# An Elementary Treatise on Optics/Chapter 7

CHAP. VII.
OF REFRACTION AT PLANE SURFACES.
53.We must here recall to the Student's recollection, the details in the introduction about the manner or law of refraction, which is, that when light enters a transparent medium, its course is bent or broken in such a manner, that the sine of the angle of incidence, bears to the sine of the angle of refraction, a certain ratio which is the same however the angle of incidence be varied. The full and correct statement of the case is, that when a ray of light passes from one medium into another, as from air into water or glass, the refraction above described takes place at their common surface.
54.In either of the instances just mentioned, the angle of refraction is less than that of incidence. If the passage of the light were from water or glass into air, the contrary would be the case, and in general it is observed that when light passes from a rarer into a denser medium, the ray is brought nearer the perpendicular to the surface bounding them: when from a denser into a rarer, the converse.
​55.There is a remarkable exception or modification to this rule in the case of combustible substances, (in which the diamond is included) which always refract much more than other substances of like densities, that is, in cases when the inclination of a ray to the perpendicular is diminished by the refraction, it becomes less for a combustible, than for an incombustible substance, but when the angle of refraction is greater than the angle of incidence, it is less increased in passing into a combustible, than into another substance of equal density.
56.Between media of equal refracting powers there is no refraction, and in general the superiority of action of one medium over another is well expressed by the invariable ratio of the sines of incidence and refraction, when a ray passes from one into the other.
57.We may here take occasion to observe that since when a ray of light passes out of a denser into a rarer medium, that is, out of one of a stronger into one of a weaker refracting power, the angle of refraction is greater than that of incidence, there is some angle of incidence for which the angle of refraction is a right angle (v. Fig. 55.) Past that point there can be no refraction, for though we might fancy an angle of incidence greater than a right angle, there is no angle whose sine is greater than the radius.
For instance, when a ray of light passes out of glass into air, the ratio of the sines of the angles of incidence and refraction is about 2 to 3. Here then we have sinφ=2/3sinφ′, or sinφ′=3/2sinφ, and since sinφ′ can not exceed 1, sinφ cannot be greater than 2/3, or φ greater than 41° 49′.
The fact is, that when the angle of incidence in the denser medium exceeds the proper limit, the light is reflected instead of being refracted, as may easily be seen by holding a glass of water above one's eye, when it will be observed that any rays of light coming from below so as to make with the surface an angle of less than 41° 24′, which in this case is the complement of the limiting angle of incidence, are strongly reflected.
This leads us to remark that as the limiting angle of incidence out of fresh water into air, is about 48° 36′, an eye placed in water, ​such as that of a flat fish, looking upwards, need only turn as much as 48° 36′ from the vertical, to see distinctly every object above the surface of the water. Such vision must, however, be rather confused as to the relative position of objects near the surface, or far from the place where the eye is, as the rays by which they are made visible must be crowded near the circumference of the cone, the half angle of which is this limit of the angles of incidence.
58.Let $QR,$ (Fig. 56.) represent a ray of light incident at $R$ on the plane surface $MN$ of a refracting medium $B$ of a different power from $A,$ that in which $QR$ lies; $RS$ the refracted ray, $XRY$ the perpendicular. Let $OP$ be a second surface parallel to $MN,$ separating $B$ from another medium $A,$ or another part of the same, as if $B$ were glass, and $A$ the air on both sides of it.
Let $ST$ be the course of the ray on passing out of $B$ into $A;\ ZSV$ the perpendicular.
Then since the difference of media at $R$ and $S$ is the same, and since the angles $YRS,\ ZSR,$ which we may call the angles of incidence, are equal, the angles of refraction $QRX,$ and $VST$ must likewise be equal, so that $QR$ and $ST$ must be parallel.
59.We may here recall to the reader’s attention a remark that was made at the end of the introductory observations, namely, that it is indifferent in which direction the light is moving along a ray or system of rays connected by reflexion or refraction. In fact, the subject is purely geometrical, and all that we have to consider is the angles made by certain lines with certain other lines according to certain laws.
60.Suppose now (Fig. 57.) that the ray passes from $S$ not immediately into the medium $A,$ but into another medium $C,$ of a power different from either of the former, and then into the medium $A,$ the surfaces being all parallel as before. In this case experiment shows that the first and last rays are parallel as before, so that the actions at $S$ and $T$ produce, when combined, the same effect as the single refraction at $S$ in the former instance.
61.It appears then that whether a ray passes from one medium into another, immediately, or through any number of intermediate ​ones, provided only the surfaces be parallel, the deviation on the whole is exactly the same.
Let $QRSTVXY$ (Fig. 58.) be the course of a beam of light through the media $A,B,C,D,E,A:Q'V'X'Y'$ that of another parallel to the former, at first, and therefore at last, passing only through the medium $A,E,A$.
Then since the refractions at $X$ and $X'$ are equal, the angles made by $VX,$ $V'X',$ with the final or the original courses of the rays, that is, the deviations, must be equal.
62. There is another thing to be observed here, which is rather remarkable. If the media $B,C,D,E,$ were separated from each other by sensible distances, that designated by $A$ pervading all the intermediate spaces, so that the ray came clear out of each before it entered the next, and if the surfaces were all as before parallel, it is evident that the last, like all the other emergent rays, would be parallel to the first incident. It appears then, that it is indifferent as to this phenomenon and its consequences, whether the surfaces touch or not, which is the more singular, as whatever be the nature of the action of substances on light, it certainly takes place only at insensible distances.
63. Prop. Given the direction in which a ray falls on a plane surface bounding a refracting medium; to find the direction of the refracted ray.
$QR$ (Fig. 59.) represents a ray refracted at $R$ into a direction $RS,$ which is produced backwards so as to intersect $AQ,$ a perpendicular to the surface through $Q,$ in $q.$
Let $m$ represent the ratio, sin inclination: sin refraction, sometimes called the index of refraction, or ratio of refraction.
$\Delta ......................AQ,$
$\Delta '......................Aq,$
$\theta .....................\angle RQA,$
$\theta '....................\angle RqA.$
Then we have $Aq=$RA$.\cot \theta '=AQ.\tan \theta .\cot \theta ',$
that is, ${\frac {\Delta '}{\Delta }}={\frac {\tan \theta }{\tan \theta '}}.$
​Now, if we suppose that the incidence of the ray $QR$ takes place nearly perpendicularly to the surface, the angles $RQA,RqA$ may be considered so small, that their sines and tangents are to all sense the same: we may therefore substitute the ratio of the sines for that of the tangents, and remembering that $RQA,RqA$ are equal to their alternate angles $QRN,qRN$ which are those of incidence and refraction, we find
${\frac {\Delta '}{\Delta }}={\frac {\sin \theta }{\sin \theta '}}=m,\quad \mathrm {or} \quad \Delta '=m\Delta .$
It follows from this, that a thin pencil of rays proceeding from a point at the distance $\Delta$ from the plane surface of a refracting medium, appears after refraction to proceed from a point $\mathrm {q}$ at the distance $\mathrm {m\Delta '.}$ In common glass, this latter distance is about half as great again as the former.
64.Now let us examine the amount of the error we commit in substituting the ratio of the sines for that of the tangents, or in other words, let us see what is the difference between the ultimate value of $\Delta '(Aq)$ and that which it actually bears when the angle $RQA$ though small, is taken into consideration: this difference, from its analogy with one of the phænomena of reflexion at curved surfaces, we will call the Aberration.
Let us then take instead of $AQ,Aq,$ the tangents of the angles $RQA,RqA$ to the radius $RA,RQ,$ and $Rq,$ which are to each other as the sines of the angles $RqA,RQq,$ that is, as $\sin \theta ',$ and $\sin \theta .$
We have then $Rq=$ M x RQ, that is, if $AR=v,$ squaring both sides,
$\Delta '^{2}+v^{2}=m^{2}(\Delta ^{2}+v^{2})$
$\therefore \quad \Delta '^{2}=m^{2}\Delta ^{2}+(m^{2}-1)v^{2}$
$=m^{2}\Delta ^{2}+(m^{2}-1)\Delta ^{2}\tan \theta '$
$=m^{2}\Delta ^{2}{\Bigl1+{\frac {m^{2}-1}{m^{2}}}\tan \theta ^{2}{\Bigr )};$
$\therefore \quad \Delta '=m\Delta {\Bigl1+{\frac {m^{2}-1}{2m^{2}}}\tan \theta ^{2}{\Bigr )}.$
​It appears then that $\Delta '$ is in general greater than its ultimate value by the quantity ${\frac {m^{2}-1}{2m}}\Delta .\tan \theta ^{2},$ supposing the angle $\theta$ to be small.
65. Prop. Supposing that a ray passes through a refracting substance bounded by two parallel plane surfaces; it is required to determine its direction after emergence.
Let $AR,A'R'$ (Fig. 60.) represent the two surfaces; $R'S$ the course of the ray after emergence, which, produced backwards, cuts $AQ$ in $V.$ We must observe, that as the refraction at $R'$ is contrary to that at $R,$ if in the latter case ${\frac {\sin \mathrm {inc^{e}.} }{\sin \mathrm {ref^{n}.} }}=m,$ in the former we must have ${\frac {\sin \mathrm {inc^{e}.} }{\sin \mathrm {ref^{n}.} }}={\frac {1}{m}}.$
Then we have $Aq=m.AQ,A'V={\frac {1}{m}}.A'q;$ that is, if we call $AA',$ the thickness of the substance, $T,$ and $AV,\Delta '',$
$\Delta ''+T={\frac {1}{m}}(T+\Delta ')={\frac {1}{m}}(T+m\Delta );$
and $\therefore \quad \Delta ''=\Delta -{\frac {m-1}{m}}T,$
whence it appears that $VQ={\frac {m-1}{m}}AA'.$
It appears from this, that a pencil of rays passing nearly perpendicularly through a refracting medium bounded by parallel planes, suffers no alteration as to convergency or divergency, only that the point of concourse of the rays is brought nearer to the surface of the medium.
66. If we take into consideration the aberrations, we shall find that the distance $Aq$ being taken too small in the calculation we have just made, the point $R'$ is rather too high, (that is, too far from $A',$) which has the effect of throwing $V$ too far out from $A',$ that is, making $A'V$ too great; but again, on account of the aberration at $R',A'V$ is too small; so that the two aberrations correct each other.
​If the ray after emergence from the refracting substance, that we have just been considering, meet with another with surfaces parallel to those of the former, whose thickness is $T'$ and for which the ratio of refraction is $m',$ its direction will, after the second emergence, be of course parallel to the former, and its intersection with $AQ$ will be removed by an additional distance ${\frac {m'-1}{m'}}T',$ so that the whole deviation will be ${\frac {m-1}{m}}T+{\frac {m'-1}{m'}}T',$ and if there were more refracting substances of the same description, their effects would, in like manner, be all added together.
67. Prop.To determine the refraction which a ray experiences in passing through a medium bounded by planes not parallel, for example, a triangular prism of glass.
We will suppose the incidence to take place in a plane perpendicular to the axis of the prism, in which case a transverse section of the prism such as $HIK,$ (Fig. 61.) will contain all the lines necessary for the figure.
Let $QR$ be the incident ray, refracted at $R$ into the direction $RS,$ and again at $S$ into the direction $ST.$
Let $\iota$ = the angle of the prism $HIK,$
$\phi$ = angle of incidence $QRx,$
$\phi '$= angle of 1st refraction $ERS,$
$\psi '$ = angle of 2$^d$ refraction $ESR,$
$\psi$ = angle of emergence $TSy.$
Then $\sin \phi =m\sin \phi ',\quad \sin \psi =m\sin \psi '.$Now $IRS+ISR+RIS=180^{o};$
that is, ${\frac {\pi }{2}}-\phi '+{\frac {\pi }{2}}-\psi '+\iota =\pi ,\quad \therefore \phi '+\psi '=\iota .$
From these equations, knowing $\iota$ and $\phi ,$ we may find successively $\phi ',\psi ',\mathrm {and} \ \psi .$
The deviation $TVq=VRS+VSR=\phi -\phi '+\psi -\psi '.$
​That is, calling the deviation $\delta ,$ we have
$\delta =\phi +\psi -\iota .$
68. If the angle of the prism, and the angles of incidence and emergence be exceedingly small, we may without great inaccuracy substitute for the sines of $\phi ,\phi ',\psi ,\psi '$ the arcs themselves which measure those angles. Then we have
$\phi =m\phi ';\quad \therefore \phi '={\frac {\phi }{m}},$
$\psi =m\psi '$
$\psi '=\iota -{\frac {\phi }{m}},$
$\psi =m\iota -\phi ,$
$\delta =m\iota -\iota =(m-1)\iota .$
69. From this equation we may find the value of $m,$ if $\iota$ be known, and $\delta$ observed, for
$m-1={\frac {\delta }{\iota }};\quad \therefore m=1+{\frac {\delta }{\iota }}={\frac {\delta +1}{\iota }}.$
70. The reader may observe, that in the figure we have been using, the ray is, by the refraction, bent away from the angle $I$ of the prism; this is universally the case, as we may easily show.
Let us take the three different cases.
(1) When the angle $IRQ$ is an obtuse angle, (Fig. 61.)
(2) When it is a right angle, (Fig. 62.)
(3) When it is an acute angle, (Fig. 63.)
In the first case, $IRS$ and $ISR$ are both acute angles, and it is clear, that the bending of $RST$ is from the perpendicular $Sy,$ that is, from the angle $I.$
In the second there is no inflexion at $R,$ but $ISK$ is an acute angle, and therefore the emergent ray is on the far side of the perpendicular from $I,$ and of course declining from it.
In the third case, the deviation is at first towards the angle, afterwards from it, and we must show that the second deviation exceeds the first.
​Supposing the ray to proceed both ways out of the prism at $R$ and $S,$ the angle of incidence at $R,\ SER,$ is less than that at $S,$ namely, $ySR,$ the latter being the exterior angle of the triangle $SER,$ the former an interior and opposite angle.
Now the greater the angle of incidence, the greater is that of deviation, for if $\phi ,\ \phi '$ be the angles of incidence and refraction, $\phi -\phi '$ is the deviation.
Then since $\sin \phi =m\sin \phi ',$
$\sin(\phi +\phi ')-\sin(\phi -\phi ')=2\cos \phi .\sin \phi '$
$=2.\cos \phi .{\frac {\sin \phi }{m}}$
$={\frac {1}{m}}.\sin 2\phi ;$
$\therefore \sin \phi -\phi '=\sin(\phi +\phi ')-{\frac {1}{m}}.\sin 2\phi .$
Now as $\phi$ increases, $\phi '$ increases also, and the sine of $\phi +\phi '$ increases faster than that of the larger angle $2\phi ,$ so that the whole of the 2nd member of this equation increases; therefore $\sin \phi -\phi '$ must increase also, to maintain the equality, and consequently $\phi -\phi ',$ the deviation must increase.
In the above case then, the deviation at $S$ being greater than that at $R,$ the inflexion of the ray is, on the whole, from the angle of the prism.
71. In making experiments with a prism through which a beam of light is made to pass in a plane perpendicular to its axis, it will be found, that if the prism be turned on its axis, the deviation of the emergent ray from the incident, will in some cases increase, in others diminish, so as to have a minimum value. Let us see to what case that value answers.
Adopting the same notation as before, we have
$\delta =\phi +\psi -\iota ;$$\therefore d\delta =d\phi +d\psi ,$
$\sin \phi =m\sin \phi ';$$\therefore \cos \phi \ d\phi =m\cos \phi '.d\phi ',$
$\sin \psi =m\sin \psi ';$$\therefore \cos \psi .d\psi =m.cos\psi '.d\psi '.$
Also $\phi '+\psi '=\iota ;\quad \therefore d\phi '+d\psi '=0$.
Eliminating therefore $d\phi '$ and $d\psi '$, we find
$\cos \psi '.\cos \phi .d\phi +\cos \phi '.\cos \psi .d\psi =0;$
$\therefore d\psi =-{\frac {\cos \phi }{\cos \phi '}}.{\frac {\cos \psi '}{\cos \psi }}d\phi ,$
and $d\delta =d\phi \left\{1-{\frac {\cos \phi }{\cos \phi '}}.{\frac {\cos \psi '}{\cos \psi }}\right\}$.
Now this will clearly be equal to nothing, when $\phi$ and $\psi$ are equal, as their cosines and those of $\phi '$ and $\psi '$ will also be equal, and therefore ${\frac {\cos \phi }{\cos \psi }}.{\frac {\cos \psi '}{\cos \phi '}}=1$.
It appears then, that the deviation is a minimum, when the incident and emergent ray make equal angles with the sides of the prism.
72.Prop.A thin conical pencil of rays pass nearly perpendicularly through both sides of a very thin prism; required the focus of the emergent rays.
Let $Q$, (Fig. 64.) be the focus of the incident rays, $QO$ a perpendicular to the surface. The focus of the rays in their passage through the prism will be $x$, a point in $OQ$, such that
$Ox=m\times OQ,$ (Art. 63.).
From $x$ draw $xy$ perpendicular to the second side of the prism, then, as $x$ is the focus of rays incident on that surface at $S$, their focus after the second refraction will be a point $q$, such that
$yx=m\times yq$.
73.To find experimentally the refracting power of any given substance, we may form a piece of it, if solid, into a prism, and observe an object through it. See Fig. 65, where $P$ is the place of the eye, $Q$ the object observed. Let the angles at $P,\ Q$ be measured; also the angle $PAH$ or $PAI,$ in order to have the angle of incidence at $A$.
​Then $QPA=\theta ,$
$PAB=\pi -\phi +\phi ',$
$ABQ=\pi -\psi +\psi ',$
$BQP=\eta ,$
${\overline {2\pi =2\pi -\phi +\phi '-\psi +\psi '+\theta +\eta ;}}$
$\therefore \ \psi =\theta +\eta -\phi +\phi '+\psi '$
$=\theta +\eta +\iota -\phi$ . . . . . . . . . . . . . . . . . . . . (1).
We have now to find $\psi ',$
$\sin \phi =m\sin(\iota -\psi '),$$$
$\sin \psi =m\sin \psi ',$
${\frac {\sin \psi -\sin \phi }{\sin \psi +\sin \phi }}={\frac {\sin \psi '-\sin(\iota -\psi '}{\sin \psi '-\sin(\iota -\psi '}}$
$\therefore \tan {\frac {\psi -\phi }{2}}.\cot {\frac {\psi +\phi }{2}}=\tan {\frac {2\psi '-\iota }{2}}\cot {\frac {\iota }{2}},$
$\tan {\Bigl\psi '-{\frac {\iota }{2}}{\Bigr )}=\tan {\frac {\iota }{2}}.\tan {\frac {\psi -\phi }{2}}.\cot {\frac {\psi +\phi }{2}};$
$\therefore \psi '={\frac {\iota }{2}}+\tan ^{-1}{\Bigl\tan {\frac {\iota }{2}}.\tan {\frac {\psi -\phi }{2}}.\cot {\frac {\psi +\phi }{2}}{\Bigl )}$
Having found $\psi$ and $\psi ',$ we have only to divide $\sin \psi$ by $\sin \psi '$ to get $m,$ the index of the refracting power required.
This process furnishes an easy method of verifying the third law, by calculating the values of $m$ from different observations, changing the position of the eye or object, and prism; they will be found all to agree.
74.In order to find the refracting power of a liquid substance, we have only to put some of it into a hollow prism of glass, having the surfaces of its sides ground very true and parallel. The refractions of the ray in passing through these will not change the, direction of the emergent ray, as far as regards the angles it makes ​with the incident ray, and that inside the liquid prism, although the relative positions of the eye and object will not be the same, as if there were no refraction but that of this latter prism.
We may sometimes save the trouble of grinding a solid substance into a prismatic form, by placing it in a fluid of the same refracting power as itself, which, in fact, amounts to using instead of the substance, another of the same refracting power, and appears to involve a petition of the point in question, namely, the refracting power of the substance; but it is not so in practice, because by placing a bit of any transparent substance in a dense fluid nearly of the same colour, and diluting this with a rarer fluid, we shall soon see when the fluid is reduced to the same density as the solid, by there being no irregular refraction caused in light passing through the liquid and the solid, which, in fact, will become in many cases quite invisible in the liquid. It is evident, that it is quite indifferent in making the optical experiment afterwards, whether the light pass through the bit of the solid substance or not.
Canada balsam, diluted with spirit, is a convenient liquid to use for solid substances of small densities.
For further particulars, I beg leave to refer the reader to Biot's Physique, vol. III. Dioptrique, Chap. I, or to Dr. Brewster's Treatise on new Philosophical Instruments.
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