Page:Optics.djvu/30

 It will be easily seen, that

$$\overline {q_1q_6}= \phi_1 + \phi_5 = 2 \phi_1-4a,$$

$$\overline {q_1q_{2m}}= \phi_1 + \phi_{2m-1} = 2 \phi_1-(2m-2) a. $$

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 nth 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.

8.. Rays meeting in a point being incident on a spherical reflecting surface; it is required to determine the directions of the reflected rays.

$$RAV,$$ Fig. 4, represent the spherical surface, which we will suppose concave, or rather a section of it by a diametric plane containing an incident ray $$QR, \ Q$$ being the point from which that and the other rays are supposed to proceed.