Question

A ray of light is incident on the plane mirror at rest. The mirror starts turning at a uniform acceleration of $2\pi rad/{s^2}$ about an axis in the plane of the mirror. The reflected ray at the end of $\dfrac{1}{4}s$ must have turned through:
A. 90
B. 45
C. 22.5
D. 11.25

Answer 1

Hint: Try to calculate the angle by which the mirror has rotated. Then, find the angle by which the reflected ray has turned around. Use the law of reflection in the sense that angle between the incident and reflected ray is 2 times the angle of incident/reflection. So, if the mirror is rotated by angle x then the reflected ray has turned through 2x.

Complete step by step answer:
According to the question, the initial angular velocity ${\omega _i} = 0$

$\alpha = 2\pi rad/{s^2} $

$t = \dfrac{1}{4}\sec  $

By Newton’s equations of motion for rotational dynamics we have,

\[\theta = {\omega _i}t + \dfrac{1}{2}\alpha {t^2} \Rightarrow \theta = 0t + \dfrac{1}{2}(2\pi ){(\dfrac{1}{2})^2} \]

\[\theta = \dfrac{\pi }{{16}}rad \]

According to the law of reflection, the angle between the incident and reflected ray is 2 times the angle of incident/reflection.

As shown in the figure, with the help of basic geometry we can show that the angle between the mirror and reflected ray changed by 2 times the angle by which the mirror rotated. So, the angle by which the reflected ray gets rotated is,
$ {\theta _{reflected}} = 2\theta = \dfrac{{2\pi }}{{16}}rad $
$ \implies {\theta _{reflected}} = {(\dfrac{{2\pi }}{{16}} \times \dfrac{{180}}{\pi })^ \circ } = {22.5^ \circ }  $
Hence, the angle by which reflected ray turns around is ${22.5^ \circ }$. So, the correct answer is “Option (C)”.

Additional Information:
 If The mirror rotates further the angle between incident angle and the mirror keeps on decreasing and after some time x becomes greater than i. Now, the incident ray falls on the back of the mirror, not on the reflective side. Therefore, no reflection takes place. Only the reflected ray gets deflected more and more in one half cycle of rotation of the mirror.

Note:
The source of incident light is not changing, it is stable.
The incident ray still makes angle I with the vertical and i-x with the mirror.
While returning the deviation angle of the reflected ray keeps decreasing until zero and then increases like before.