Question

A beam of plane polarized light falls normally on a polarizer of cross sectional area $3\times {10}^{–4}{\text{m}}^2$. Flux of energy of incident ray is ${10}^{–3}$ Watt. The polarizer rotates with an angular frequency of $31.4{\text{rad s}}^{–1}$. The energy of light passing through the polarizer per revolution will be -

• A. ${10}^{-4}\text{Joule}$
• B. ${10}^{-3}\text{Joule}$
• C. ${10}^{-2}\text{Joule}$
• D. ${10}^{-1}\text{Joule}$

Answer 1

The correct option is A ${10}^{-4}\text{Joule}$

Using Matus law, $I=I_0{\cos }^2\theta$
As here polariser is rotating that means all the values of $\theta$ are possible.
$I_{av}=\frac{1}{2\pi }\underset{0}{\overset{2\pi }{\int }}Id\theta =\frac{1}{2\pi }\underset{0}{\overset{2\pi }{\int }}{I}_0{\cos }^2\theta d\theta$
On integrating, we get $I_{av}=\frac{I_0}{2}$
where
$I_0=\frac{\text{Energy}}{\text{Area}\times \text{Time}}=\frac{p}{A}=\frac{{10}^{-3}}{3\times {10}^{-4}}=\frac{10\text{Watt}}{3{\text{m}}^2}$
$\Rightarrow I_{av}=\frac{1}{2}\times \frac{10}{3}=\frac{5}{3}\frac{\text{Watt}}{m^2}$
and Time period $T=\frac{2\pi }{\omega }=\frac{2\times 3.14}{31.4}=\frac{1}{5}\text{sec}$
So, energy of light passing through the polariser per revolution.
$=I_{av}\times \text{Area}\times T=\frac{5}{3}\times 3\times {10}^{-4}\times \frac{1}{5}={10}^{-4}\text{J}$