Question

A ray of light is incident on the plane mirror at rest. The mirror starts turning at a uniform angular acceleration of $\pi$ rad.$s^{-2}$. The reflected ray at the end of $\frac{1}{4} s$ must have turned through

• A. $90^{\circ}$
• B. $45^{\circ}$
• C. $22.5^{\circ}$
• D. $11.25^{\circ}$

Answer 1

Answer: (C)

As we know, if a mirror is rotated
Given, angular acceleration, $\alpha=\pi\, rad / s^{2}$
Initial angular velocity $= \omega_{0} =0$
Time, $t=\frac{1}{4}\,s$
Now, as $\theta=\omega_{0}t+\frac{1}{2} \alpha t^{2}$
$=0\times\frac{1}{4}+\frac{1}{2}\times\pi\times\left(\frac{1}{4}\right)^{2}$
$=\frac{180}{2\times16}=5.625^{\circ}$
Since, mirror is turned by $5.625^{\circ}$, therefore reflected ray will be turned by $2\times5.625=11.25^{\circ}$