Find the net angle of deviation if the angle of incidence is ${30}^{\circ }$ as shown in the figure.

The correct option is D ${180}^{\circ }$ clockwise
Given,
Angle of incidence of light ray onto mirror$-1$ $\left(i_1\right)={30}^{\circ }$

Angular deviation of incident light rays from Mirror$-1$

${\delta }_1=180-2i_1={120}^{\circ }\circlearrowright$

From the diagram, we can deduce that, the angle of incidence of light ray on mirror $-2$ is $i_2={90}^{\circ }-{30}^{\circ }={60}^{\circ }$

Angular deviation of incident light rays from Mirror$-2$

${\delta }_2=180-2i_2={\delta }_2={60}^{\circ }\circlearrowright$

Total deviation of light ray with respect to the incident light ray

${\delta }_{net}={\delta }_1+{\delta }_2={120}^{\circ }\circlearrowright +{60}^{\circ }\circlearrowright ={180}^{\circ }\circlearrowright \text{or}{180}^{\circ }\circlearrowleft$




Alternate method :

Let $\theta$ be the angle between two mirrors.

As we know, ${\delta }_{net}$ is independent of $i$ (Angle of incidence).

We can write that, ${\delta }_{net}=2\theta \circlearrowleft$
[since $\theta$ is anticlockwise from mirror 1 to mirror 2]

$\therefore {\delta }_{net}=2\times {90}^{\circ }={180}^{\circ }\circlearrowleft \text{or}{180}^{\circ }\circlearrowright$


$\begin{array}{c}\text{Why this Question}?\\ \text{This question is aimed at enhancing the conceptual understanding of angle of deviation}\\ \text{when two plane mirrors are placedpe rpendicular to each other}\end{array}$