Page:Optics.djvu/29

5

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]}.

$\angle gkR=kRT+kTR$

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

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

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

Errata

1. Original: TV was amended to TU: detail