JEE Main 2020 — Wave Optics Question with Solution

Given,

Intensity of plane polarized light of large cross-sectional area is $I_0=3.3 \mathrm{W} {\mathrm{m}}^{-2}$

Area of cross-section of polarizer is $A=3\times {10}^{-4 }{\mathrm{m}}^2$

Angular speed of rotation is $\omega =31.4 \mathrm{rad} {\mathrm{s}}^{-1}$

The intensity of light is the given by the definition,

 $\mathrm{I}=\frac{\mathrm{E}}{AT} ...\left(1\right)$

Intensity of light passing through the polarizer as polarizer rotates is given by

$I_{avg}=I_0\frac{1}{T}{\int }_0^T{\cos }^2\left(\omega t\right)dt$

$\Rightarrow I_{avg}=I_0\frac{1}{T}{\int }_0^T\left[\frac{1+\cos \left(2\omega t\right)}{2}\right]dt$

$\Rightarrow I_{avg}=I_0\frac{1}{2T}\left[{\int }_0^{T}dt+{\int }_0^T\cos \left(2\omega t\right)dt\right]$

$\Rightarrow I_{avg}=I_0\frac{1}{2T}\left[T+0\right]$

$\Rightarrow I_{avg}=\frac{I_0}{2} ...\left(2\right)$

Using equation (1) and (2) we get energy per revolution is given by

$E=\frac{I_{0}A}{2}T=\frac{I_{0}A}{2}\frac{2\pi }{\omega }$

$E=\frac{3.3\times 3\times {10}^{-4}}{2}\frac{3.14}{31.4}$

$E\simeq 1\times {10}^{-4} \mathrm{J}$