Show that when light reflects from two mirrors that meet each other at a right angle, the outgoing ray is parallel to the incoming ray, as illustrated in the following figure.

When light (or other kinds of electromagnetic radiation) waves collide with a surface or other boundary that does not interact the energy of the radiation, the waves bounce away from the surface.

Rename the angles made by the first and the second incident rays on the mirrors and use the property of the parallel lines that the sum of their co-interior angles is \({\rm{18}}{{\rm{0}}^{\rm{o}}}\) .

As displayed in the figure, ${\theta }_1+\varphi ={90}^{\circ }$, and ${\theta }_2+\sigma ={90}^{\circ }$. But since $\sigma$and $\varphi$are angles in a right-angle triangle, then $\sigma +\varphi ={90}^{\circ }$. Accordingly, if the first two equations are added and substituted using the third equation, it is obtained that –

${\theta }_1+{\theta }_2={90}^{\circ }$

If we multiply the equation by 2, the sum of the co-interior angles is ${180}^{\circ }$, which implies that the reflected ray is parallel to the incident ray.

Therefore, the outgoing ray is parallel to the incoming ray.