, mn, will have the same ratio to lB as IA to [Pg 76]KB. Whence it is evident that this wave of semi-diameter mn, and the other of semi-diameter IA will have the same tangent BA. And similarly for all the partial spherical waves which will be formed outside the crystal by the impact of all the points of the wave IK against the surface of the Ether IB. It is then precisely the tangent BA which will be the continuation of the wave IK, outside the crystal, when the piece K has reached B. And in consequence IA, which is perpendicular to BA, will be the refraction of the ray CI on emerging from the crystal. Now it is clear that IA is parallel to the incident ray RC, since IB is equal to CK, and IA equal to KO, and the angles A and O are right angles.

Let’s assume the same conditions as before; that is, let CO, which is perpendicular to CR, represent a segment of a wave that continues in the crystal as IK, so the segment C will extend in a straight line CI, while O moves to K. Now, if we take another time period that’s equal to the first, the segment K of the wave IK will, in this second time period, have moved along the straight line KB, which is equal and parallel to CI, because every part of the wave CO, upon reaching the surface CK, should continue in the crystal just like the segment C; and during this same time, a partial spherical wave will form in the air from point I, with a radius IA equal to KO, as KO has been covered in equal time. Similarly, if we look at another point of the wave IK, like h, it will move along hm, parallel to CI, to meet the surface IB, while point K travels Kl equal to hm; and while this completes the remaining lB, a partial wave will start from point m with a radius mn that has the same ratio to lB as IA does to [Pg 76]KB. Therefore, it’s clear that this wave with radius mn and the other with radius IA will share the same tangent BA. The same applies to all the partial spherical waves that will form outside the crystal from the impact of all the points of the wave IK against the surface of the Ether IB. Thus, the tangent BA will be the continuation of the wave IK outside the crystal when the segment K reaches B. Consequently, IA, which is perpendicular to BA, will represent the refraction of ray CI as it exits the crystal. It is evident that IA is parallel to the incident ray RC, since IB equals CK, and IA equals KO, and the angles A and O are right angles.

It is seen then that, according to our hypothesis, the reciprocal relation of refraction holds good in this crystal as well as in ordinary transparent bodies; as is thus in fact found by observation.

It is clear that, according to our hypothesis, the reciprocal relationship of refraction applies to this crystal just like it does in regular transparent materials; this has been confirmed by observation.

36. I pass now to the consideration of other sections of the crystal, and of the refractions there produced, on which, as will be seen, some other very remarkable phenomena depend.

36. I will now move on to examining other parts of the crystal and the refractions that occur there, which, as we will see, are related to some other very interesting phenomena.

Let ABH be a parallelepiped of crystal, and let the top surface AEHF be a perfect rhombus, the obtuse angles of which are equally divided by the straight line EF, and the acute angles by the straight line AH perpendicular to FE.

Let ABH be a crystal parallelepiped, and let the top surface AEHF be a perfect rhombus, with the obtuse angles equally divided by the straight line EF, and the acute angles divided by the straight line AH, which is perpendicular to FE.

The section which we have hitherto considered is that which passes through the lines EF, EB, and which at the same time cuts the plane AEHF at right angles. Refractions in this section have this in common with the refractions in ordinary media that the plane which is drawn through the incident ray and which also intersects the [Pg 77]surface of the crystal at right angles, is that in which the refracted ray also is found. But the refractions which appertain to every other section of this crystal have this strange property that the refracted ray always quits the plane of the incident ray perpendicular to the surface, and turns away towards the side of the slope of the crystal. For which fact we shall show the reason, in the first place, for the section through AH; and we shall show at the same time how one can determine the refraction, according to our hypothesis. Let there be, then, in the plane which passes through AH, and which is perpendicular to the plane AFHE, the incident ray RC; it is required to find its refraction in the crystal.

The section we've looked at so far goes through the lines EF and EB and cuts the plane AEHF at right angles. Refractions in this section are similar to those in regular materials in that the plane drawn through the incident ray, which also intersects the surface of the crystal at a right angle, is where the refracted ray is located. However, the refractions in all other sections of this crystal have a unique characteristic: the refracted ray always leaves the plane of the incident ray perpendicularly to the surface and shifts toward the side of the crystal's slope. We will first explain the reason for this in the section through AH, and at the same time, show how to determine the refraction based on our hypothesis. So, in the plane that passes through AH and is perpendicular to the plane AFHE, let’s consider the incident ray RC; we need to find its refraction in the crystal.

[Pg 78]37. About the centre C, which I suppose to be in the intersection of AH and FE, let there be imagined a hemi-spheroid QGqgM, such as the light would form in spreading in the crystal, and let its section by the plane AEHF form the Ellipse QGqg, the major diameter of which Qq, which is in the line AH, will necessarily be one of the major diameters of the spheroid; because the axis of the spheroid being in the plane through FEB, to which QC is perpendicular, it follows that QC is also perpendicular to the axis of the spheroid, and consequently QCq one of its major diameters. But the minor diameter of this Ellipse, Gg, will bear to Qq the proportion which has been defined previously, Article 27, between CG and the major semi-diameter of the spheroid, CP, namely, that of 98,779 to 105,032.

[Pg 78]37. Imagine a semi-spheroid QGqgM centered at point C, where I assume C is at the intersection of lines AH and FE. The intersection of this semi-spheroid with the plane AEHF will create an ellipse QGqg. The longer diameter of this ellipse, Qq, which lies along line AH, will definitely be one of the major diameters of the semi-spheroid. This is because the axis of the semi-spheroid lies in the plane through FEB, and since QC is perpendicular to that plane, QC is also perpendicular to the axis of the semi-spheroid. Therefore, QCq is one of its major diameters. The shorter diameter of this ellipse, Gg, will have the same ratio to Qq that was defined earlier in Article 27, which compares CG to the major semi-diameter of the semi-spheroid, CP, specifically a ratio of 98,779 to 105,032.

Let the line N be the length of the travel of light in air during the time in which, within the crystal, it makes, from the centre C, the spheroid QCqgM. Then having drawn CO perpendicular to the ray CR and situate in the plane through CR and AH, let there be adjusted, across the angle ACO, the straight line OK equal to N and perpendicular to CO, and let it meet the straight line AH at K. Supposing consequently that CL is perpendicular to the surface of the crystal AEHF, and that CM is the refraction of the ray which falls perpendicularly on this same surface, let there be drawn a plane through the line CM and through KCH, making in the spheroid the semi-ellipse QMq, which will be given, since the angle MCL is given of value 6 degrees 40 minutes. And it is certain, according to what has been explained above, Article 27, that a plane which would touch the spheroid at the point M, where I suppose the [Pg 79]straight line CM to meet the surface, would be parallel to the plane QGq. If then through the point K one now draws KS parallel to Gg, which will be parallel also to QX, the tangent to the Ellipse QGq at Q; and if one conceives a plane passing through KS and touching the spheroid, the point of contact will necessarily be in the Ellipse QMq, because this plane through KS, as well as the plane which touches the spheroid at the point M, are parallel to QX, the tangent of the spheroid: for this consequence will be demonstrated at the end of this Treatise. Let this point of contact be at I, then making KC, QC, DC proportionals, draw DI parallel to CM; also join CI. I say that CI will be the required refraction of the ray RC. This will be manifest if, in considering CO, which is perpendicular to the ray RC, as a portion of the wave of light, we can demonstrate that the continuation of its piece C will be found in the crystal at I, when O has arrived at K.

Let the line N represent the distance light travels in air during the time it forms the spheroid QCqgM from the center C within the crystal. Next, draw CO perpendicular to the ray CR and place it in the plane through CR and AH. Across angle ACO, adjust the straight line OK equal to N and perpendicular to CO, meeting line AH at K. Assuming CL is perpendicular to the crystal surface AEHF and CM is the refraction of the ray hitting this surface perpendicularly, draw a plane through line CM and through KCH, creating the semi-ellipse QMq, which is defined since angle MCL is 6 degrees 40 minutes. As explained in Article 27, a plane touching the spheroid at point M, where I assume the straight line CM meets the surface, would be parallel to plane QGq. If we then draw KS from point K, parallel to Gg and also to QX, the tangent to the ellipse QGq at Q, and imagine a plane passing through KS that touches the spheroid, the point of contact must be on the ellipse QMq. This is because the plane through KS and the plane that touches the spheroid at point M are both parallel to QX, the tangent of the spheroid. We will demonstrate this at the end of this Treatise. Let the point of contact be at I, and then make KC, QC, DC proportional, drawing DI parallel to CM and joining CI. I claim that CI will be the required refraction of ray RC. This will be clear if we consider CO, which is perpendicular to ray RC, as part of the wave of light, and demonstrate that the extension of its segment C will be found in the crystal at I when O reaches K.

38. Now as in the Chapter on Reflexion, in demonstrating that the incident and reflected rays are always in the same plane perpendicular to the reflecting surface, we considered the breadth of the wave of light, so, similarly, we must here consider the breadth of the wave CO in the diameter Gg. Taking then the breadth Cc on the side toward the angle E, let the parallelogram COoc be taken as a portion of a wave, and let us complete the parallelograms CKkc, CIic, Klik, OKko. In the time then that the line Oo arrives at the surface of the crystal at Kk, all the points of the wave COoc will have arrived at the rectangle Kc along lines parallel to OK; and from the points of their incidences there will originate, beyond that, in the crystal partial hemi-spheroids, similar to the [Pg 80]hemi-spheroid QMq, and similarly disposed. These hemi-spheroids will necessarily all touch the plane of the parallelogram KIik at the same instant that Oo has reached Kk. Which is easy to comprehend, since, of these hemi-spheroids, all those which have their centres along the line CK, touch this plane in the line KI (for this is to be shown in the same way as we have demonstrated the refraction of the oblique ray in the principal section through EF) and all those which have their centres in the line Cc will touch the same plane KI in the line Ii; all these being similar to the hemi-spheroid QMq. Since then the parallelogram Ki is that which touches all these spheroids, this same parallelogram will be precisely the continuation of the wave COoc in the crystal, when Oo has arrived at Kk, because it forms the termination of the movement and because of the quantity of movement which occurs more there than anywhere else: and thus it appears that the piece C of the wave COoc has its continuation at I; that is to say, that the ray RC is refracted as CI.

38. Just like we discussed in the Chapter on Reflection, where we showed that the incident and reflected rays are always in the same plane perpendicular to the reflective surface, we now need to consider the breadth of the wave of light CO within the diameter Gg. Taking the breadth Cc on the side towards the angle E, let’s represent the parallelogram COoc as a segment of a wave, and we’ll complete the parallelograms CKkc, CIic, Klik, OKko. By the time the line Oo reaches the surface of the crystal at Kk, all points of the wave COoc will have arrived at the rectangle Kc along lines parallel to OK; and from where they hit, partial hemi-spheroids, like the hemi-spheroid QMq, will form in the crystal, similarly arranged. These hemi-spheroids will all touch the plane of the parallelogram KIik at the same moment that Oo reaches Kk. This is easy to understand because all the hemi-spheroids that have their centers along the line CK touch this plane at the line KI (we can show this the same way we demonstrated the refraction of the oblique ray in the main section through EF), and all those with centers in the line Cc will touch the same plane KI at the line Ii; all of these are similar to the hemi-spheroid QMq. Since the parallelogram Ki touches all these spheroids, it will serve as the precise continuation of the wave COoc in the crystal once Oo reaches Kk, as it marks the end of the movement and has the most movement occurring there: thus it follows that the piece C of the wave COoc transitions to I; meaning that the ray RC is refracted as CI.

From this it is to be noted that the proportion of the refraction for this section of the crystal is that of the line N to the semi-diameter CQ; by which one will easily find the refractions of all incident rays, in the same way as we have shown previously for the case of the section through FE; and the demonstration will be the same. But it appears that the said proportion of the refraction is less here than in the section through FEB; for it was there the same as the ratio of N to CG, that is to say, as 156,962 to 98,779, very nearly as 8 to 5; and here it is the ratio of N to CQ the major semi-diameter of the spheroid, that is to say, as 156,962 to 105,032, very nearly [Pg 81]as 3 to 2, but just a little less. Which still agrees perfectly with what one finds by observation.

From this, it's clear that the refraction ratio for this part of the crystal is the line N to the semi-diameter CQ; this allows us to easily calculate the refractions of all incoming rays, just like we showed earlier for the section through FE, and the proof will be the same. However, it seems that this refraction proportion is less here than in the section through FEB; there it was the same as the ratio of N to CG, which is about 156,962 to 98,779, roughly equal to 8 to 5; and here it is the ratio of N to CQ, the larger semi-diameter of the spheroid, meaning about 156,962 to 105,032, approximately [Pg 81]like 3 to 2, but slightly less. This still matches perfectly with what we observe.

39. For the rest, this diversity of proportion of refraction produces a very singular effect in this Crystal; which is that when it is placed upon a sheet of paper on which there are letters or anything else marked, if one views it from above with the two eyes situated in the plane of the section through EF, one sees the letters raised up by this irregular refraction more than when one puts one's eyes in the plane of section through AH: and the difference of these elevations appears by comparison with the other ordinary refraction of the crystal, the proportion of which is as 5 to 3, and which always raises the letters equally, and higher than the irregular refraction does. For one sees the letters and the paper on which they are written, as on two different stages at the same time; and in the first position of the eyes, namely, when they are in the plane through AH these two stages are four times more distant from one another than when the eyes are in the plane through EF.

39. For the rest, this variety in how light is bent creates a unique effect in this crystal; when it’s placed on a sheet of paper with letters or any markings, if you look down at it with both eyes positioned in the plane of the section through EF, you see the letters appearing raised more than when you position your eyes in the plane of section through AH. The difference in these elevations becomes evident when compared to the regular refraction of the crystal, which is at a ratio of 5 to 3, and consistently raises the letters to the same height, which is higher than the irregular refraction does. This means you see the letters and the paper they’re written on as if they’re on two different levels at the same time; and when your eyes are in the plane through AH, these two levels are four times farther apart than when your eyes are in the plane through EF.

We will show that this effect follows from the refractions; and it will enable us at the same time to ascertain the apparent place of a point of an object placed immediately under the crystal, according to the different situation of the eyes.

We will demonstrate that this effect comes from the refractions; and it will allow us to determine the apparent position of a point of an object placed directly beneath the crystal, based on the different positions of the eyes.

40. Let us see first by how much the irregular refraction of the plane through AH ought to lift the bottom of the crystal. Let the plane of this figure represent separately the section through Qq and CL, in which section there is also the ray RC, and let the semi-elliptic plane through Qq and CM be inclined to the former, as previously, by an angle of 6 degrees 40 minutes; and in this plane CI is then the refraction of the ray RC.[Pg 82]

40. Let's first look at how much the uneven bending of the plane through AH should raise the bottom of the crystal. The plane in this diagram shows the section through Qq and CL, which also includes the ray RC. The semi-elliptical plane through Qq and CM is tilted from the first one, as before, at an angle of 6 degrees and 40 minutes; in this plane, CI represents the bending of the ray RC.[Pg 82]

If now one considers the point I as at the bottom of the crystal, and that it is viewed by the rays ICR, Icr, refracted equally at the points Cc, which should be equally distant from D, and that these rays meet the two eyes at Rr; it is certain that the point I will appear raised to S where the straight lines RC, rc, meet; which point S is in DP, perpendicular to Qq. And if upon DP there is drawn the perpendicular IP, which will lie at the bottom of the crystal, the length SP will be the apparent elevation of the point I above the bottom.

If we consider point I at the bottom of the crystal and view it through the rays ICR, Icr, which are refracted equally at points Cc that are the same distance from D, these rays meet both eyes at Rr. It's clear that point I will appear to be elevated to point S, where the straight lines RC, rc, intersect; point S is on line DP, which is perpendicular to Qq. If we draw a perpendicular line IP from DP that lies at the bottom of the crystal, the length SP will represent the apparent height of point I above the bottom.

Let there be described on Qq a semicircle cutting the ray CR at B, from which BV is drawn perpendicular to Qq; and let the proportion of the refraction for this section be, as before, that of the line N to the semi-diameter CQ.

Let there be described on Qq a semicircle intersecting the ray CR at B, from which BV is drawn perpendicular to Qq; and let the ratio of the refraction for this section be, as before, that of the line N to the semi-diameter CQ.

Then as N is to CQ so is VC to CD, as appears by the method of finding the refraction which we have shown above, Article 31; but as VC is to CD, so is VB to DS. Then as N is to CQ, so is VB to DS. Let ML be perpendicular to CL. And because I suppose the eyes Rr to be distant about a foot or so from the crystal, and consequently the angle RSr very small, VB may be considered as equal to the semi-diameter CQ, and DP as equal to CL; then as N is to [Pg 83]CQ so is CQ to DS. But N is valued at 156,962 parts, of which CM contains 100,000 and CQ 105,032. Then DS will have 70,283. But CL is 99,324, being the sine of the complement of the angle MCL which is 6 degrees 40 minutes; CM being supposed as radius. Then DP, considered as equal to CL, will be to DS as 99,324 to 70,283. And so the elevation of the point I by the refraction of this section is known.

Then, just as N relates to CQ, VC relates to CD, as shown by the method for finding refraction detailed above in Article 31. Similarly, as VC relates to CD, VB relates to DS. Thus, just as N relates to CQ, VB relates to DS. Let ML be perpendicular to CL. Since I assume the eyes Rr are about a foot away from the crystal, making the angle RSr very small, VB can be considered equal to the semi-diameter CQ, and DP can be considered equal to CL. Therefore, just as N relates to [Pg 83]CQ, CQ relates to DS. N is valued at 156,962 parts, with CM containing 100,000 and CQ 105,032. This means DS will amount to 70,283. CL is 99,324, representing the sine of the complement of the angle MCL, which is 6 degrees 40 minutes, with CM treated as the radius. Then DP, being equal to CL, will relate to DS as 99,324 to 70,283. Thus, the elevation of point I, due to the refraction in this section, is determined.

41. Now let there be represented the other section through EF in the figure before the preceding one; and let CMg be the semi-ellipse, considered in Articles 27 and 28, which is made by cutting a spheroidal wave having centre C. Let the point I, taken in this ellipse, be imagined again at the bottom of the Crystal; and let it be viewed by the refracted rays ICR, Icr, which go to the two eyes; CR and cr being equally inclined to the surface of the crystal Gg. This being so, if one draws ID parallel to CM, which I suppose to be the refraction of the perpendicular ray incident at the point C, the distances DC, Dc, will be equal, as is easy to see by that which has been demonstrated in Article 28. Now it is certain that the point I should appear at S where the straight lines RC, rc, meet when prolonged; and that this point will fall in the [Pg 84]line DP perpendicular to Gg. If one draws IP perpendicular to this DP, it will be the distance PS which will mark the apparent elevation of the point I. Let there be described on Gg a semicircle cutting CR at B, from which let BV be drawn perpendicular to Gg; and let N to GC be the proportion of the refraction in this section, as in Article 28. Since then CI is the refraction of the radius BC, and DI is parallel to CM, VC must be to CD as N to GC, according to what has been demonstrated in Article 31. But as VC is to CD so is BV to DS. Let ML be drawn perpendicular to CL. And because I consider, again, the eyes to be distant above the crystal, BV is deemed equal to the semi-diameter CG; and hence DS will be a third proportional to the lines N and CG: also DP will be deemed equal to CL. Now CG consisting of 98,778 parts, of which CM contains 100,000, N is taken as 156,962. Then DS will be 62,163. But CL is also determined, and contains 99,324 parts, as has been said in Articles 34 and 40. Then the ratio of PD to DS will be as 99,324 to 62,163. And thus one knows the elevation of the point at the bottom I by the refraction of this section; and it appears that this elevation is greater than that by the refraction of the preceding section, since the ratio of PD to DS was there as 99,324 to 70,283.

41. Now let’s represent the other section through EF in the figure from before; and let CMg be the semi-ellipse discussed in Articles 27 and 28, created by cutting a spheroidal wave centered at C. Imagine point I, located in this ellipse, again at the bottom of the crystal; it will be viewed by the refracted rays ICR, Icr, that go to both eyes; CR and cr being equally tilted toward the surface of the crystal Gg. Given this, if we draw ID parallel to CM, which I assume to be the refraction of the perpendicular ray hitting at point C, the distances DC and Dc will be equal, as shown in Article 28. It’s clear that point I should appear at S where the straight lines RC and rc intersect when extended; and this point will lie along the [Pg 84]line DP, which is perpendicular to Gg. If we draw IP perpendicular to this DP, it will mark the distance PS, representing the apparent height of point I. We can draw a semicircle on Gg that intersects CR at B, from which we can draw BV perpendicular to Gg; let N be the refraction ratio in this section, as per Article 28. Since CI is the refraction of the radius BC, and DI is parallel to CM, VC must relate to CD in the same way N relates to GC, as demonstrated in Article 31. Just as VC relates to CD, BV relates to DS. Let’s draw ML perpendicular to CL. Considering again that the eyes are positioned above the crystal, BV is treated as equal to the semi-diameter CG; therefore, DS will be a third proportional to the lines N and CG: DP will also be treated as equal to CL. Now, CG consists of 98,778 parts, with CM containing 100,000, and N is taken as 156,962. Thus, DS will be 62,163. CL is also established, containing 99,324 parts, as mentioned in Articles 34 and 40. Consequently, the ratio of PD to DS will be as 99,324 to 62,163. This demonstrates the elevation of point I at the bottom through the refraction of this section; it appears that this elevation is greater than that from the refraction of the previous section since the ratio of PD to DS there was 99,324 to 70,283.

But by the regular refraction of the crystal, of which we have above said that the proportion is 5 to 3, the elevation of the point I, or P, from the bottom, will be 2/5 of the height DP; as appears by this figure, where the point P being viewed by the rays PCR, Pcr, refracted equally [Pg 85]at the surface Cc, this point must needs appear to be at S, in the perpendicular PD where the lines RC, rc, meet when prolonged: and one knows that the line PC is to CS as 5 to 3, since they are to one another as the sine of the angle CSP or DSC is to the sine of the angle SPC. And because the ratio of PD to DS is deemed the same as that of PC to CS, the two eyes Rr being supposed very far above the crystal, the elevation PS will thus be 2/5 of PD.

But through the regular refraction of the crystal, which we've noted has a ratio of 5 to 3, the height of point I, or P, from the bottom will be 2/5 of the height DP. This is shown in the figure, where point P is observed by the rays PCR and Pcr, refracted equally [Pg 85]at the surface Cc. Therefore, this point must appear at S along the perpendicular PD where the lines RC and rc intersect when extended. It's known that the line PC is to CS as 5 is to 3, since they relate to one another as the sine of angle CSP or DSC is to the sine of angle SPC. Also, because the ratio of PD to DS is considered the same as that of PC to CS, and assuming the two eyes Rr are situated very high above the crystal, the elevation PS will thus be 2/5 of PD.

42. If one takes a straight line AB for the thickness of the crystal, its point B being at the bottom, and if one divides it at the points C, D, E, according to the proportions of the elevations found, making AE 3/5 of AB, AB to AC as 99,324 to 70,283, and AB to AD as 99,324 to 62,163, these points will divide AB as in this figure. And it will be found that this agrees perfectly with experiment; that is to say by placing the eyes above in the plane which cuts the crystal according to the shorter diameter of the rhombus, the regular refraction will lift up the letters to E; and one will see the bottom, and the letters over which it is placed, lifted up to D by the irregular refraction. But by placing the eyes above in the plane which cuts the crystal according to the longer diameter of the rhombus, the regular refraction will lift the letters to E as before; but the irregular refraction will make them, at the same time, appear lifted up only to C; and in such a way that the interval CE will be quadruple the interval ED, which one previously saw.

42. If you take a straight line AB as the thickness of the crystal, with point B at the bottom, and then divide it at points C, D, and E based on the proportions of the heights measured, making AE 3/5 of AB, and AB to AC as 99,324 to 70,283, and AB to AD as 99,324 to 62,163, these points will divide AB as shown in this figure. It will be found that this aligns perfectly with the experiment; that is, by placing your eyes above in the plane that cuts the crystal along the shorter diameter of the rhombus, the regular refraction will raise the letters to E; and you will see the bottom, and the letters sitting over it will be lifted to D due to the irregular refraction. However, when you place your eyes above in the plane that cuts the crystal along the longer diameter of the rhombus, the regular refraction will still raise the letters to E as before; but the irregular refraction will make them appear lifted only to C at the same time, such that the distance CE will be four times the distance ED, which was observed earlier.

43. I have only to make the remark here that in both the positions of the eyes the images caused by the irregular refraction do not appear directly below those which proceed [Pg 86]from the regular refraction, but they are separated from them by being more distant from the equilateral solid angle of the Crystal. That follows, indeed, from all that has been hitherto demonstrated about the irregular refraction; and it is particularly shown by these last demonstrations, from which one sees that the point I appears by irregular refraction at S in the perpendicular line DP, in which line also the image of the point P ought to appear by regular refraction, but not the image of the point I, which will be almost directly above the same point, and higher than S.

43. I just want to note here that in both eye positions, the images created by irregular refraction don’t appear directly below those produced by regular refraction. Instead, they are offset because they are further away from the equilateral solid angle of the Crystal. This conclusion follows from everything that has been demonstrated about irregular refraction so far, and it's particularly illustrated by these latest demonstrations. From these, you can see that the point I shows up at S along the perpendicular line DP, where the image of point P should also appear via regular refraction, but not the image of point I, which will be almost directly above it and higher than S.

But as to the apparent elevation of the point I in other positions of the eyes above the crystal, besides the two positions which we have just examined, the image of that point by the irregular refraction will always appear between the two heights of D and C, passing from one to the other as one turns one's self around about the immovable crystal, while looking down from above. And all this is still found conformable to our hypothesis, as any one can assure himself after I shall have shown here the way of finding the irregular refractions which appear in all other sections of the crystal, besides the two which we have considered. Let us suppose one of the faces of the crystal, in which let there be the Ellipse HDE, the centre C of which is also the centre of the spheroid HME in which the light spreads, and of which the said Ellipse is the section. And let the incident ray be RC, the refraction of which it is required to find.

But regarding the apparent elevation of point I in other positions of the eyes above the crystal, in addition to the two positions we've just examined, the image of that point will always appear between the two heights of D and C, shifting from one to the other as one turns around the stationary crystal while looking down from above. This still aligns with our hypothesis, as anyone can confirm once I show the method for finding the irregular refractions that appear in all other sections of the crystal, besides the two we've considered. Let’s consider one of the faces of the crystal, which includes the Ellipse HDE, where the center C is also the center of the spheroid HME that the light spreads across, and of which the mentioned Ellipse is the section. Let the incident ray be RC, and we need to find its refraction.

Let there be taken a plane passing through the ray RC and which is perpendicular to the plane of the ellipse HDE, cutting it along the straight line BCK; and having in the same plane through RC made CO perpendicular to CR, [Pg 87]let OK be adjusted across the angle OCK, so as to be perpendicular to OC and equal to the line N, which I suppose to measure the travel of the light in air during the time that it spreads in the crystal through the spheroid HDEM. Then in the plane of the Ellipse HDE let KT be drawn, through the point K, perpendicular to BCK. Now if one conceives a plane drawn through the straight line KT and touching the spheroid HME at I, the straight line CI will be the refraction of the ray RC, as is easy to deduce from that which has been demonstrated in Article 36.

Let’s take a plane that goes through the ray RC and is perpendicular to the plane of the ellipse HDE, cutting it along the straight line BCK. In the same plane through RC, make CO perpendicular to CR, [Pg 87]and let OK be set across the angle OCK, so that it is perpendicular to OC and equal to the line N, which I assume measures the light's travel in air while it spreads in the crystal through the spheroid HDEM. Then, in the plane of the ellipse HDE, draw KT through the point K, perpendicular to BCK. Now, if we imagine a plane drawn through the straight line KT that touches the spheroid HME at I, the straight line CI will represent the refraction of the ray RC, as can be easily derived from what was proven in Article 36.

But it must be shown how one can determine the point of contact I. Let there be drawn parallel to the line KT a line HF which touches the Ellipse HDE, and let this point of contact be at H. And having drawn a straight line along CH to meet KT at T, let there be imagined a plane passing through the same CH and through CM (which I suppose to be the refraction of the perpendicular ray), which makes in the spheroid the elliptical section HME. It is certain that the plane which will pass through the straight line KT, and which will touch the spheroid, will touch it at a point in the Ellipse HME, according to the Lemma which will be demonstrated at the end of the [Pg 88]Chapter. Now this point is necessarily the point I which is sought, since the plane drawn through TK can touch the spheroid at one point only. And this point I is easy to determine, since it is needful only to draw from the point T, which is in the plane of this Ellipse, the tangent TI, in the way shown previously. For the Ellipse HME is given, and its conjugate semi-diameters are CH and CM; because a straight line drawn through M, parallel to HE, touches the Ellipse HME, as follows from the fact that a plane taken through M, and parallel to the plane HDE, touches the spheroid at that point M, as is seen from Articles 27 and 23. For the rest, the position of this ellipse, with respect to the plane through the ray RC and through CK, is also given; from which it will be easy to find the position of CI, the refraction corresponding to the ray RC.

But we need to explain how to find the point of contact I. First, draw a line HF parallel to the line KT that touches the ellipse HDE at point H. Then, draw a straight line from point C to meet KT at point T. Now, imagine a plane that passes through line CH and line CM (which I assume is the refraction of the perpendicular ray), creating the elliptical section HME within the spheroid. It's clear that the plane passing through the straight line KT and touching the spheroid will make contact at a point on the ellipse HME, according to the lemma that will be explained at the end of the [Pg 88]Chapter. This point is necessarily the point I we're looking for, as the plane created with TK can touch the spheroid at only one point. It’s straightforward to determine point I since we only need to draw a tangent TI from point T, which lies in the plane of this ellipse, as previously shown. The ellipse HME is given, with its conjugate semi-diameters CH and CM; because a straight line drawn through M, parallel to HE, touches the ellipse HME, as indicated by the fact that a plane taken through M, parallel to the plane HDE, touches the spheroid at point M, as noted in Articles 27 and 23. Additionally, the position of this ellipse in relation to the plane through rays RC and CK is already defined; therefore, it will be easy to find the position of CI, the refraction that corresponds to ray RC.

Now it must be noted that the same ellipse HME serves to find the refractions of any other ray which may be in the plane through RC and CK. Because every plane, parallel to the straight line HF, or TK, which will touch the spheroid, will touch it in this ellipse, according to the Lemma quoted a little before.

Now it should be noted that the same ellipse HME is used to find the refractions of any other ray that may be in the plane through RC and CK. This is because every plane parallel to the straight line HF or TK, which will touch the spheroid, will do so at this ellipse, according to the lemma mentioned a little earlier.

I have investigated thus, in minute detail, the properties of the irregular refraction of this Crystal, in order to see whether each phenomenon that is deduced from our hypothesis accords with that which is observed in fact. And this being so it affords no slight proof of the truth of our suppositions and principles. But what I am going to add here confirms them again marvellously. It is this: that there are different sections of this Crystal, the surfaces of which, thereby produced, give rise to refractions precisely such as they ought to be, and as I had foreseen them, according to the preceding Theory.

I have carefully examined the properties of the irregular refraction of this crystal to determine whether each phenomenon we inferred from our hypothesis matches what we actually observe. This provides strong evidence for the validity of our assumptions and principles. Furthermore, what I am about to add reinforces them remarkably. Specifically, there are different sections of this crystal, and the resulting surfaces produce refractions exactly as they should, just as I predicted according to the earlier theory.

[Pg 89]In order to explain what these sections are, let ABKF be the principal section through the axis of the crystal ACK, in which there will also be the axis SS of a spheroidal wave of light spreading in the crystal from the centre C; and the straight line which cuts SS through the middle and at right angles, namely PP, will be one of the major diameters.

[Pg 89]To explain what these sections are, let ABKF be the main section through the axis of the crystal ACK, where the axis SS will contain a spheroidal light wave spreading from the center C in the crystal. The straight line that intersects SS at its midpoint and at a right angle, which we will call PP, will be one of the major diameters.

{Section ABKF}

Now as in the natural section of the crystal, made by a plane parallel to two opposite faces, which plane is here represented by the line GG, the refraction of the surfaces which are produced by it will be governed by the hemi-spheroids GNG, according to what has been explained in the preceding Theory. Similarly, cutting the Crystal through NN, by a plane perpendicular to the parallelogram ABKF, the refraction of the surfaces will be governed by the hemi-spheroids NGN. And if one cuts it through PP, perpendicularly to the said parallelogram, the refraction of the surfaces ought to be governed by the hemi-spheroids PSP, and so for others. But I saw that if the plane NN was almost perpendicular to the plane GG, making the angle NCG, which is on the side A, an angle of 90 degrees 40 minutes, the hemi-spheroids NGN would become similar to the hemi-spheroids GNG, since the planes NN and GG were equally inclined by an angle of 45 degrees 20 minutes to the axis SS. In consequence it must needs be, if our theory is true, that the surfaces which the section through [Pg 90]NN produces should effect the same refractions as the surfaces of the section through GG. And not only the surfaces of the section NN but all other sections produced by planes which might be inclined to the axis at an angle equal to 45 degrees 20 minutes. So that there are an infinitude of planes which ought to produce precisely the same refractions as the natural surfaces of the crystal, or as the section parallel to any one of those surfaces which are made by cleavage.

Now, just like in the natural part of the crystal, created by a plane that runs parallel to two opposite faces, which is represented here by the line GG, the refraction of the surfaces created by this will be controlled by the hemi-spheroids GNG, as explained in the previous theory. Similarly, if we cut the crystal through NN with a plane that is perpendicular to the parallelogram ABKF, the refraction of those surfaces will be governed by the hemi-spheroids NGN. And if we cut it through PP, also perpendicular to that parallelogram, the refraction of those surfaces should be governed by the hemi-spheroids PSP, and the same goes for others. However, I observed that if plane NN is nearly perpendicular to plane GG, creating an angle NCG, on the side A, of 90 degrees 40 minutes, the hemi-spheroids NGN would become similar to the hemi-spheroids GNG, since both planes NN and GG are inclined at an angle of 45 degrees 20 minutes to the axis SS. Therefore, if our theory is correct, the surfaces produced by the section through [Pg 90]NN should produce the same refractions as the surfaces from the section through GG. This applies not only to the surfaces of section NN but to all other sections created by planes that are inclined to the axis at an angle of 45 degrees 20 minutes. Thus, there are countless planes that should produce exactly the same refractions as the natural surfaces of the crystal, or as the sections parallel to any of those surfaces made by cleavage.

I saw also that by cutting it by a plane taken through PP, and perpendicular to the axis SS, the refraction of the surfaces ought to be such that the perpendicular ray should suffer thereby no deviation; and that for oblique rays there would always be an irregular refraction, differing from the regular, and by which objects placed beneath the crystal would be less elevated than by that other refraction.

I also realized that if you slice it with a plane through PP, perpendicular to the axis SS, the perpendicular ray shouldn’t change direction at all; however, for angled rays, there would always be an irregular refraction that's different from the usual one, meaning that objects positioned underneath the crystal would appear lower than they would with that other type of refraction.

That, similarly, by cutting the crystal by any plane through the axis SS, such as the plane of the figure is, the perpendicular ray ought to suffer no refraction; and that for oblique rays there were different measures for the irregular refraction according to the situation of the plane in which the incident ray was.

That, in the same way, by slicing the crystal with any plane through the axis SS, like the plane shown in the figure, the perpendicular ray should experience no refraction; and that for oblique rays, there were different measurements for the irregular refraction based on the position of the plane where the incident ray was.

Now these things were found in fact so; and, after that, I could not doubt that a similar success could be met with everywhere. Whence I concluded that one might form from this crystal solids similar to those which are its natural forms, which should produce, at all their surfaces, the same regular and irregular refractions as the natural surfaces, and which nevertheless would cleave in quite other ways, and not in directions parallel to any of their faces. That out of it one would be able to fashion pyramids, having their base square, pentagonal, hexagonal, or with as many sides [Pg 91]as one desired, all the surfaces of which should have the same refractions as the natural surfaces of the crystal, except the base, which will not refract the perpendicular ray. These surfaces will each make an angle of 45 degrees 20 minutes with the axis of the crystal, and the base will be the section perpendicular to the axis.

Now, these things were actually found to be true; and after that, I couldn't doubt that a similar success could be achieved anywhere. So, I concluded that one could create solid shapes from this crystal similar to its natural forms, which would produce both the same regular and irregular refractions as the natural surfaces. However, they would break in completely different ways, not along directions parallel to any of their faces. From it, one would be able to shape pyramids with square, pentagonal, hexagonal bases, or with as many sides [Pg 91]as desired, all of which would have the same refractions as the natural surfaces of the crystal, except for the base, which would not refract the perpendicular ray. Each of these surfaces will make an angle of 45 degrees 20 minutes with the axis of the crystal, and the base will be the section perpendicular to the axis.

That, finally, one could also fashion out of it triangular prisms, or prisms with as many sides as one would, of which neither the sides nor the bases would refract the perpendicular ray, although they would yet all cause double refraction for oblique rays. The cube is included amongst these prisms, the bases of which are sections perpendicular to the axis of the crystal, and the sides are sections parallel to the same axis.

That, ultimately, you could also create triangular prisms or prisms with as many sides as you want, where neither the sides nor the bases would bend the perpendicular ray, even though they would still cause double refraction for oblique rays. The cube is part of these prisms, with bases that are sections perpendicular to the axis of the crystal and sides that are sections parallel to the same axis.

From all this it further appears that it is not at all in the disposition of the layers of which this crystal seems to be composed, and according to which it splits in three different senses, that the cause resides of its irregular refraction; and that it would be in vain to wish to seek it there.

From all this, it also seems that the cause of its irregular refraction doesn't lie in the arrangement of the layers that make up this crystal, which is why it splits in three different directions, and it would be pointless to look for it there.

But in order that any one who has some of this stone may be able to find, by his own experience, the truth of what I have just advanced, I will state here the process of which I have made use to cut it, and to polish it. Cutting is easy by the slicing wheels of lapidaries, or in the way in which marble is sawn: but polishing is very difficult, and by employing the ordinary means one more often depolishes the surfaces than makes them lucent.

But so that anyone who has some of this stone can discover the truth of what I've just mentioned through their own experience, I will explain the process I used to cut and polish it. Cutting is easy with the slicing wheels of gem cutters or like how marble is sawed, but polishing is quite challenging, and when using standard methods, it's more common to dull the surfaces than to make them shiny.

After many trials, I have at last found that for this service no plate of metal must be used, but a piece of mirror glass made matt and depolished. Upon this, with fine sand and water, one smoothes the crystal little by little, in the same [Pg 92]way as spectacle glasses, and polishes it simply by continuing the work, but ever reducing the material. I have not, however, been able to give it perfect clarity and transparency; but the evenness which the surfaces acquire enables one to observe in them the effects of refraction better than in those made by cleaving the stone, which always have some inequality.

After many attempts, I've finally discovered that for this task, I shouldn't use a metal plate, but rather a piece of frosted mirror glass. On this, using fine sand and water, you smooth the crystal gradually, just like how you would with eyeglass lenses, continuing to polish it while gradually reducing the material. However, I haven't been able to achieve perfect clarity and transparency; still, the smoothness of the surfaces allows for better observation of the refraction effects than those made by splitting the stone, which always have some unevenness.

Even when the surface is only moderately smoothed, if one rubs it over with a little oil or white of egg, it becomes quite transparent, so that the refraction is discerned in it quite distinctly. And this aid is specially necessary when it is wished to polish the natural surfaces to remove the inequalities; because one cannot render them lucent equally with the surfaces of other sections, which take a polish so much the better the less nearly they approximate to these natural planes.

Even when the surface is only somewhat smoothed, if you rub it with a bit of oil or egg white, it becomes quite clear, so that you can see the refraction in it very clearly. This technique is especially important when you want to polish the natural surfaces to get rid of the imperfections; because you can't make them transparent in the same way as other sections, which polish much better the less they resemble these natural planes.

Before finishing the treatise on this Crystal, I will add one more marvellous phenomenon which I discovered after having written all the foregoing. For though I have not been able till now to find its cause, I do not for that reason wish to desist from describing it, in order to give opportunity to others to investigate it. It seems that it will be necessary to make still further suppositions besides those which I have made; but these will not for all that cease to keep their probability after having been confirmed by so many tests.

Before wrapping up the discussion on this Crystal, I want to share one more amazing phenomenon that I found after writing everything above. Even though I haven't been able to figure out its cause yet, I still want to describe it to encourage others to look into it. It appears that we will need to make some additional assumptions beyond the ones I've already laid out; however, these will still remain plausible after being verified through numerous tests.

The phenomenon is, that by taking two pieces of this crystal and applying them one over the other, or rather holding them with a space between the two, if all the sides of one are parallel to those of the other, then a ray of light, such as AB, is divided into two in the first piece, namely into BD and BC, following the two refractions, [Pg 93]regular and irregular. On penetrating thence into the other piece each ray will pass there without further dividing itself in two; but that one which underwent the regular refraction, as here DG, will undergo again only a regular refraction at GH; and the other, CE, an irregular refraction at EF. And the same thing occurs not only in this disposition, but also in all those cases in which the principal section of each of the pieces is situated in one and the same plane, without it being needful for the two neighbouring surfaces to be parallel. Now it is marvellous why the rays CE and DG, incident from the air on the lower crystal, do not divide themselves the same as the first ray AB. One would say that it must be that the ray DG in passing through the upper piece has lost something which is necessary to move the matter which serves for the irregular refraction; and that likewise CE has lost that which [Pg 94]was necessary to move the matter which serves for regular refraction: but there is yet another thing which upsets this reasoning. It is that when one disposes the two crystals in such a way that the planes which constitute the principal sections intersect one another at right angles, whether the neighbouring surfaces are parallel or not, then the ray which has come by the regular refraction, as DG, undergoes only an irregular refraction in the lower piece; and on the contrary the ray which has come by the irregular refraction, as CE, undergoes only a regular refraction.

The phenomenon is that when you take two pieces of this crystal and place them one over the other, or rather keep them apart with a space in between, if all the sides of one align with those of the other, then a ray of light, like AB, is split into two in the first piece, specifically into BD and BC, following the two types of refractions, [Pg 93]regular and irregular. As it then enters the second piece, each ray will pass through without further splitting; but the one that went through regular refraction, as shown by DG, will only experience regular refraction again at GH, while the other, CE, will have irregular refraction at EF. This also happens not just in this arrangement, but in all instances where the main section of each piece is in the same plane, even if the neighboring surfaces are not parallel. It’s curious why the rays CE and DG, which come from the air to the lower crystal, do not divide like the initial ray AB. It would seem that ray DG, by passing through the upper piece, loses something essential for the irregular refraction; similarly, CE must lose what [Pg 94]is needed for the regular refraction. However, another factor complicates this reasoning. When the two crystals are arranged so that the planes of their main sections intersect at right angles, whether or not the neighboring surfaces are parallel, the ray that has gone through regular refraction, like DG, will only have irregular refraction in the lower piece; conversely, the ray that went through irregular refraction, like CE, will have only regular refraction.

But in all the infinite other positions, besides those which I have just stated, the rays DG, CE, divide themselves anew each one into two, by refraction in the lower crystal so that from the single ray AB there are four, sometimes of equal brightness, sometimes some much less bright than others, according to the varying agreement in the positions of the crystals: but they do not appear to have all together more light than the single ray AB.

But in all the countless other positions, besides the ones I just mentioned, the rays DG and CE split again, each into two, due to refraction in the lower crystal. This means that from the single ray AB, there are now four rays, sometimes of equal brightness and sometimes with one much dimmer than the others, depending on how the crystals are positioned. However, they don't seem to have more total brightness than the single ray AB.

When one considers here how, while the rays CE, DG, remain the same, it depends on the position that one gives to the lower piece, whether it divides them both in two, or whether it does not divide them, and yet how the ray AB above is always divided, it seems that one is obliged to conclude that the waves of light, after having passed through the first crystal, acquire a certain form or disposition in virtue of which, when meeting the texture of the second crystal, in certain positions, they can move the two different kinds of matter which serve for the two species of refraction; and when meeting the second crystal in another position are able to move only one of these kinds of matter. But to tell how this occurs, I have hitherto found nothing which satisfies me.

When you think about how the rays CE and DG stay the same, it depends on where you place the lower piece—whether it splits both rays in half or doesn’t split them at all. Still, the ray AB above is always split. This suggests that the light waves, after passing through the first crystal, take on a specific form or arrangement that allows them to interact with the structure of the second crystal. Depending on the orientation, they can affect both types of matter used for the two types of refraction, but in a different position, they might only influence one of those types. However, I haven't found a clear explanation of how this process works so far.

[Pg 95]Leaving then to others this research, I pass to what I have to say touching the cause of the extraordinary figure of this crystal, and why it cleaves easily in three different senses, parallel to any one of its surfaces.

[Pg 95]Leaving this research to others, I’ll move on to what I need to say about the reason behind the unique shape of this crystal and why it splits easily in three different directions, parallel to any of its surfaces.

There are many bodies, vegetable, mineral, and congealed salts, which are formed with certain regular angles and figures. Thus among flowers there are many which have their leaves disposed in ordered polygons, to the number of 3, 4, 5, or 6 sides, but not more. This well deserves to be investigated, both as to the polygonal figure, and as to why it does not exceed the number 6.

There are many forms, including plants, minerals, and solidified salts, that are shaped with specific angles and patterns. For example, among flowers, many have their leaves arranged in organized polygons, with sides numbering 3, 4, 5, or 6, but never more. This is definitely worth exploring, both in terms of the polygonal shape and why it doesn’t go beyond 6 sides.

Rock Crystal grows ordinarily in hexagonal bars, and diamonds are found which occur with a square point and polished surfaces. There is a species of small flat stones, piled up directly upon one another, which are all of pentagonal figure with rounded angles, and the sides a little folded inwards. The grains of gray salt which are formed from sea water affect the figure, or at least the angle, of the cube; and in the congelations of other salts, and in that of sugar, there are found other solid angles with perfectly flat faces. Small snowflakes almost always fall in little stars with 6 points, and sometimes in hexagons with straight sides. And I have often observed, in water which is beginning to freeze, a kind of flat and thin foliage of ice, the middle ray of which throws out branches inclined at an angle of 60 degrees. All these things are worthy of being carefully investigated to ascertain how and by what artifice nature there operates. But it is not now my intention to treat fully of this matter. It seems that in general the regularity which occurs in these productions comes from the arrangement of the small invisible equal particles of which they are composed. And, coming to our Iceland Crystal, I say [Pg 96]that if there were a pyramid such as ABCD, composed of small rounded corpuscles, not spherical but flattened spheroids, such as would be made by the rotation of the ellipse GH around its lesser diameter EF (of which the ratio to the greater diameter is very nearly that of 1 to the square root of 8)—I say that then the solid angle of the point D would be equal to the obtuse and equilateral angle of this Crystal. I say, further, that if these corpuscles were lightly stuck together, on breaking this pyramid it would break along faces parallel to those that make its point: and by this means, as it is easy to see, it would produce prisms similar to those of the same crystal as this other figure represents. The reason is that when broken in this fashion a whole layer separates easily from its neighbouring layer since each spheroid has to be detached only from the three spheroids of the next layer; of which three there is but one which touches it on its flattened surface, and the other two at the edges. And the reason why the surfaces separate sharp and polished is that if any spheroid of the neighbouring surface would come out by attaching itself to the surface which is being separated, it would be needful for it to detach itself from six other spheroids which hold it locked, and four of which press it by these flattened surfaces. Since then not only the angles of our crystal but also the manner in which it splits agree precisely with what is observed in the assemblage composed of such spheroids, there is great reason to believe that the particles are shaped and ranged in the same way.

Rock crystal usually grows in hexagonal bars, and diamonds are found with a square point and polished surfaces. There’s a type of small flat stones that stack directly on top of each other, all in a pentagonal shape with rounded corners and slightly inward-folded sides. The grains of gray salt that form from seawater affect the shape, or at least the angles, of the cube; and in the freezing of other salts, as well as sugar, there are other solid angles with perfectly flat faces. Small snowflakes almost always fall as tiny stars with 6 points, and sometimes as hexagons with straight sides. I've often observed that in water starting to freeze, there's a kind of flat and thin layer of ice, where the middle ray has branches that extend at a 60-degree angle. All these things deserve careful investigation to understand how and by what means nature operates. However, I don’t intend to fully discuss this topic right now. It seems that in general, the regularity seen in these formations comes from the arrangement of the tiny invisible equal particles they’re made of. Now, regarding our Iceland crystal, I say [Pg 96]that if there were a pyramid like ABCD, made of small rounded particles—not spherical but flattened spheroids, created by rotating the ellipse GH around its shorter diameter EF (which is nearly in the ratio of 1 to the square root of 8)—I propose that the solid angle at point D would be equal to the obtuse and equilateral angle of this crystal. Furthermore, if these particles were loosely bonded together, breaking this pyramid would result in a fracture along faces parallel to those that form its apex: and this would naturally produce prisms similar to those of the same crystal as shown in the other figure. The reason is that when broken this way, an entire layer separates easily from its adjacent layer since each spheroid only needs to detach from the three spheroids of the next layer; of those three, only one touches it on its flattened surface, while the other two do so at the edges. The reason the surfaces break off sharp and polished is that if any spheroid from the neighboring surface tries to stick to the separating surface, it would need to detach from six other spheroids that hold it in place, four of which press against it with their flattened surfaces. Since both the angles of our crystal and how it splits align perfectly with what’s observed in an arrangement made from such spheroids, there’s strong reason to believe that the particles are shaped and organized in the same manner.

{Pyramid and section of spheroids}

[Pg 97]There is even probability enough that the prisms of this crystal are produced by the breaking up of pyramids, since Mr. Bartholinus relates that he occasionally found some pieces of triangularly pyramidal figure. But when a mass is composed interiorly only of these little spheroids thus piled up, whatever form it may have exteriorly, it is certain, by the same reasoning which I have just explained, that if broken it would produce similar prisms. It remains to be seen whether there are other reasons which confirm our conjecture, and whether there are none which are repugnant to it.

[Pg 97]There’s a good chance that the prisms of this crystal are formed from the breakdown of pyramids, since Mr. Bartholinus reported occasionally finding pieces in a triangular pyramidal shape. However, if a mass is made up entirely of these small spheroids stacked together, regardless of its external shape, it's clear—following the reasoning I just explained—that if it were broken, it would create similar prisms. We need to see if there are additional reasons that support our idea and whether there are any that contradict it.

{paralleloid arrangement of spheroids with planes of potential cleavage}

It may be objected that this crystal, being so composed, might be capable of cleavage in yet two more fashions; one of which would be along planes parallel to the base of the pyramid, that is to say to the triangle ABC; the other would be parallel to a plane the trace of which is marked by the lines GH, HK, KL. To which I say that both the one and the other, though practicable, are more difficult than those which were parallel to any one of the three planes of the pyramid; and that therefore, when striking on the crystal in order to break it, it ought always to split rather along these three planes than along the two others. When one has a number of spheroids of the form above described, and ranges them in a pyramid, one sees why the two methods of division are more difficult. For in the case of that division which would be parallel to the base, [Pg 98]each spheroid would be obliged to detach itself from three others which it touches upon their flattened surfaces, which hold more strongly than the contacts at the edges. And besides that, this division will not occur along entire layers, because each of the spheroids of a layer is scarcely held at all by the 6 of the same layer that surround it, since they only touch it at the edges; so that it adheres readily to the neighbouring layer, and the others to it, for the same reason; and this causes uneven surfaces. Also one sees by experiment that when grinding down the crystal on a rather rough stone, directly on the equilateral solid angle, one verily finds much facility in reducing it in this direction, but much difficulty afterwards in polishing the surface which has been flattened in this manner.

It could be argued that this crystal, being structured this way, might be able to split in two additional ways: one would be along planes parallel to the base of the pyramid, specifically to the triangle ABC; the other would be parallel to a plane marked by the lines GH, HK, KL. However, I maintain that both methods, while possible, are harder than those that align with any of the three planes of the pyramid. Thus, when striking the crystal to break it, it will typically split along these three planes instead of the two others. When you have several spheroids shaped as described and arrange them in a pyramid, it becomes clear why the two splitting methods are more challenging. For the splitting parallel to the base, each spheroid has to separate from three others it touches at their flat surfaces, which connect more strongly than the edges. Moreover, this splitting won’t happen along complete layers because each spheroid in a layer is only slightly held by the six others in that layer that surround it, as they only touch at the edges. So, it easily adheres to the neighboring layer, and vice versa, leading to uneven surfaces. Also, experiments show that when grinding down the crystal on a rough stone directly on the equilateral solid angle, it is quite easy to reduce it this way, but difficult afterward to polish the surface that has been flattened.

As for the other method of division along the plane GHKL, it will be seen that each spheroid would have to detach itself from four of the neighbouring layer, two of which touch it on the flattened surfaces, and two at the edges. So that this division is likewise more difficult than that which is made parallel to one of the surfaces of the crystal; where, as we have said, each spheroid is detached from only three of the neighbouring layer: of which three there is one only which touches it on the flattened surface, and the other two at the edges only.

As for the other way to divide along the plane GHKL, it's clear that each spheroid would need to separate from four neighboring layers, two of which touch it on the flat surfaces and two at the edges. This means that this division is also more challenging than the one made parallel to one of the crystal surfaces; in that case, as we've mentioned, each spheroid only separates from three neighboring layers, and out of those three, only one touches it on the flat surface, while the other two only touch it at the edges.

However, that which has made me know that in the crystal there are layers in this last fashion, is that in a piece weighing half a pound which I possess, one sees that it is split along its length, as is the above-mentioned prism by the plane GHKL; as appears by colours of the Iris extending throughout this whole plane although the two pieces still hold together. All this proves then that the composition of the crystal is such as we have stated. To [Pg 99]which I again add this experiment; that if one passes a knife scraping along any one of the natural surfaces, and downwards as it were from the equilateral obtuse angle, that is to say from the apex of the pyramid, one finds it quite hard; but by scraping in the opposite sense an incision is easily made. This follows manifestly from the situation of the small spheroids; over which, in the first manner, the knife glides; but in the other manner it seizes them from beneath almost as if they were the scales of a fish.

However, what has made me realize that there are layers in the crystal in this way is that in a piece I have weighing half a pound, you can see it split along its length, just like the prism mentioned above at the plane GHKL; the colors of the rainbow extend throughout this entire plane, even though the two pieces still hold together. All this proves that the composition of the crystal is as we've stated. Additionally, I will add this experiment: if you scrape a knife along any of the natural surfaces, moving downward from the equilateral obtuse angle, which is to say from the top of the pyramid, you find it quite hard; but if you scrape in the opposite direction, making an incision is easy. This is clearly due to the arrangement of the small spheres; in the first direction, the knife glides over them, but in the other direction, it catches them from underneath, almost as if they were the scales of a fish.

I will not undertake to say anything touching the way in which so many corpuscles all equal and similar are generated, nor how they are set in such beautiful order; whether they are formed first and then assembled, or whether they arrange themselves thus in coming into being and as fast as they are produced, which seems to me more probable. To develop truths so recondite there would be needed a knowledge of nature much greater than that which we have. I will add only that these little spheroids could well contribute to form the spheroids of the waves of light, here above supposed, these as well as those being similarly situated, and with their axes parallel.

I won't attempt to explain how so many identical and similar particles are created, or how they're organized so beautifully; whether they're formed first and then grouped together, or if they automatically arrange themselves as they're produced, which seems more likely to me. To uncover such deep truths would require a much greater understanding of nature than what we currently possess. I’ll only add that these tiny spheres could very well help form the spheres of the light waves mentioned above, as both are similarly positioned, with their axes parallel.

Calculations which have been supposed in this Chapter.

Calculations in this Chapter.

Mr. Bartholinus, in his treatise of this Crystal, puts at 101 degrees the obtuse angles of the faces, which I have stated to be 101 degrees 52 minutes. He states that he measured these angles directly on the crystal, which is difficult to do with ultimate exactitude, because the edges such as CA, CB, in this figure, are generally worn, and not quite straight. For more certainty, therefore, I preferred to measure actually the obtuse angle by which the faces [Pg 100]CBDA, CBVF, are inclined to one another, namely the angle OCN formed by drawing CN perpendicular to FV, and CO perpendicular to DA. This angle OCN I found to be 105 degrees; and its supplement CNP, to be 75 degrees, as it should be.

Mr. Bartholinus, in his work on this crystal, lists the obtuse angles of the faces at 101 degrees, while I have noted them as 101 degrees 52 minutes. He mentions that he measured these angles directly on the crystal, which is tricky to do with complete accuracy because the edges such as CA and CB, as shown in this figure, are usually worn and not perfectly straight. To be more certain, I preferred to measure the obtuse angle formed by the faces [Pg 100]CBDA, CBVF, which is the angle OCN created by drawing CN perpendicular to FV and CO perpendicular to DA. I found this angle OCN to be 105 degrees, and its supplement CNP to be 75 degrees, as expected.

To find from this the obtuse angle BCA, I imagined a sphere having its centre at C, and on its surface a spherical triangle, formed by the intersection of three planes which enclose the solid angle C. In this equilateral triangle, which is ABF in this other figure, I see that each of the angles should be 105 degrees, namely equal to the angle OCN; and that each of the sides should be of as many degrees as the angle ACB, or ACF, or BCF. Having then drawn the arc FQ perpendicular to the side AB, which it divides equally at Q, the triangle FQA has a right angle at Q, the angle A 105 degrees, and F half as much, namely 52 degrees 30 minutes; whence the hypotenuse AF is found to be 101 degrees 52 minutes. And this arc AF is the measure of the angle ACF in the figure of the crystal.

To find the obtuse angle BCA, I envisioned a sphere centered at C, with a spherical triangle on its surface formed by the intersection of three planes that create solid angle C. In this equilateral triangle, which is ABF in the other figure, each of the angles is 105 degrees, equal to the angle OCN; and each side has the same degree measurement as angle ACB, ACF, or BCF. I then drew the arc FQ perpendicular to side AB, which it divides equally at Q. The triangle FQA has a right angle at Q, angle A at 105 degrees, and F at half that, which is 52 degrees 30 minutes; thus, the hypotenuse AF measures 101 degrees 52 minutes. This arc AF represents the measure of angle ACF in the crystal figure.

In the same figure, if the plane CGHF cuts the crystal so that it divides the obtuse angles ACB, MHV, in the middle, it is stated, in Article 10, that the angle CFH is 70 degrees 57 minutes. This again is easily shown in the [Pg 101]same spherical triangle ABF, in which it appears that the arc FQ is as many degrees as the angle GCF in the crystal, the supplement of which is the angle CFH. Now the arc FQ is found to be 109 degrees 3 minutes. Then its supplement, 70 degrees 57 minutes, is the angle CFH.

In the same figure, if the plane CGHF cuts the crystal so that it divides the obtuse angles ACB and MHV evenly, Article 10 states that the angle CFH is 70 degrees 57 minutes. This can be easily demonstrated in the [Pg 101]same spherical triangle ABF, where it is evident that the arc FQ is equal in degrees to the angle GCF in the crystal, the supplement of which is the angle CFH. The arc FQ measures 109 degrees 3 minutes. Therefore, its supplement, 70 degrees 57 minutes, corresponds to the angle CFH.

It was stated, in Article 26, that the straight line CS, which in the preceding figure is CH, being the axis of the crystal, that is to say being equally inclined to the three sides CA, CB, CF, the angle GCH is 45 degrees 20 minutes. This is also easily calculated by the same spherical triangle. For by drawing the other arc AD which cuts BF equally, and intersects FQ at S, this point will be the centre of the triangle. And it is easy to see that the arc SQ is the measure of the angle GCH in the figure which represents the crystal. Now in the triangle QAS, which is right-angled, one knows also the angle A, which is 52 degrees 30 minutes, and the side AQ 50 degrees 56 minutes; whence the side SQ is found to be 45 degrees 20 minutes.

It was stated in Article 26 that the straight line CS, which is CH in the previous figure and serves as the axis of the crystal, is equally inclined to the three sides CA, CB, and CF, making the angle GCH 45 degrees 20 minutes. This can also be easily calculated using the same spherical triangle. By drawing the other arc AD, which intersects BF equally and meets FQ at S, this point will be the center of the triangle. It's clear that the arc SQ measures the angle GCH in the figure representing the crystal. In the right-angled triangle QAS, we also know that angle A is 52 degrees 30 minutes, and side AQ is 50 degrees 56 minutes; therefore, side SQ is determined to be 45 degrees 20 minutes.

In Article 27 it was required to show that PMS being an ellipse the centre of which is C, and which touches the straight line MD at M so that the angle MCL which CM makes with CL, perpendicular on DM, is 6 degrees 40 minutes, and its semi-minor axis CS making with CG (which is parallel to MD) an angle GCS of 45 degrees 20 minutes, it was required to show, I say, that, CM being 100,000 parts, PC the semi-major diameter of this ellipse is 105,032 parts, and CS, the semi-minor diameter, 93,410.

In Article 27, it was necessary to demonstrate that PMS is an ellipse centered at C, which touches the straight line MD at M. The angle MCL that CM makes with CL, which is perpendicular to DM, is 6 degrees 40 minutes. Additionally, the semi-minor axis CS forms an angle GCS of 45 degrees 20 minutes with CG, which is parallel to MD. It was required to show that if CM is 100,000 parts, then PC, the semi-major diameter of this ellipse, is 105,032 parts, and CS, the semi-minor diameter, is 93,410.

Let CP and CS be prolonged and meet the tangent DM at D and Z; and from the point of contact M let MN and MO be drawn as perpendiculars to CP and CS. Now because the angles SCP, GCL, are right angles, the [Pg 102]angle PCL will be equal to GCS which was 45 degrees 20 minutes. And deducting the angle LCM, which is 6 degrees 40 minutes, from LCP, which is 45 degrees 20 minutes, there remains MCP, 38 degrees 40 minutes. Considering then CM as a radius of 100,000 parts, MN, the sine of 38 degrees 40 minutes, will be 62,479. And in the right-angled triangle MND, MN will be to ND as the radius of the Tables is to the tangent of 45 degrees 20 minutes (because the angle NMD is equal to DCL, or GCS); that is to say as 100,000 to 101,170: whence results ND 63,210. But NC is 78,079 of the same parts, CM being 100,000, because NC is the sine of the complement of the angle MCP, which was 38 degrees 40 minutes. Then the whole line DC is 141,289; and CP, which is a mean proportional between DC and CN, since MD touches the Ellipse, will be 105,032.

Let CP and CS be extended and meet the tangent DM at points D and Z; from the contact point M, draw perpendiculars MN and MO to CP and CS. Since the angles SCP and GCL are right angles, the angle PCL will equal GCS, which measures 45 degrees 20 minutes. By subtracting the angle LCM, which is 6 degrees 40 minutes, from LCP, which is 45 degrees 20 minutes, we find MCP is 38 degrees 40 minutes. Considering CM as a radius of 100,000 units, the sine of 38 degrees 40 minutes, MN, will be 62,479. In the right triangle MND, MN relates to ND just as the radius of the tables relates to the tangent of 45 degrees 20 minutes (since angle NMD is equal to DCL, or GCS); that is, as 100,000 is to 101,170, resulting in ND being 63,210. However, NC is 78,079 of the same units, with CM being 100,000, because NC represents the sine of the complement of angle MCP, which is 38 degrees 40 minutes. Therefore, the total line DC measures 141,289, and CP, being a mean proportional between DC and CN, will amount to 105,032 since MD touches the ellipse.

Similarly, because the angle OMZ is equal to CDZ, or LCZ, which is 44 degrees 40 minutes, being the complement of GCS, it follows that, as the radius of the Tables is to the tangent of 44 degrees 40 minutes, so will OM 78,079 be to OZ 77,176. But OC is 62,479 of these same parts of which CM is 100,000, because it is equal to MN, the sine of the angle MCP, which is 38 degrees 40 minutes. Then the whole line CZ is 139,655; and CS, which is a mean proportional between CZ and CO will be 93,410.

Similarly, since the angle OMZ is equal to CDZ or LCZ, which is 44 degrees and 40 minutes, being the complement of GCS, it follows that the ratio of the radius of the Tables to the tangent of 44 degrees and 40 minutes is the same as the ratio of OM 78,079 to OZ 77,176. But OC is 62,479 of those same parts, while CM is 100,000, because it is equal to MN, the sine of the angle MCP, which is 38 degrees and 40 minutes. Then the total length of line CZ is 139,655, and CS, which is a mean proportional between CZ and CO, will be 93,410.

[Pg 103]At the same place it was stated that GC was found to be 98,779 parts. To prove this, let PE be drawn in the same figure parallel to DM, and meeting CM at E. In the right-angled triangle CLD the side CL is 99,324 (CM being 100,000), because CL is the sine of the complement of the angle LCM, which is 6 degrees 40 minutes. And since the angle LCD is 45 degrees 20 minutes, being equal to GCS, the side LD is found to be 100,486: whence deducting ML 11,609 there will remain MD 88,877. Now as CD (which was 141,289) is to DM 88,877, so will CP 105,032 be to PE 66,070. But as the rectangle MEH (or rather the difference of the squares on CM and CE) is to the square on MC, so is the square on PE to the square on Cg; then also as the difference of the squares on DC and CP to the square on CD, so also is the square on PE to the square on gC. But DP, CP, and PE are known; hence also one knows GC, which is 98,779.

[Pg 103]At the same location, it was noted that GC was found to be 98,779 parts. To demonstrate this, let PE be drawn in the same figure parallel to DM, and intersecting CM at E. In the right-angled triangle CLD, the side CL is 99,324 (with CM being 100,000), because CL is the sine of the complement of the angle LCM, which is 6 degrees 40 minutes. And since the angle LCD is 45 degrees 20 minutes, equal to GCS, the side LD is determined to be 100,486; thus, by subtracting ML 11,609, MD will be 88,877. Now, as CD (which was 141,289) relates to DM 88,877, so does CP 105,032 relate to PE 66,070. Additionally, as the rectangle MEH (or rather the difference of the squares on CM and CE) relates to the square on MC, so does the square on PE relate to the square on Cg; similarly, as the difference of the squares on DC and CP relates to the square on CD, so does the square on PE relate to the square on gC. However, DP, CP, and PE are known; therefore, GC is also known, which is 98,779.

Lemma which has been supposed.

Assumed lemma.

If a spheroid is touched by a straight line, and also by two or more planes which are parallel to this line, though not parallel to one another, all the points of contact of the line, as well as of the planes, will be in one and the same ellipse made by a plane which passes through the centre of the spheroid.

If a sphere is touched by a straight line and also by two or more planes that are parallel to this line, but not parallel to each other, all the points where the line and the planes touch will be on the same ellipse formed by a plane that goes through the center of the sphere.

Let LED be the spheroid touched by the line BM at the point B, and also by the planes parallel to this line at the points O and A. It is required to demonstrate that the points B, O, and A are in one and the same Ellipse made in the spheroid by a plane which passes through its centre.[Pg 104]

Let LED be the spheroid that the line BM touches at point B, and also intersects with the planes parallel to this line at points O and A. We need to show that points B, O, and A all lie on the same ellipse created in the spheroid by a plane that goes through its center.[Pg 104]

Through the line BM, and through the points O and A, let there be drawn planes parallel to one another, which, in cutting the spheroid make the ellipses LBD, POP, QAQ; which will all be similar and similarly disposed, and will have their centres K, N, R, in one and the same diameter of the spheroid, which will also be the diameter of the ellipse made by the section of the plane that passes through the centre of the spheroid, and which cuts the planes of the three said Ellipses at right angles: for all this is manifest by proposition 15 of the book of Conoids and Spheroids of Archimedes. Further, the two latter planes, which are drawn through the points O and A, will also, by cutting the planes which touch the spheroid in these same points, generate straight lines, as OH and AS, which will, as is easy to see, be parallel to BM; and all three, BM, OH, AS, will touch the Ellipses LBD, POP, QAQ in these points, B, O, A; since they are in the planes of these ellipses, and at the same time in the planes which touch the spheroid. If now from these points B, O, A, there are drawn the straight lines BK, ON, AR, through the centres of the same ellipses, and if through these centres there are drawn also the diameters LD, PP, QQ, parallel to the tangents BM, OH, AS; these will be conjugate to the aforesaid BK, ON, AR. And because the three ellipses are similar and similarly [Pg 105]disposed, and have their diameters LD, PP, QQ parallel, it is certain that their conjugate diameters BK, ON, AR, will also be parallel. And the centres K, N, R being, as has been stated, in one and the same diameter of the spheroid, these parallels BK, ON, AR will necessarily be in one and the same plane, which passes through this diameter of the spheroid, and, in consequence, the points R, O, A are in one and the same ellipse made by the intersection of this plane. Which was to be proved. And it is manifest that the demonstration would be the same if, besides the points O, A, there had been others in which the spheroid had been touched by planes parallel to the straight line BM.

Through the line BM, and through points O and A, we can draw planes that are parallel to each other. These planes intersect the spheroid to create the ellipses LBD, POP, and QAQ; all of which will be similar and arranged in the same way, with their centers K, N, and R located on the same diameter of the spheroid. This diameter will also be that of the ellipse formed by the section of the plane that goes through the center of the spheroid and cuts the three ellipses at right angles. This is evident from proposition 15 in Archimedes' book on Conoids and Spheroids. Additionally, the two planes drawn through points O and A will also generate straight lines, OH and AS, by intersecting the planes that touch the spheroid at those points, and it is clear that these will be parallel to BM. All three lines, BM, OH, and AS, will touch the ellipses LBD, POP, and QAQ at points B, O, and A since they lie in the planes of the ellipses and at the same time in the planes that touch the spheroid. If we draw straight lines BK, ON, and AR from these points B, O, and A through the centers of the ellipses, and also draw diameters LD, PP, and QQ through these centers that are parallel to the tangents BM, OH, and AS; these will be conjugate to the lines BK, ON, and AR. Since the three ellipses are similar and similarly disposed, having their diameters LD, PP, and QQ parallel, it follows that their conjugate diameters BK, ON, and AR will also be parallel. The centers K, N, and R, being on the same diameter of the spheroid, means that the lines BK, ON, and AR must lie in the same plane that contains this diameter of the spheroid. Consequently, the points R, O, and A are in the same ellipse formed by the intersection of this plane. This was to be proved. It is also clear that the demonstration would remain the same if, in addition to points O and A, there had been other points where the spheroid was touched by planes parallel to the straight line BM.


CHAPTER VI

ON THE FIGURES OF THE TRANSPARENT BODIES

Which serve for Refraction and for Reflexion.

A

fter having explained how the properties of reflexion and refraction follow from what we have supposed concerning the nature of light, and of opaque bodies, and of transparent media, I will here set forth a very easy and natural way of deducing, from the same principles, the true figures which serve, either by reflexion or by refraction, to collect or disperse the rays of light, as may be desired. For though I do not see yet that there are means of making use of these figures, so far as relates to Refraction, not only because of the difficulty of shaping the glasses of Telescopes with the requisite[Pg 106] exactitude according to these figures, but also because there exists in refraction itself a property which hinders the perfect concurrence of the rays, as Mr. Newton has very well proved by experiment, I will yet not desist from relating the invention, since it offers itself, so to speak, of itself, and because it further confirms our Theory of refraction, by the agreement which here is found between the refracted ray and the reflected ray. Besides, it may occur that some one in the future will discover in it utilities which at present are not seen.

After explaining how the properties of reflection and refraction stem from our understanding of light, opaque objects, and transparent materials, I will now present a straightforward and natural method for deriving the true shapes that can either focus or spread light rays, depending on what we want. Although I currently don’t believe there are practical ways to use these shapes in relation to refraction, not only due to the difficulty of shaping telescope lenses with the necessary precision based on these forms, but also because refraction itself has a quality that prevents perfect alignment of the rays, as Mr. Newton has effectively demonstrated through experiments, I will still share the invention since it seems to arise naturally and further supports our theory of refraction through the correlation between the refracted ray and the reflected ray. Additionally, it’s possible that someone in the future will find uses for it that we cannot yet see.

To proceed then to these figures, let us suppose first that it is desired to find a surface CDE which shall reassemble at a point B rays coming from another point A; and that the summit of the surface shall be the given point D in the straight line AB. I say that, whether by reflexion or by refraction, it is only necessary to make this surface such that the path of the light from the point A to all points of the curved line CDE, and from these to the point of concurrence (as here the path along the straight lines AC, CB, along AL, LB, and along AD, DB), shall be everywhere traversed in equal times: by which principle the finding of these curves becomes very easy.[Pg 107]

To move on to these figures, let's first assume we want to find a surface CDE that can gather rays coming from a point A at another point B; and that the peak of the surface will be at the given point D on the straight line AB. I propose that, whether through reflection or refraction, it's necessary to shape this surface so that the light travels from point A to all points on the curved line CDE, and then from these points to the point of intersection (as shown by the paths along the straight lines AC, CB, as well as AL, LB, and AD, DB), in equal amounts of time: using this principle makes it quite simple to determine these curves.[Pg 107]

So far as relates to the reflecting surface, since the sum of the lines AC, CB ought to be equal to that of AD, DB, it appears that DCE ought to be an ellipse; and for refraction, the ratio of the velocities of waves of light in the media A and B being supposed to be known, for example that of 3 to 2 (which is the same, as we have shown, as the ratio of the Sines in the refraction), it is only necessary to make DH equal to 3/2 of DB; and having after that described from the centre A some arc FC, cutting DB at F, then describe another from centre B with its semi-diameter BX equal to 2/3 of FH; and the point of intersection of the two arcs will be one of the points required, through which the curve should pass. For this point, having been found in this fashion, it is easy forthwith to demonstrate that the time along AC, CB, will be equal to the time along AD, DB.

As for the reflecting surface, since the total lengths AC and CB should equal the total lengths AD and DB, it seems that DCE should be an ellipse. For refraction, if we assume the ratio of the speeds of light in media A and B is known, for example 3 to 2 (which, as we have shown, is the same as the ratio of the Sines in refraction), we just need to set DH to be 3/2 of DB. After that, draw an arc FC from center A that intersects DB at F, and then draw another from center B with a radius BX equal to 2/3 of FH. The point where the two arcs intersect will be one of the points that the curve should pass through. Once this point is found, it's straightforward to demonstrate that the time taken along AC and CB will equal the time taken along AD and DB.

For assuming that the line AD represents the time which the light takes to traverse this same distance AD in air, it is evident that DH, equal to 3/2 of DB, will represent the time of the light along DB in the medium, because it needs here more time in proportion as its speed is slower. Therefore the whole line AH will represent the time along AD, DB. Similarly the line AC or AF will represent the time along AC; and FH being by construction equal to 3/2 of CB, it will represent the time along CB in the medium; and in consequence the whole line AH will represent also the time along AC, CB. Whence it appears that the [Pg 108]time along AC, CB, is equal to the time along AD, DB. And similarly it can be shown if L and K are other points in the curve CDE, that the times along AL, LB, and along AK, KB, are always represented by the line AH, and therefore equal to the said time along AD, DB.

Assuming that the line AD shows the time it takes for light to travel the same distance AD in air, it’s clear that DH, which is equal to 3/2 of DB, represents the time it takes for light to travel along DB in the medium, since it takes longer where the speed is slower. Thus, the entire line AH represents the time along AD and DB. Likewise, the line AC or AF will represent the time along AC; and since FH is constructed to be equal to 3/2 of CB, it represents the time along CB in the medium. Consequently, the entire line AH also represents the time along AC and CB. This indicates that the [Pg 108]time along AC and CB is equal to the time along AD and DB. Similarly, it can be demonstrated that if L and K are other points on the curve CDE, the times along AL, LB, and along AK, KB, are always represented by the line AH, and therefore equal to the time along AD and DB.

In order to show further that the surfaces, which these curves will generate by revolution, will direct all the rays which reach them from the point A in such wise that they tend towards B, let there be supposed a point K in the curve, farther from D than C is, but such that the straight line AK falls from outside upon the curve which serves for the refraction; and from the centre B let the arc KS be described, cutting BD at S, and the straight line CB at R; and from the centre A describe the arc DN meeting AK at N.

To further demonstrate that the surfaces generated by these curves through revolution will direct all the rays coming from point A towards point B, let's imagine a point K on the curve, which is further from D than C but positioned such that the straight line AK approaches the curve for refraction from outside. From point B, let’s draw arc KS, which intersects line BD at point S and straight line CB at point R. Next, from point A, draw arc DN, which meets line AK at point N.

Since the sums of the times along AK, KB, and along AC, CB are equal, if from the former sum one deducts the time along KB, and if from the other one deducts the time along RB, there will remain the time along AK as equal to the time along the two parts AC, CR. Consequently in the time that the light has come along AK it will also have come along AC and will in addition have made, in the medium from the centre C, a partial spherical wave, having a semi-diameter equal to CR. And this wave will necessarily touch the circumference KS at R, since CB cuts this circumference at right angles. Similarly, having taken any other point L in the curve, one can show that in the same time as the light passes along AL it will also have come along AL and in addition will have made a partial wave, from the centre L, which will touch the same circumference KS. And so with all other points of the curve CDE. Then at the moment that the light reaches K the arc KRS will be the termination [Pg 109]of the movement, which has spread from A through DCK. And thus this same arc will constitute in the medium the propagation of the wave emanating from A; which wave may be represented by the arc DN, or by any other nearer the centre A. But all the pieces of the arc KRS are propagated successively along straight lines which are perpendicular to them, that is to say, which tend to the centre B (for that can be demonstrated in the same way as we have proved above that the pieces of spherical waves are propagated along the straight lines coming from their centre), and these progressions of the pieces of the waves constitute the rays themselves of light. It appears then that all these rays tend here towards the point B.

Since the times taken along AK, KB, and along AC, CB are equal, if you subtract the time along KB from the first sum and the time along RB from the second sum, the time along AK will equal the time along the two parts AC and CR. Therefore, in the time that light travels along AK, it will also travel along AC, and in addition, it will create a partial spherical wave in the medium from the center C, with a radius equal to CR. This wave will necessarily touch the circumference KS at R, since CB intersects this circumference at right angles. Similarly, if we choose any other point L on the curve, we can show that while light travels along AL, it will also travel along AL and will generate a partial wave from the center L, which will also touch the same circumference KS. The same applies to all other points on the curve CDE. Then, when the light reaches K, the arc KRS will mark the end [Pg 109]of the movement that has spread from A through DCK. Thus, this same arc will represent the wave propagation in the medium coming from A; this wave could be illustrated by the arc DN or any other closer to the center A. However, all segments of the arc KRS travel successively along straight lines that are perpendicular to them, which means they head toward the center B (this can be shown in the same way we proved earlier that segments of spherical waves propagate along straight lines emanating from their center), and these movements of the wave segments constitute the rays of light. It seems that all these rays are directed toward point B.

One might also determine the point C, and all the others, in this curve which serves for the refraction, by dividing DA at G in such a way that DG is 2/3 of DA, and describing from the centre B any arc CX which cuts BD at N, and another from the centre A with its semi-diameter AF equal to 3/2 of GX; or rather, having described, as before, the arc CX, it is only necessary to make DF equal to 3/2 of DX, and from-the centre A to strike the arc FC; for these two constructions, as may be easily known, come back to the first one which was shown before. And it is manifest by the last method that this curve is the same that Mr. Des Cartes has given in his Geometry, and which he calls the first of his Ovals.

One can also find point C and all the others on this curve used for refraction by dividing DA at G so that DG is 2/3 of DA, and then drawing arc CX from center B, which intersects BD at N. Additionally, from center A, draw another arc with its semi-diameter AF equal to 3/2 of GX; or, after drawing arc CX as before, just make DF equal to 3/2 of DX and draw arc FC from center A. These two methods, as is clear, will return to the first one we discussed earlier. It is evident from the last method that this curve is the same one that Mr. Des Cartes has presented in his Geometry, which he refers to as the first of his Ovals.

It is only a part of this oval which serves for the refraction, namely, the part DK, ending at K, if AK is the tangent. As to the, other part, Des Cartes has remarked that it could serve for reflexions, if there were some material of a mirror of such a nature that by its [Pg 110]means the force of the rays (or, as we should say, the velocity of the light, which he could not say, since he held that the movement of light was instantaneous) could be augmented in the proportion of 3 to 2. But we have shown that in our way of explaining reflexion, such a thing could not arise from the matter of the mirror, and it is entirely impossible.

Only a part of this oval is used for refraction, specifically the segment DK, which ends at K, if AK is the tangent. Regarding the other part, Des Cartes noted that it could be used for reflection if there were a type of mirror material that could enhance the force of the rays (or, as we would say, the speed of light, which he couldn't say since he believed that light moved instantaneously) in a ratio of 3 to 2. However, we have demonstrated that in our explanation of reflection, this enhancement cannot be produced by the material of the mirror, and it is completely impossible.

From what has been demonstrated about this oval, it will be easy to find the figure which serves to collect to a point incident parallel rays. For by supposing just the same construction, but the point A infinitely distant, giving parallel rays, our oval becomes a true Ellipse, the [Pg 111]construction of which differs in no way from that of the oval, except that FC, which previously was an arc of a circle, is here a straight line, perpendicular to DB. For the wave of light DN, being likewise represented by a straight line, it will be seen that all the points of this wave, travelling as far as the surface KD along lines parallel to DB, will advance subsequently towards the point B, and will arrive there at the same time. As for the Ellipse which served for reflexion, it is evident that it will here become a parabola, since its focus A may be regarded as infinitely distant from the other, B, which is here the focus of the parabola, towards which all the reflexions of rays parallel to AB tend. And the demonstration of these effects is just the same as the preceding.

Based on what we've shown about this oval, it will be straightforward to find the shape that focuses parallel rays to a point. By imagining the same setup, but with point A infinitely far away, giving us parallel rays, our oval transforms into a true ellipse. The [Pg 111]construction of this ellipse is no different from that of the oval, except that FC, which was once a circular arc, is now a straight line, perpendicular to DB. Since the wave of light DN is also represented by a straight line, we can see that all the points of this wave, moving to the surface KD along lines parallel to DB, will subsequently move towards point B and will all arrive at the same time. As for the ellipse used for reflection, it's clear that it will now become a parabola, since its focus A can be treated as infinitely far away from the other focus, B, which is the focus of the parabola where all reflections of rays parallel to AB converge. The demonstration of these effects is exactly the same as before.

But that this curved line CDE which serves for refraction is an Ellipse, and is such that its major diameter is to the distance between its foci as 3 to 2, which is the proportion of the refraction, can be easily found by the calculus of Algebra. For DB, which is given, being called a; its undetermined perpendicular DT being called x; and TC y; FB will be a - y; CB will be sqrt(xx + aa - 2ay + yy). But the nature of the curve is such that 2/3 of TC together with CB is equal to DB, as was stated in the last construction: then the equation will be between (2/3)y + sqrt(xx + aa - 2ay + yy) and a; which being reduced, gives (6/5)ay - yy equal to (9/5)xx; that is to say that having made DO equal to 6/5 of DB, the rectangle DFO is equal to 9/5 of the square on FC. Whence it is seen that DC is an ellipse, of which the axis DO is to the parameter as 9 to 5; and therefore the square on DO is to the square of the distance between the foci as 9 to 9 - 5, that is to say 4; and finally the line DO will be to this distance as 3 to 2.[Pg 112]

But the curved line CDE that is used for refraction is an ellipse, with its major diameter related to the distance between its foci in the ratio of 3 to 2, which corresponds to the proportion of the refraction. This can be easily determined using algebra. Let DB be a, its undetermined perpendicular DT be x, and TC y; then FB will be a - y; and CB will be sqrt(xx + aa - 2ay + yy). The nature of the curve shows that 2/3 of TC plus CB equals DB, as stated in the previous construction. Therefore, the equation will relate (2/3)y + sqrt(xx + aa - 2ay + yy) to a; simplifying gives (6/5)ay - yy equal to (9/5)xx; meaning that when DO is set to 6/5 of DB, the rectangle DFO equals 9/5 of the square on FC. Thus, it follows that DC is an ellipse, where the axis DO is to the parameter as 9 to 5; consequently, the square on DO is to the square of the distance between the foci in the ratio of 9 to 4; and ultimately, the line DO will be to this distance as 3 to 2.[Pg 112]

Again, if one supposes the point B to be infinitely distant, in lieu of our first oval we shall find that CDE is a true Hyperbola; which will make those rays become parallel which come from the point A. And in consequence also those which are parallel within the transparent body will be collected outside at the point A. Now it must be remarked that CX and KS become straight lines perpendicular to BA, because they represent arcs of circles the centre of which is infinitely distant. And the intersection of the perpendicular CX with the arc FC will give the point C, one of those through which the curve ought to pass. And this operates so that all the parts of the wave of light DN, coming to meet the surface KDE, will advance thence along parallels to KS and will arrive at this straight line at the same time; of which the proof is again the same as that which served for the first oval. Besides one finds by a calculation as easy as the preceding one, that CDE is here a hyperbola of which the axis DO [Pg 113]is 4/5 of AD, and the parameter equal to AD. Whence it is easily proved that DO is to the distance between the foci as 3 to 2.

Again, if we assume point B is infinitely far away, instead of our first oval, we find that CDE is a true hyperbola; this means that the rays coming from point A will become parallel. Consequently, the rays that are parallel within the transparent body will converge outside at point A. It should also be noted that CX and KS become straight lines perpendicular to BA because they represent arcs of circles whose center is infinitely distant. The intersection of the perpendicular CX with the arc FC will produce point C, one of the points that the curve should pass through. This means that all parts of the light wave DN that meet the surface KDE will advance along parallel lines to KS and will reach this straight line at the same time; the proof is the same as the one used for the first oval. Additionally, we can easily calculate, just like before, that CDE is a hyperbola where the axis DO [Pg 113] is 4/5 of AD, and the parameter equals AD. From this, it's easy to show that DO is to the distance between the foci as 3 to 2.

These are the two cases in which Conic sections serve for refraction, and are the same which are explained, in his Dioptrique, by Des Cartes, who first found out the use of these lines in relation to refraction, as also that of the Ovals the first of which we have already set forth. The second oval is that which serves for rays that tend to a given point; in which oval, if the apex of the surface which receives the rays is D, it will happen that the other apex will be situated between B and A, or beyond A, according as the ratio of AD to DB is given of greater or lesser value. And in this latter case it is the same as that which Des Cartes calls his 3rd oval.

These are the two situations where conic sections are used for refraction, the same ones that Descartes describes in his Dioptrique. He was the first to uncover the application of these shapes in relation to refraction, along with that of the ovals we have already discussed. The second oval is used for rays that converge to a specific point; in this oval, if the top point of the surface that receives the rays is D, the other top point will be located between B and A, or beyond A, depending on whether the ratio of AD to DB is greater or lesser. In this latter case, it is the same as what Descartes refers to as his 3rd oval.

Now the finding and construction of this second oval is [Pg 114]the same as that of the first, and the demonstration of its effect likewise. But it is worthy of remark that in one case this oval becomes a perfect circle, namely when the ratio of AD to DB is the same as the ratio of the refractions, here as 3 to 2, as I observed a long time ago. The 4th oval, serving only for impossible reflexions, there is no need to set it forth.

Now the discovery and construction of this second oval is [Pg 114]the same as that of the first, along with the demonstration of its effect. However, it's worth noting that in one case, this oval turns into a perfect circle, specifically when the ratio of AD to DB matches the ratio of the refractions, which here is 3 to 2, as I noted long ago. The 4th oval, which only deals with impossible reflections, doesn't require further explanation.

As for the manner in which Mr. Des Cartes discovered these lines, since he has given no explanation of it, nor any one else since that I know of, I will say here, in passing, what it seems to me it must have been. Let it be proposed to find the surface generated by the revolution of the curve KDE, which, receiving the incident rays coming to it from the point A, shall deviate them toward the point B. Then considering this other curve as already known, and that its apex D is in the straight line AB, let us divide it up into an infinitude of small pieces by the points G, C, F; and having drawn from each of these points, straight lines towards A to represent the incident rays, and other straight lines towards B, let there also be described with centre A the arcs GL, CM, FN, DO, cutting the rays that come from A at L, M, N, O; and from the points K, G, C, F, [Pg 115]let there be described the arcs KQ, GR, CS, FT cutting the rays towards B at Q, R, S, T; and let us suppose that the straight line HKZ cuts the curve at K at right-angles.

As for how Mr. Des Cartes discovered these lines, since he hasn’t explained it himself, and no one else has that I know of, I’ll briefly say what I think it must have been. Let’s propose to find the surface created by the revolution of the curve KDE, which, upon receiving the incoming rays from point A, will redirect them toward point B. Now, considering this other curve as already known, with its peak D located on the straight line AB, let’s break it down into an infinite number of small segments at points G, C, and F; and from each of these points, let’s draw straight lines toward A to represent the incoming rays, and other straight lines toward B. Additionally, let’s draw arcs GL, CM, FN, DO, centered at A, which intersect with the rays coming from A at points L, M, N, O; and from points K, G, C, and F, let’s also draw arcs KQ, GR, CS, FT that intersect the rays aimed at B at points Q, R, S, T; and let’s assume that the straight line HKZ intersects the curve at K at a right angle.

Then AK being an incident ray, and KB its refraction within the medium, it needs must be, according to the law of refraction which was known to Mr. Des Cartes, that the sine of the angle ZKA should be to the sine of the angle HKB as 3 to 2, supposing that this is the proportion of the refraction of glass; or rather, that the sine of the angle KGL should have this same ratio to the sine of the angle GKQ, considering KG, GL, KQ as straight lines because of their smallness. But these sines are the lines KL and GQ, if GK is taken as the radius of the circle. Then LK ought to be to GQ as 3 to 2; and in the same ratio MG to CR, NC to FS, OF to DT. Then also the sum of all the antecedents to all the consequents would be as 3 to 2. Now by prolonging the arc DO until it meets AK at X, KX is the sum of the antecedents. And by prolonging the arc KQ till it meets AD at Y, the sum of [Pg 116]the consequents is DY. Then KX ought to be to DY as 3 to 2. Whence it would appear that the curve KDE was of such a nature that having drawn from some point which had been assumed, such as K, the straight lines KA, KB, the excess by which AK surpasses AD should be to the excess of DB over KB, as 3 to 2. For it can similarly be demonstrated, by taking any other point in the curve, such as G, that the excess of AG over AD, namely VG, is to the excess of BD over DG, namely DP, in this same ratio of 3 to 2. And following this principle Mr. Des Cartes constructed these curves in his Geometric; and he easily recognized that in the case of parallel rays, these curves became Hyperbolas and Ellipses.

Then AK is an incoming ray, and KB is its refraction within the medium. According to the law of refraction that Mr. Descartes knew, the sine of angle ZKA must be to the sine of angle HKB as 3 to 2, assuming this is the ratio of glass refraction. Alternatively, the sine of angle KGL should have the same ratio to the sine of angle GKQ, considering KG, GL, and KQ as straight lines due to their small size. These sines correspond to the lines KL and GQ if GK is taken as the radius of the circle. Thus, LK should relate to GQ as 3 to 2; the same applies to MG to CR, NC to FS, and OF to DT. Furthermore, the total of all the antecedents compared to all the consequents would also be in the ratio of 3 to 2. By extending arc DO until it meets AK at X, KX represents the total of the antecedents. Similarly, by extending arc KQ until it intersects AD at Y, DY is the total of the consequents. Therefore, KX should relate to DY as 3 to 2. This suggests that the curve KDE behaves in such a way that from an assumed point like K, the straight lines KA and KB indicate that the excess of AK over AD must compare to the excess of DB over KB in a ratio of 3 to 2. This can also be demonstrated by choosing any other point on the curve, such as G, where the excess of AG over AD, which is VG, compares to the excess of BD over DG, which is DP, in the same 3 to 2 ratio. Following this principle, Mr. Descartes constructed these curves in his Geometry; he recognized that when parallel rays were involved, these curves became hyperbolas and ellipses.

Let us now return to our method and let us see how it leads without difficulty to the finding of the curves which one side of the glass requires when the other side is of a given figure; a figure not only plane or spherical, or made by one of the conic sections (which is the restriction with which Des Cartes proposed this problem, leaving the solution to those who should come after him) but generally any figure whatever: that is to say, one made by the revolution of any given curved line to which one must merely know how to draw straight lines as tangents.

Let’s go back to our method and see how it easily helps us find the curves needed on one side of the glass when the other side has a specific shape. This shape can be anything, not just flat, spherical, or created by one of the conic sections (which is the limitation Des Cartes set for this problem, leaving the solution for future thinkers). In fact, it can be any figure at all: essentially, one created by rotating any given curved line, as long as we know how to draw straight lines as tangents.

Let the given figure be that made by the revolution of some curve such as AK about the axis AV, and that this side of the glass receives rays coming from the point L. Furthermore, let the thickness AB of the middle of the glass be given, and the point F at which one desires the rays to be all perfectly reunited, whatever be the first refraction occurring at the surface AK.

Let the figure created by rotating a curve like AK around the axis AV be established, and let this side of the glass receive rays coming from point L. Additionally, let the thickness AB of the center of the glass be specified, along with point F, where we want all the rays to perfectly converge, regardless of the first refraction happening at the surface AK.

I say that for this the sole requirement is that the outline BDK which constitutes the other surface shall be [Pg 117]such that the path of the light from the point L to the surface AK, and from thence to the surface BDK, and from thence to the point F, shall be traversed everywhere in equal times, and in each case in a time equal to that which the light employs, to pass along the straight line LF of which the part AB is within the glass.

I say that the only requirement for this is that the outline BDK, which makes up the other surface, should be [Pg 117]so that the light travels from point L to surface AK, then from there to surface BDK, and finally to point F, taking the same amount of time everywhere. In each case, the time should be equal to the time it takes for light to travel along the straight line LF, where part AB is inside the glass.

Let LG be a ray falling on the arc AK. Its refraction GV will be given by means of the tangent which will be drawn at the point G. Now in GV the point D must be found such that FD together with 3/2 of DG and the straight line [Pg 118]GL, may be equal to FB together with 3/2 of BA and the straight line AL; which, as is clear, make up a given length. Or rather, by deducting from each the length of LG, which is also given, it will merely be needful to adjust FD up to the straight line VG in such a way that FD together with 3/2 of DG is equal to a given straight line, which is a quite easy plane problem: and the point D will be one of those through which the curve BDK ought to pass. And similarly, having drawn another ray LM, and found its refraction MO, the point N will be found in this line, and so on as many times as one desires.

Let LG be a ray hitting the arc AK. Its refraction GV will be determined by the tangent drawn at point G. Now in GV, we need to find point D such that FD, along with 3/2 of DG and the straight line [Pg 118]GL, is equal to FB plus 3/2 of BA and the straight line AL; which, as is clear, adds up to a specific length. Alternatively, by subtracting LG's length, which is also specified, we only need to adjust FD to align with straight line VG so that FD plus 3/2 of DG equals a given straight line, which is a relatively simple plane problem: and point D will be one of those through which the curve BDK should pass. Similarly, after drawing another ray LM and determining its refraction MO, point N can be found on this line, and this can be repeated as many times as desired.

To demonstrate the effect of the curve, let there be described about the centre L the circular arc AH, cutting LG at H; and about the centre F the arc BP; and in AB let AS be taken equal to 2/3 of HG; and SE equal to GD. Then considering AH as a wave of light emanating from the point L, it is certain that during the time in which its piece H arrives at G the piece A will have advanced within the transparent body only along AS; for I suppose, as above, the proportion of the refraction to be as 3 to 2. Now we know that the piece of wave which is incident on G, advances thence along the line GD, since GV is the refraction of the ray LG. Then during the time that this piece of wave has taken from G to D, the other piece which was at S has reached E, since GD, SE are equal. But while the latter will advance from E to B, the piece of wave which was at D will have spread into the air its partial wave, the semi-diameter of which, DC (supposing this wave to cut the line DF at C), will be 3/2 of EB, since the velocity of light outside the medium is to that inside as 3 to 2. Now it is easy to show that this wave will touch the arc BP at this point C. For since, by construction, FD + [Pg 119]3/2 DG + GL are equal to FB + 3/2 BA + AL; on deducting the equals LH, LA, there will remain FD + 3/2 DG + GH equal to FB + 3/2 BA. And, again, deducting from one side GH, and from the other side 3/2 of AS, which are equal, there will remain FD with 3/2 DG equal to FB with 3/2 of BS. But 3/2 of DG are equal to 3/2 of ES; then FD is equal to FB with 3/2 of BE. But DC was equal to 3/2 of EB; then deducting these equal lengths from one side and from the other, there will remain CF equal to FB. And thus it appears that the wave, the semi-diameter of which is DC, touches the arc BP at the moment when the light coming from the point L has arrived at B along the line LB. It can be demonstrated similarly that at this same moment the light that has come along any other ray, such as LM, MN, will have propagated the movement which is terminated at the arc BP. Whence it follows, as has been often said, that the propagation of the wave AH, after it has passed through the thickness of the glass, will be the spherical wave BP, all the pieces of which ought to advance along straight lines, which are the rays of light, to the centre F. Which was to be proved. Similarly these curved lines can be found in all the cases which can be proposed, as will be sufficiently shown by one or two examples which I will add.

To show the effect of the curve, let's describe the circular arc AH around the center L, cutting LG at H; and the arc BP around the center F. In AB, let AS be equal to 2/3 of HG, and SE equal to GD. Now, considering AH as a wave of light coming from point L, it’s clear that by the time piece H reaches G, piece A will have moved only along AS within the transparent body, assuming, as before, the ratio of refraction is 3 to 2. We know that the wave piece incident on G moves along the line GD, because GV represents the refraction of the ray LG. While this wave piece travels from G to D, the piece at S has reached E since GD and SE are equal. However, as the latter moves from E to B, the wave piece at D will have expanded into the air its partial wave, the semi-diameter of which, DC (assuming this wave intersects line DF at C), will be 3/2 of EB since the speed of light in air compared to the medium is 3 to 2. It’s straightforward to show that this wave will touch the arc BP at point C. Because, as constructed, FD + [Pg 119]3/2 DG + GL equals FB + 3/2 BA + AL; after subtracting the equal lengths LH, LA, we get FD + 3/2 DG + GH equal to FB + 3/2 BA. Next, if we subtract GH from one side and 3/2 of AS from the other, which are equal, we’ll have FD + 3/2 DG equal to FB + 3/2 of BS. But 3/2 of DG equals 3/2 of ES; therefore, FD equals FB plus 3/2 of BE. Since DC equals 3/2 of EB, when we subtract these equal lengths from both sides, we find CF equal to FB. Thus, it's clear that the wave, with a semi-diameter of DC, touches the arc BP at the moment the light from point L arrives at B along the line LB. It can similarly be demonstrated that at this moment, light traveling along any other ray, like LM or MN, will have produced the movement that meets the arc BP. This leads to the conclusion, as often stated, that the propagation of wave AH after passing through the thickness of the glass will become the spherical wave BP, with all parts of it moving in straight lines, which are the rays of light, toward the center F. This was to be proved. Likewise, we can find these curved lines in all proposed cases, as will be sufficiently demonstrated by one or two examples I will add.

Let there be given the surface of the glass AK, made by the revolution about the axis BA of the line AK, which may be straight or curved. Let there be also given in the axis the point L and the thickness BA of the glass; and let it be required to find the other surface KDB, which receiving rays that are parallel to AB will direct them in such wise that after being again refracted at the given surface AK they will all be reassembled at the point L.

Let’s consider the surface of glass AK, created by rotating the line AK around the axis BA, which can be straight or curved. Also, let’s include point L on the axis and the thickness BA of the glass. The goal is to find the other surface KDB, which will take rays parallel to AB and direct them in such a way that after they pass through the given surface AK, they will all converge at point L.

From the point L let there be drawn to some point of [Pg 120]the given line AK the straight line LG, which, being considered as a ray of light, its refraction GD will then be found. And this line being then prolonged at one side or the other will meet the straight line BL, as here at V. Let there then be erected on AB the perpendicular BC, which will represent a wave of light coming from the infinitely distant point F, since we have supposed the rays to be parallel. Then all the parts of this wave BC must arrive at the same time at the point L; or rather all the parts of a wave emanating from the point L must arrive at the same time at the straight line BC. And for that, it is necessary to find in the line VGD the point D such that having drawn DC parallel to AB, the sum of CD, plus 3/2 of DG, plus GL may be equal to 3/2 of AB, plus AL: or rather, on deducting from both sides GL, which is given, CD plus 3/2 of DG must be equal to a given length; which is a still easier problem than the preceding construction. The point D thus found will be one of those through which the curve ought to pass; and the proof will be the same as before. And by this it will be proved that the waves which come from the point L, after having passed through the glass KAKB, will take [Pg 121]the form of straight lines, as BC; which is the same thing as saying that the rays will become parallel. Whence it follows reciprocally that parallel rays falling on the surface KDB will be reassembled at the point L.

From point L, draw a line to some point of [Pg 120] the given line AK, called the straight line LG, which can be thought of as a ray of light; its refraction GD can then be determined. This line, extended on either side, will intersect the straight line BL at point V. Next, erect a perpendicular BC on AB, representing a wave of light coming from the infinitely distant point F, since we are assuming that the rays are parallel. All parts of this wave BC must reach point L simultaneously; or, more accurately, all parts of a wave starting from point L must arrive at the straight line BC together. To achieve this, we need to find point D on line VGD such that if we draw DC parallel to AB, the sum of CD, plus 3/2 of DG, plus GL equals 3/2 of AB, plus AL. Alternatively, if we deduct GL from both sides (since it's a given length), then CD plus 3/2 of DG must equal a specified length, which is a simpler problem than the previous construction. The point D we find will be one of those points through which the curve should pass, and the proof will be consistent with what we've done before. Thus, it will be shown that the waves coming from point L, after passing through glass KAKB, will form straight lines, like BC, which means that the rays will become parallel. Consequently, we can conclude that parallel rays striking the surface KDB will converge at point L.

Again, let there be given the surface AK, of any desired form, generated by revolution about the axis AB, and let the thickness of the glass at the middle be AB. Also let the point L be given in the axis behind the glass; and let it be supposed that the rays which fall on the surface AK tend to this point, and that it is required to find the surface BD, which on their emergence from the glass turns them as if they came from the point F in front of the glass.

Again, let’s consider the surface AK, of any shape we want, created by rotating around the axis AB, and let the thickness of the glass in the middle be AB. Also, let point L be placed on the axis behind the glass; and let’s assume that the rays hitting the surface AK are directed towards this point, and we need to find the surface BD, which, upon exiting the glass, causes the rays to appear as if they originated from point F, in front of the glass.

Having taken any point G in the line AK, and drawing the straight line IGL, its part GI will represent one of the incident rays, the refraction of which, GV, will then be found: and it is in this line that we must find the point D, one of those through which the curve DG ought to pass. Let us suppose that it has been found: and about L as centre let there be described GT, the arc of a circle cutting the straight line AB at T, in case the distance LG is greater than LA; for otherwise the arc AH must be described about the same centre, cutting the straight line LG at H. This arc GT (or AH, in the other case) will represent an incident wave of light, the rays of which [Pg 122]tend towards L. Similarly, about the centre F let there be described the circular arc DQ, which will represent a wave emanating from the point F.

Having chosen any point G on the line AK and drawing the straight line IGL, the segment GI will represent one of the incoming rays, the refraction of which, GV, will then be determined. It is along this line that we need to find point D, one of the points through which the curve DG should pass. Let’s assume we’ve found it: and around point L as the center, draw the arc GT of a circle that intersects the straight line AB at T, in case the distance LG is greater than LA; otherwise, the arc AH should be drawn around the same center, intersecting the straight line LG at H. This arc GT (or AH, in the other scenario) will represent an incoming wave of light, the rays of which [Pg 122]move towards L. Similarly, around point F, draw the circular arc DQ, which will represent a wave originating from point F.

Then the wave TG, after having passed through the glass, must form the wave QD; and for this I observe that the time taken by the light along GD in the glass must be equal to that taken along the three, TA, AB, and BQ, of which AB alone is within the glass. Or rather, having taken AS equal to 2/3 of AT, I observe that 3/2 of GD ought to be equal to 3/2 of SB, plus BQ; and, deducting both of them from FD or FQ, that FD less 3/2 of GD ought to be equal to FB less 3/2 of SB. And this last difference is a given length: and all that is required is to draw the straight line FD from the given point F to meet VG so that it may be thus. Which is a problem quite similar to that which served for the first of these constructions, where FD plus 3/2 of GD had to be equal to a given length.

Then the wave TG, after passing through the glass, has to create the wave QD; and for this, I notice that the time taken by the light along GD in the glass must equal the time taken along the three segments, TA, AB, and BQ, where only AB is inside the glass. Or rather, if I make AS equal to 2/3 of AT, I see that 3/2 of GD should equal 3/2 of SB plus BQ; and by subtracting both from FD or FQ, FD minus 3/2 of GD should equal FB minus 3/2 of SB. This last difference is a specific length: all that remains is to draw the straight line FD from the given point F to intersect VG as specified. This is a problem very similar to the one used for the first of these constructions, where FD plus 3/2 of GD had to equal a specific length.

In the demonstration it is to be observed that, since the arc BC falls within the glass, there must be conceived an arc RX, concentric with it and on the other side of QD. Then after it shall have been shown that the piece G of the wave GT arrives at D at the same time that the piece T arrives at Q, which is easily deduced from the construction, it will be evident as a consequence that the partial wave generated at the point D will touch the arc RX at the moment when the piece Q shall have come to R, and that thus this arc will at the same moment be the termination of the movement that comes from the wave TG; whence all the rest may be concluded.

In the demonstration, it should be noted that since the arc BC is inside the glass, we need to think of an arc RX that is concentric with it but on the other side of QD. After it has been shown that the wave piece G from GT reaches D at the same time as piece T arrives at Q—which is easy to deduce from the setup—it will be clear that the partial wave created at point D will touch the arc RX just as piece Q reaches R. Therefore, this arc will simultaneously mark the end of the movement coming from the wave TG, allowing us to draw the rest of the conclusions.

Having shown the method of finding these curved lines which serve for the perfect concurrence of the rays, [Pg 123]there remains to be explained a notable thing touching the uncoordinated refraction of spherical, plane, and other surfaces: an effect which if ignored might cause some doubt concerning what we have several times said, that rays of light are straight lines which intersect at right angles the waves which travel along them.

Having demonstrated how to find these curved lines that allow the rays to align perfectly, [Pg 123]we still need to discuss an important aspect regarding the uncoordinated refraction of spherical, plane, and other surfaces: an effect that, if overlooked, could create confusion about what we've repeatedly stated, which is that rays of light are straight lines that intersect at right angles to the waves traveling along them.

For in the case of rays which, for example, fall parallel upon a spherical surface AFE, intersecting one another, after refraction, at different points, as this figure represents; what can the waves of light be, in this transparent body, which are cut at right angles by the converging rays? For they can not be spherical. And what will these waves become after the said rays begin to intersect one another? It will be seen in the solution of this difficulty that something very remarkable comes to pass herein, and that the waves do not cease to persist though they do not continue entire, as when they cross the glasses designed according to the construction we have seen.

For rays that fall parallel onto a spherical surface AFE and cross each other after refraction at different points, as shown in this figure, what can the light waves be in this transparent medium that are intersected by the converging rays at right angles? They can't be spherical. And what happens to these waves once the rays start to intersect? It will become clear in solving this issue that something quite remarkable occurs here, and that the waves continue to exist even though they don't remain whole, similar to how they interact with the lenses designed according to the construction we've observed.

[Pg 124]According to what has been shown above, the straight line AD, which has been drawn at the summit of the sphere, at right angles to the axis parallel to which the rays come, represents the wave of light; and in the time taken by its piece D to reach the spherical surface AGE at E, its other parts will have met the same surface at F, G, H, etc., and will have also formed spherical partial waves of which these points are the centres. And the surface EK which all those waves will touch, will be the continuation of the wave AD in the sphere at the moment when the piece D has reached E. Now the line EK is not an arc of a circle, but is a curved line formed as the evolute of another curve ENC, which touches all the rays HL, GM, FO, etc., that are the refractions of the parallel rays, if we imagine laid over the convexity ENC a thread which in unwinding describes at its end E the said curve EK. For, supposing that this curve has been thus described, we will show that the said waves formed from the centres F, G, H, etc., will all touch it.

[Pg 124]Based on what we've seen, the straight line AD drawn at the top of the sphere, perpendicular to the axis that the rays come in along, represents the light wave. While point D takes time to reach the surface of the sphere at point E, its other parts will have hit the same surface at points F, G, H, etc., and will have also created spherical partial waves with these points as their centers. The surface EK, which all these waves will touch, will be the continuation of the wave AD within the sphere at the moment when point D reaches E. Now, the line EK isn’t an arc of a circle; it’s a curved line created as the evolute of another curve ENC, which touches all the rays HL, GM, FO, etc., that are the refractions of the parallel rays. If we imagine laying a thread over the curve ENC that unwinds and traces the curve EK at its end E. Assuming that this curve has been traced this way, we will demonstrate that the waves formed from the centers F, G, H, etc., will all touch it.

It is certain that the curve EK and all the others described by the evolution of the curve ENC, with different lengths of thread, will cut all the rays HL, GM, FO, etc., at right angles, and in such wise that the parts of them intercepted between two such curves will all be equal; for this follows from what has been demonstrated in our treatise de Motu Pendulorum. Now imagining the incident rays as being infinitely near to one another, if we consider two of them, as RG, TF, and draw GQ perpendicular to RG, and if we suppose the curve FS which intersects GM at P to have been described by evolution from the curve NC, beginning at F, as far as which the thread is supposed to extend, we may assume the small piece FP as a straight line perpendicular [Pg 125]to the ray GM, and similarly the arc GF as a straight line. But GM being the refraction of the ray RG, and FP being perpendicular to it, QF must be to GP as 3 to 2, that is to say in the proportion of the refraction; as was shown above in explaining the discovery of Des Cartes. And the same thing occurs in all the small arcs GH, HA, etc., namely that in the quadrilaterals which enclose them the side parallel to the axis is to the opposite side as 3 to 2. Then also as 3 to 2 will the sum of the one set be to the sum of the other; that is to say, TF to AS, and DE to AK, and BE to SK or DV, supposing V to be the intersection of the curve EK and the ray FO. But, making FB perpendicular to DE, the ratio of 3 to 2 is also that of BE to the semi-diameter of the spherical wave which emanated from the point F while the light outside the transparent body traversed the space BE. Then it appears that this wave will intersect the ray FM at the same point V where it is intersected at right angles by the curve EK, and consequently that the wave will touch this curve. In the same way it can be proved that the same will apply to all the other waves above mentioned, originating at the points G, H, etc.; to wit, that they will touch the curve EK at the moment when the piece D of the wave ED shall have reached E.

It’s clear that the curve EK and all the others described by the evolution of the curve ENC, with different thread lengths, will intersect all the rays HL, GM, FO, etc., at right angles, in such a way that the segments of these rays between two such curves will all be equal; this follows from what we've demonstrated in our work de Motu Pendulorum. If we visualize the incoming rays as being infinitely close to each other and take two of them, like RG and TF, and draw GQ perpendicular to RG, and we assume the curve FS intersects GM at point P and has evolved from the curve NC, starting at F, where the thread is assumed to extend, we can treat the small segment FP as a straight line perpendicular [Pg 125]to ray GM, and similarly consider the arc GF as a straight line. Since GM is the refraction of ray RG and FP is perpendicular to it, QF must relate to GP in a 3 to 2 ratio, which reflects the proportion of refraction; as was previously demonstrated when explaining Des Cartes' findings. The same applies to all the small arcs GH, HA, etc., meaning that in the quadrilaterals surrounding them, the side parallel to the axis is to the opposite side as 3 to 2. Furthermore, the total of one set will also maintain this 3 to 2 ratio with the other; specifically, TF to AS, DE to AK, and BE to SK or DV, assuming V is where the curve EK intersects ray FO. By making FB perpendicular to DE, the ratio of 3 to 2 also applies to BE compared to the semi-diameter of the spherical wave originating from point F while the light outside the transparent object passes through the space BE. Consequently, this wave will intersect ray FM at point V, which is where it is intersected at right angles by curve EK, meaning that the wave will touch this curve. Similarly, it can be shown that all the previously mentioned waves originating from points G, H, etc., will touch curve EK at the moment when segment D of wave ED reaches point E.

Now to say what these waves become after the rays have begun to cross one another: it is that from thence they fold back and are composed of two contiguous parts, one being a curve formed as evolute of the curve ENC in one sense, and the other as evolute of the same curve in the opposite sense. Thus the wave KE, while advancing toward the meeting place becomes abc, whereof the part ab is made by the evolute bC, a portion of the curve [Pg 126]ENC, while the end C remains attached; and the part bc by the evolute of the portion bE while the end E remains attached. Consequently the same wave becomes def, then ghk, and finally CY, from whence it subsequently spreads without any fold, but always along curved lines which are evolutes of the curve ENC, increased by some straight line at the end C.

Now to explain what these waves become after the rays start to cross each other: they fold back and consist of two adjacent parts, one being a curve formed as the evolute of the curve ENC in one direction, and the other as the evolute of the same curve in the opposite direction. So the wave KE, while moving toward the meeting point, becomes abc, where the part ab is created by the evolute bC, a segment of the curve [Pg 126]ENC, while the end C stays attached; and the part bc is made by the evolute of the segment bE while the end E remains attached. As a result, the same wave turns into def, then ghk, and finally CY, from which it later spreads out without any folds, but always along curved lines that are evolutes of the curve ENC, extended by some straight line at the end C.

There is even, in this curve, a part EN which is straight, N being the point where the perpendicular from the centre X of the sphere falls upon the refraction of the ray DE, which I now suppose to touch the sphere. The folding of the waves of light begins from the point N up to the end of the curve C, which point is formed by taking AC to CX in the proportion of the refraction, as here 3 to 2.

There is even a straight section EN in this curve, with N being the point where the perpendicular from the center X of the sphere intersects the refraction of the ray DE, which I now assume touches the sphere. The bending of the light waves starts from point N and continues to the end of the curve C, which is determined by taking AC to CX in the ratio of the refraction, here 3 to 2.

As many other points as may be desired in the curve NC are found by a Theorem which Mr. Barrow has demonstrated in section 12 of his Lectiones Opticae, though for another purpose. And it is to be noted that a straight line equal in length to this curve can be given. For since it together with the line NE is equal to the line CK, which is known, since DE is to AK in the proportion of the refraction, it appears that by deducting EN from CK the remainder will be equal to the curve NC.

As many other points as needed on the curve NC can be found using a theorem that Mr. Barrow proved in section 12 of his Lectiones Opticae, even though it was for a different purpose. It's also important to note that a straight line equal in length to this curve can be created. Since this line, along with line NE, is equal to line CK, which is known, and because DE is proportional to AK based on the refraction, it turns out that by subtracting EN from CK, the result will be equal to the curve NC.

Similarly the waves that are folded back in reflexion by a concave spherical mirror can be found. Let ABC be the section, through the axis, of a hollow hemisphere, the centre of which is D, its axis being DB, parallel to which I suppose the rays of light to come. All the reflexions of those rays which fall upon the quarter-circle AB will touch a curved line AFE, of which line the end E is at the focus of the hemisphere, that is to say, at the point which divides the semi-diameter BD into two equal parts. [Pg 127]The points through which this curve ought to pass are found by taking, beyond A, some arc AO, and making the arc OP double the length of it; then dividing the chord OP at F in such wise that the part FP is three times the part FO; for then F is one of the required points.

Similarly, the waves that bounce back in reflection from a concave spherical mirror can be observed. Let ABC represent the section through the axis of a hollow hemisphere, with the center at D and its axis being DB, which I assume the rays of light will come parallel to. All the reflections of those rays that strike the quarter-circle AB will meet a curved line AFE, with the endpoint E located at the focus of the hemisphere, meaning at the point that splits the semi-diameter BD into two equal parts. [Pg 127]The points through which this curve should pass can be found by taking an arc AO beyond A and making the arc OP twice its length; then divide the chord OP at F so that the segment FP is three times the length of segment FO; this way, F becomes one of the necessary points.

And as the parallel rays are merely perpendiculars to the waves which fall on the concave surface, which waves are parallel to AD, it will be found that as they come successively to encounter the surface AB, they form on reflexion folded waves composed of two curves which originate from two opposite evolutions of the parts of the curve AFE. So, taking AD as an incident wave, when the part AG shall have met the surface AI, that is to say when the piece G shall have reached I, it will be the curves HF, FI, generated as evolutes of the curves FA, FE, both beginning at F, which together constitute the propagation of the part AG. And a little afterwards, when the part AK has met the surface AM, the piece K having come to M, then the curves LN, NM, will together constitute the propagation of that part. And thus this folded wave will continue to advance until the point N has reached the focus E. The curve AFE can be seen in smoke, or in flying dust, when a concave mirror is held opposite the sun. And it should be known that it is none other than that curve which is described [Pg 128]by the point E on the circumference of the circle EB, when that circle is made to roll within another whose semi-diameter is ED and whose centre is D. So that it is a kind of Cycloid, of which, however, the points can be found geometrically.

And since the parallel rays are just perpendicular lines to the waves hitting the concave surface, with those waves running parallel to AD, it turns out that as they sequentially hit the surface AB, they create reflected, folded waves made up of two curves that stem from two opposite evolutions of the curve AFE. So, taking AD as an incoming wave, when the part AG meets the surface AI, meaning when point G reaches I, the curves HF and FI, generated as evolutes from the curves FA and FE, both starting at F, will together make up the movement of part AG. A little later, when the part AK meets the surface AM, with point K arriving at M, the curves LN and NM will collectively represent the movement of that part. This folded wave will keep moving until point N reaches the focus E. The curve AFE can be observed in smoke or flying dust when a concave mirror is held facing the sun. It’s important to note that this is exactly the curve that is described [Pg 128]by the point E on the edge of the circle EB when that circle rolls inside another circle with a semi-diameter of ED and center at D. So, it's a type of Cycloid, whose points can indeed be found geometrically.

Its length is exactly equal to 3/4 of the diameter of the sphere, as can be found and demonstrated by means of these waves, nearly in the same way as the mensuration of the preceding curve; though it may also be demonstrated in other ways, which I omit as outside the subject. The area AOBEFA, comprised between the arc of the quarter-circle, the straight line BE, and the curve EFA, is equal to the fourth part of the quadrant DAB.

Its length is exactly 3/4 of the diameter of the sphere, which can be shown using these waves, similar to how we measured the previous curve; although there are other ways to demonstrate it, which I will leave out as they are not relevant to the topic. The area AOBEFA, enclosed by the arc of the quarter-circle, the straight line BE, and the curve EFA, is equal to one-fourth of the quadrant DAB.

END.


[Pg 129]

INDEX

Archimedes, 104.
Atmospheric refraction, 45.

Barrow, Isaac, 126.
Bartholinus, Erasmus, 53, 54, 57, 60, 97, 99.
Boyle, Hon. Robert, 11.

Cassini, Jacques, iii.
Caustic Curves, 123.
Crystals, see Iceland Crystal, Rock Crystal.
Crystals, configuration of, 95.

Descartes, Rénê, 3, 5, 7, 14, 22, 42, 43, 109, 113.
Double Refraction, discovery of, 54, 81, 93.

Elasticity, 12, 14.
Ether, the, or Ethereal matter, 11, 14, 16, 28.
Extraordinary refraction, 55, 56.

Fermat, principle of, 42.
Figures of transparent bodies, 105.

Hooke, Robert, 20.

Iceland Crystal, 2, 52 sqq.
Iceland Crystal, Cutting and Polishing of, 91, 92, 98.

Leibnitz, G.W., vi.
Light, nature of, 3.
Light, velocity of, 4, 15.

Molecular texture of bodies, 27, 95.

Newton, Sir Isaac, vi, 106.

Opacity, 34.
Ovals, Cartesian, 107, 113.

Pardies, Rev. Father, 20.
Rays, definition of, 38, 49.
Reflexion, 22.
Refraction, 28, 34.
Rock Crystal, 54, 57, 62, 95.
Römer, Olaf, v, 7.
Roughness of surfaces, 27.

Sines, law of, 1, 35, 38, 43.
Spheres, elasticity of, 15.
Spheroidal waves in crystals, 63.
Spheroids, lemma about, 103.
Sound, speed of, 7, 10, 12.

Telescopes, lenses for, 62, 105.
Torricelli's experiment, 12, 30.
Transparency, explanation of, 28, 31, 32.

Waves, no regular succession of, 17.
Waves, principle of wave envelopes, 19, 24.
Waves, principle of elementary wave fronts, 19.
Waves, propagation of light as, 16, 63.

Archimedes, 104.
Atmospheric refraction, 45.

Barrow, Isaac, 126.
Bartholinus, Erasmus, 53, 54, 57, 60, 97, 99.
Boyle, Hon. Robert, 11.

Cassini, Jacques, iii.
Caustic Curves, 123.
Crystals, see Iceland Crystal, Rock Crystal.
Crystals, configuration of, 95.

Descartes, Rénê, 3, 5, 7, 14, 22, 42, 43, 109, 113.
Double Refraction, discovery of, 54, 81, 93.

Elasticity, 12, 14.
Ether, the, or Ethereal matter, 11, 14, 16, 28.
Extraordinary refraction, 55, 56.

Fermat, principle of, 42.
Figures of transparent bodies, 105.

Hooke, Robert, 20.

Iceland Crystal, 2, 52 sqq.
Iceland Crystal, Cutting and Polishing of, 91, 92, 98.

Leibnitz, G.W., vi.
Light, nature of, 3.
Light, velocity of, 4, 15.

Molecular texture of bodies, 27, 95.

Newton, Sir Isaac, vi, 106.

Opacity, 34.
Ovals, Cartesian, 107, 113.

Pardies, Rev. Father, 20.
Rays, definition of, 38, 49.
Reflexion, 22.
Refraction, 28, 34.
Rock Crystal, 54, 57, 62, 95.
Römer, Olaf, v, 7.
Roughness of surfaces, 27.

Sines, law of, 1, 35, 38, 43.
Spheres, elasticity of, 15.
Spheroidal waves in crystals, 63.
Spheroids, lemma about, 103.
Sound, speed of, 7, 10, 12.

Telescopes, lenses for, 62, 105.
Torricelli's experiment, 12, 30.
Transparency, explanation of, 28, 31, 32.

Waves, no regular succession of, 17.
Waves, principle of wave envelopes, 19, 24.
Waves, principle of elementary wave fronts, 19.
Waves, propagation of light as, 16, 63.

 


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