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.