Question

A small mirror of an area $a(m{)}^2$ and mass $m\left(kg\right)$ is, suspended by means of mass-less thread in vertical plane. When a beam of light of intenisty $1W{m}^{-2}$ is made an incident normally on the mirror, it gets displaced so that the thread makes angle $\theta$ with the vertical. Assuming the mirror is perfectly reflecting, the value of $\theta$ (Consider it very small) is

• A. $\frac{2Im}{cag}$
• B. $\frac{2Ic}{mag}$
• C. $\frac{2Ig}{cam}$
• D. $\frac{2Ia}{cmg}$

Answer 1

The correct option is D $\frac{2Ia}{cmg}$

Force on the mirror due to incident light beam is $\frac{2Ia}{c}$.
Balancing the torque about the hinge, assuming length of string be $L$ we get,
$\frac{2Ia}{c}lcos\left(\theta \right)=mglsin\left(\theta \right)$,
So $tan\left(\theta \right)=\frac{2Ia}{cmg}$.
Now because deflection the mirror about hinge due to light is very small, we can approximate $tan\left(\theta \right)$ by $\theta$.
Thus the angle $\theta =\frac{2Ia}{cmg}$.