Question

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 in the figure, then the angle turned by the reflected ray will be

• A. ${30}^{\circ }$ clockwise
• B. ${30}^{\circ }$ counter-clockwise
• C. ${50}^{\circ }$ clockwise
• D. ${50}^{\circ }$ counter-clockwise

Answer 1

The correct option is A ${30}^{\circ }$ clockwise

Superposing the effect due to rotation of incident ray and rotation of mirror.

Keeping the incident ray fixed, if mirror is rotated by ${20}^{\circ }$ clockwise, then reflected ray will rotate by ${40}^{\circ }$ clockwise.

$\Delta {\theta }_1={40}^{\circ }CW$

Again, keeping mirror fixed and rotating incident ray by ${10}^{\circ }$ clockwise, the reflected ray will rotate by ${10}^{\circ }$ counter-clockwise.

[since Angle of incidence = Angle of reflection]

$\Delta {\theta }_2={10}^{\circ }CCW=-{10}^{\circ }CW$

$\therefore$ Total rotation of reflected ray $\Delta \theta =\Delta {\theta }_1+\Delta {\theta }_2={40}^{\circ }CW-{10}^{\circ }CW={30}^{\circ }CW$

$\begin{array}{c}\text{Why this Question}?\\ \text{This question helps you understand the concept of rotation of mirror, incident and reflected ray.}\\ \text{Note-1 :- Direction of rotation should be carefully noted.}\\ \text{Note -2 :-The concept discussed in this question is important for JEE preparation}\end{array}$