Page:Optics.djvu/29

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

The reader will find no difficulty in following these equations,

$$\phi_1 - \phi_2 = a,$$

$$\phi_2 - \phi_3 = a,$$

$$\phi_3 - \phi_4 = a,$$

$$\phi_1 - \phi_n = \overline {n-1}\ a,$$

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.

$$ \angle gkR=kRT + kTR$$

$$=\frac{\pi}{2} - \phi_1 + \frac{\pi}{2} - \phi_3$$

$$=\pi - (\phi_1 + \phi_3);$$

$$\therefore \overline {q_1 q_4} = \phi_1 + \phi_3 = 2 \phi_1 - 2a.$$

Errata