An Elementary Treatise on Optics/Chapter 10

by Henry Coddington

CHAP. X.

REFRACTION AT CURVED SURFACES NOT SPHERICAL.

102. In like manner as in Chap. IV. we found that though a spherical surface is not capable of reflecting light accurately, those belonging to the Conic Sections have that property, so here we shall find that by means of a spheroidal or hyperboloidal surface, rays may be refracted so as to meet accurately in one point without any aberration.

This will easily be seen from simple Geometrical considerations.

Let $AM$ (Fig. 99.) be the axis major of an ellipse $ARM;\ S,\ H$ the foci, $nRN$ a normal, $QR$ parallel to $AM$.

Then since the angles $HRN,$ $SRN$ are equal,

${\frac {\sin QRn}{\sin NRS}}={\frac {\sin RNH}{\sin NRS}}={\frac {\sin RNS+\sin RNH}{\sin NRS+\sin NRH}}$

$={\frac {RS+RH}{NS+NH}}$

$={\frac {AM}{SH}}.$

From this it appears that if a transparent spheroid have the ratio of its axis major to its eccentricity equal to its retracting power, rays entering it in a direction parallel to the axis major will be refracted accurately to the farther focus. Moreover if the surface of the ellipsoid be cut by a spherical surface, having that focus for its centre, a lens will be formed which will refract parallel rays accurately to the focus as there will be no refraction in their passage through the spherical surface.

For the same reason, rays diverging from the focus of this lens will be refracted so as to become parallel to the axis.

103.Again, let $AM,$ (Fig. 100.) be the axis major of a pair of hyperbolas, $QR$ a line parallel to it, $S,\ H$ the foci, $RN$ a normal,

${\frac {\sin QRN}{\sin nRS}}={\frac {\sin RNS}{\sin NRS}}={\frac {\sin RNS-\sin RNH}{\sin NRS-\sin NRH}}$

$={\frac {RS-RH}{SN-NH}}$

$={\frac {AM}{SH}}$ as before.

Hence it appears that a plano-convex lens having its convex surface hyperboloidal, will refract parallel rays accurately to a point which is the focus of the opposite hyperboloid, and conversely, that rays diverging from a point may be refracted by such a lens so as to become parallel.

104.Sir I. Newton has given in the 14th section of the first volume of his Principia, two curious propositions relating to the present subject, which I will insert here, to save the trouble of referring to the book itself.

The first is:

To find the form of the surface of a medium, which will refract rays diverging from a point without it accurately to a point within itself.

Let $A$ (Fig. 101.) be the focus of the incident rays; $B$ that of the refracted; $CD$ the section of the surface; $AD,\ DB$ an incident and refracted ray, $DE$ an infinitely small portion of the curve; $EF,\ EG$ perpendiculars on $AD,\ DB.$

Now ${\frac {DF}{DE}}=\cos EDF=\mathrm {sin~incid.}$

${\frac {DG}{DE}}=\cos EDG=\mathrm {sin~refr.}$

therefore if $m$ be as usual the refracting power, $DF=m.DG;$ but $DF,\ DG$ are the corresponding increments of $AD,\ BD$ so that if we call these $u,\ v,$ we have this differential equation to the curve,

$du=-m.dv,\quad \mathrm {or~~} du+mdv=0$

When $B$ is removed to an infinite distance, the equation becomes $du=mdx$ if $x=AN,$ $DN$ being perpendicular to $AC;$ now we shall see that this equation belongs to an hyperbola, or an ellipse, according as $m$ is greater or less than unity,

$du=mdx$ gives, by integration, $u=mx+n,$

that is, $x^{2}+y^{2}=(mx+n)^{2};$

$\therefore y^{2}=(m^{2}-1)x^{2}+2mx+n^{2}.$

Now the equation to an hyperbola, when the abscissa is measured from the farther focus, is

y2	=(e2−1) {(x−ae)2−a2}
	=(e2−1) {x2−2aex+(e2−1)a2},

which agrees with that above, if

m=e,

mn/m2−1=	−ae; ∴ a=−n/m2−1,
n2/m2−1=	(e2−1)a2, or a2=n2/(m2−1)2, as before.

If m be less than unity, the integrated equation becomes

y²=n²+2mnx−(1−m²)x²,

which agrees with the equation to the ellipse,

y2	=(1−e2) {a2−(x−ae)2},
or y2	=(1−e2) {(1−e2)a2+2aex−x2}.

The general equation integrated gives

u+mv=n,

that is, √x²+y²+m√(c−x)²+y²=n.

The curve to which this belongs is a certain oval, which Descartes has described.

105. Newton's second proposition is:

To find the form of a convex lens, that shall refract light accurately from one point to another.

He supposes the first surface given, and determines the second thus, (Fig. 102.)

Let A be the focus of incident, B of refracted rays; ADFB the course of a ray; CP, ER, circular arcs with centers A, B; CQ, ES, orthogonal trajectories to DF.

Let AB, AD, DF, be produced so that BG=(m−1)CE, AH=AG; DK=1/mDH. Join KB, and let a circle with centre D, and radius DH cut it in L. Draw BF parallel to DL. In the first place PD=m·DQ; FR=Fn·FS. For let AD, AE (Fig. 103.) be two incident rays inclined to each other at an infinitely small angle, DK, EL, the refracted rays. Let EF, EG be perpendicular to AD, DK. Then PF=Ep, QG=Eq; therefore DF=d·DP, DG=d·DQ. Now it was proved in the former proposition that DF=m·DG, and DP, DQ are evidently integrals of these differentials beginning together from nothing.

To return then to our problem,

m=	PD/DQ,
but m=	DL/DK=FB/FK	=PH−PD−FB/FQ−QD;
∴ m=	PH−FB/FQ	=CE+EG−BR−FR/CE−FS
		=CE+BG−FR/CE−FS
		=CE+(m−1)·CE−FR/CE−FS;
	∴ m	=FR/FS,

and therefore the ray ADF is refracted to B.