Question

A light ray $I$ is incident on a plane mirror $M$. The mirror is rotated in the direction shown in the figure at frequency $\frac{9}{\pi }\text{rps}$. The light reflected by the mirror is received on the wall $W$ at a distance $10\text{m}$ from the axis of rotation. When the angle of incidence becomes ${37}^{\circ }$, the speed of the spot on the wall is

• A. $10\text{m/s}$
• B. $1000\text{m/s}$
• C. $500\text{m/s}$
• D. None of these

Answer 1

The correct option is B $1000\text{m/s}$

We know that,

The angle turned by the reflected ray is twice the angle turned by the mirror in the same sense.

When mirror is rotated with angular speed $\omega$, the reflected ray rotates with angular speed ${\omega }^{\prime }=2\omega =2\times \frac{9}{\pi }\times 2\pi =36{\text{rad s}}^{-1}$ in the same sense.

From the figure, ${\omega }^{\prime }=\frac{d\theta }{dt}$


Let $h$ be the height on the screen where the spot is observed.
Then, $h=10\cot \theta$

Speed of the spot is given by,

$\left|\frac{dh}{dt}\right|=\left|\frac{d}{dt}\left(10\cot \theta \right)\right|=|-10{\text{cosec}}^2\theta \frac{d\theta }{dt}|$

From the data given in the question,
when $\theta ={37}^{\circ },\text{cosec}\theta =\frac{1}{0.6}$

Hence, speed of the spot
$\left|\frac{dh}{dt}\right|=|\frac{10}{{\left(0.6\right)}^2}\times 36|=1000{\text{ms}}^{-1}$