A beam of plane-polarized light falls normally on a polarizer (cross-sectional area $3\times {10}^{-4}{m}^2$) which rotates about the axis of the ray with an angular velocity of  $31.4rad{s}^{-1}$. Find the energy of light passing through the polarizer per revolution, if the flux of energy of the incident ray is ${10}^{-3}$ W.

Cross-sectional area of polaroid,  $A=3\times {10}^{-4}{m}^2$
Angular velocity,  $\omega =3.14rad{s}^{-1}$
Time taken to complete one revolution, $T=\frac{2\pi }{\omega }=\frac{2\times 3.14}{31.4}=0.2s$
(Energy incident/sec)  $={10}^{-3}W$
So, intensity of incident polarized beam is given by  ${\text{I}}_0=\frac{\text{Energy incident/sec}}{\text{Area}}=\frac{{10}^{-3}}{3\times {10}^{-4}}=\frac{10}{3}W{m}^{-2}$
Since, $I=I_0{\cos }^2\theta$ where  $⟨{\cos }^2\theta ⟩=\frac{1}{2}$
So, average intensity transmitted is $I_{av}=\frac{I_0}{2}=\frac{10}{3\times 2}=1.67W{m}^{-2}$
Light energy passing through polarizer per revolution is given by
$E=I_{av}AT=\frac{10}{6}(3\times {10}^{-4})\left(0.2\right)={10}^{-4}J$