Question

A light ray is incident on a horizontal plane mirror at an angle of ${45}^o$. At what angle should a second plane mirror be placed in order that the reflected ray finally be reflected horizontally from the second mirror, as shown in figure:

Answer 1

From parallel line theory we can say that

$\angle CBP=\angle BCD$

As CD is parallel to PQ.

SO,

$3\times {45}^{\circ }=2\alpha$

$\alpha =\frac{3}{2}\left(45\right)={67.5}^{\circ }$

Now the inclination of mirror with horizontal so that the ray is reflected horizontally is.

${90}^{\circ }-{67.5}^{\circ }={22.5}^{\circ }$