Two mirrors are kept at an angle ${60}^{\circ }$ between them as shown in the fig. A ray of light strikes the first mirror and after reflection strikes the second mirror. If the final reflected ray goes parallel to the first mirror, calculate the angle of incidence.

The correct option is C

${30}^{\circ }$

Lets draw the ray diagram as mentioned in the problem.

$\angle 1={90}^{\circ }-\theta$ (Angle of incidence = angle of reflection) ...1)
$\angle 1=\angle 3+\angle 4$ (Alternate opposite angles) ...2)
$\angle 3=\angle 4$ (Angle of incidence = angle of reflection) ...3)
From above three we get,
$\angle 3={45}^{\circ }-\frac{\theta }{2}.$
$\angle 3+\angle 2={90}^{\circ }.$
(as BD is the normal)
Hence
$\angle 2={45}^{\circ }+\frac{\theta }{2}$
Now, in triangle, ABO
$\angle 1+\angle 2+{60}^{\circ }={180}^{\circ }$
Substituting $\angle 1\&\angle 2$, we get
${90}^{\circ }-\theta +{45}^{\circ }+\frac{\theta }{2}={120}^{\circ }$
From above equation
$\frac{\theta }{2}={15}^{\circ }$
Or,
$\theta ={30}^{\circ }$