**Question:**

A beam of plane polarised light of large cross sectional area and uniform intensity of $3.3 \mathrm{Wm}^{-2}$ falls normally on a polariser (cross sectional area $3 \times 10^{-4} \mathrm{~m}^{2}$ ) which rotates about its axis with an angular speed of $31.4 \mathrm{rad} / \mathrm{s}$. The energy of light passing through the polariser per revolution, is close to :

Correct Option: , 3

**Solution:**

Intensity, $\mathrm{I}=3.3 \mathrm{Wm}^{-2}$

Area, $\mathrm{A}=3 \times 10-4 \mathrm{~m}^{2}$

Angular speed, $\omega=31.4 \mathrm{rad} / \mathrm{s}$

$\because<\cos ^{2} \theta>=\frac{1}{2}$, in one time period

$\therefore$ Average energ $\mathrm{y}=\mathrm{I}_{0} \mathrm{~A} \times \frac{1}{2}$

$=\frac{(3.3)\left(3 \times 10^{-4}\right)}{2}$

$\simeq 5 \times 10^{-4} \mathrm{~J}$