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 $\text{s}^{- 2}$ . 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: (D)

As we know, if a mirror is rotated.
Given, angular acceleration, $\alpha =\pi \, \text{r} \text{a} \text{d} / \text{s}^{2}$
Initial angular velocity $=\omega _{0}=0$
Time, $t=\frac{1}{4}\text{ s}$
Now, as $\theta =\omega _{0}t+\frac{1}{2}\alpha t^{2}$
$=0\times \frac{1}{4}+\frac{1}{2}\times \pi \times \left(\frac{1}{4}\right)^{2}$
$=\frac{180}{2 \times 16}=5.625^{\text{o}}$
Since, mirror is turned by $5.625^{\text{o}}$ , therefore reflected ray will be turned by $2\times 5.625=11.25^{\text{o}}$ .
Hence, option $\left(\text{4}\right)$ is correct.