An Elementary Treatise on Optics/Chapter 1

by Henry Coddington

CHAP. I.

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REFLEXION AT PLANE SURFACES.

6. Prop. To find the direction which a ray of light, emanating from a given point, takes after reflexion at a plane mirror in a given position.

Let QR, (Fig. 2.) represent a ray of light, proceeding from the point Q; XY, the section of the reflecting surface by a plane perpendicular to it containing the line QR; RS, the reflected ray making with XY an angle SRX equal to the angle QRY which QR makes with the same line: let QA be perpendicular to XY, and let SR meet it in q.

Then since the angle QRA is equal to SRX, that is, to qRA, the right-angled triangles QAR, qAR, having the side AR in common, are equal in all respects. Therefore qA is equal to AQ.

Any other reflected ray R′S′ will of course intersect QA in the same point q; so that if several incident rays proceed from Q, the reflected rays will all appear to proceed from q, which as we have seen is at the same distance behind the mirror as Q is before it.

7. Suppose now that a ray QR (Fig. 3.) reflected into the direction RS by a plane mirror HI, meet in S another mirror IK inclined to the former at an angle I. It will of course undergo a second reflexion, and returning to meet the first mirror be reflected again, and so on; so that the course of the light will be the broken line QRSTUVX.

Let perpendiculars be drawn to HI, IK, at the points R, S, T, V, … meeting each other successively in L, M, N, O, … Each of the angles at these points will of course be equal to the angle at I.

Let the angle of incidence at	R be called	φ1,
that at	S……	φ2,

and so on.

Let also $a$ represent the angle at $I.$

The reader will find no difficulty in following these equations,

$\phi _{1}-\phi _{2}=a,$

$\phi _{2}-\phi _{3}=a,$

$\phi _{3}-\phi _{4}=a,$

::$\quad \quad \quad \quad$

$\phi _{1}-\phi _{n}={\overline {n-1}}\ a,$

or $\phi _{n}-\phi _{1}-{\overline {n-1}}\ a.$

If now $\phi _{1}$ be any multiple of $a,$ as ${\overline {n-1}}\ a,$ we shall have somewhere $\phi _{n}=0;$ that is, some reflected ray will be perpendicular to one of the mirrors, and these of course will end the series of reflexions.

If $\phi _{1}$ be not a multiple of $a,$ some value of $n$ will make ${\overline {n-1}}\ a,$ greater than $\phi _{1},$ and then $\phi _{n}$ will become negative. The geometrical fact indicated by this is that the broken line $QRST...$ will at length be turned back upon itself, and the light after coming down the angle $HIK$ will go up again.

Let $a$ be the intersection of $QR$ and $ST,$

$b$ . . . . . . . . . . . . . . . . . . $ST$ and $UV,$

$g$ . . . . . . . . . . . . . . . . . . $QR$ and $UV.$

Then it will immediately be seen that the value of the angle at $a$ is $2\phi _{1}-2\phi _{2},$ or $2a;$ that at $b$ is the same; that at $g$ is double of these or $4a:$ so that if we represent the lines $QR,\ RS,\ ST,...$ by $q_{1},\ q_{2},\ q_{3},...$ and the angles between them by ${\overline {q_{1}q_{2}}},\ {\overline {q_{1}q_{3}}},$ &c., we shall have

${\overline {q_{1}q_{3}}}={\overline {q_{3}q_{5}}}=...{\overline {q_{n-1}q_{n+1}}}=2a,$

${\overline {q_{m}q_{n+m}}}=2na,$

(provided $m$ be an odd number.)

Let $k$ be the intersection of $QR$ and TU^{[errata 1]}.

$\angle gkR=kRT+kTR$

$={\frac {\pi }{2}}-\phi _{1}+{\frac {\pi }{2}}-\phi _{3}$

$=\pi -(\phi _{1}+\phi _{3});$

$\therefore {\overline {q_{1}q_{4}}}=\phi _{1}+\phi _{3}=2\phi _{1}-2a.$

It will be easily seen, that

${\overline {q_{1}q_{6}}}=\phi _{1}+\phi _{5}=2\phi _{1}-4a,$

${\overline {q_{1}q_{2m}}}=\phi _{1}+\phi _{2m-1}=2\phi _{1}-(2m-2)a.$

Let us resume now the first equation

$\phi _{n}=\phi _{1}-{\overline {n-1}}\ a.$

Suppose $a={\frac {2\pi }{n-1}};$ then $\phi _{n}=\phi _{1}-2\pi ,$ or $-\phi _{n}=2\pi -\phi _{1}.$

The first and n^th angles will then be equal.

We must observe, that there is a limit to the angle of incidence after it becomes negative, namely, the double right angle; if it becomes exactly equal to this, the last ray will be parallel to one of the mirrors; if greater, it would meet it if produced backwards.

If $-\phi _{n}=(n-1)a-\phi _{1}=\pi ,\quad n-1={\frac {\pi -\phi _{1}}{a}};$

that is, if $\phi _{1}^{0},\ a^{0},$ represent the number of degrees in $\phi _{1},\ a,\ {\frac {180+\phi _{1}^{0}}{a^{0}}}$ must be a whole number.

Errata

1. Original: TV was amended to TU: detail