It will be easily seen, that

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

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

Let us resume now the first equation

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

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

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

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

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

Let ${\displaystyle RAV,}$ 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 ${\displaystyle QR,\ Q}$ being the point from which that and the other rays are supposed to proceed.