Figure shows a plane mirror onto which a light ray is incident. If the incident light ray is turned by ${10}^{\circ }$ and the mirror by ${20}^{\circ }$ as shown, then the angle turned by the reflected ray is

The correct option is A ${30}^{\circ }$ clockwise
${\theta }_1=$ angle turned by reflected ray due to rotation of plane mirror by ${\theta }_m$ angle.
${\theta }_2=$ angle turned by reflected ray due to rotation of incident ray by ${\theta }_i$ angle.

Superposing the effect, due to rotation of incident ray and rotation of mirror.
Keeping the incident ray fixed, if the mirror will rotate by ${\theta }_m={20}^{\circ }CW$.
$\Rightarrow {\theta }_1=2{\theta }_m={40}^{\circ }CW$
Again, keeping mirror fixed and rotating incident ray by ${\theta }_i={10}^{\circ }CW$, the reflected ray will also rotate by ${10}^{\circ }$ but in anticlockwise $\left(ACW\right)$.
$\Rightarrow {\theta }_2=-{\theta }_i=-{10}^{\circ }CW$
$\therefore$ Net rotation of reflected ray $={40}^{\circ }CW-{10}^{\circ }CW={30}^{\circ }CW$
Hence, option $\left(a\right)$ is the correct answer.