Question

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

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

Answer 1

Answer: (A)

Using Malus law, $I=I_0{\cos }^2\theta$
As here polariser is rotating, i.e., all the values of $\theta$ are possible.
$I_{\alpha v}=\frac{1}{2\pi }\underset{0}{\overset{2\pi }{\int }}Id\theta =\frac{1}{2\pi }\underset{0}{\overset{2\pi }{\int }}{I}_0{\cos }^2\theta d\theta$
On integration we get $I_{av}=\frac{I_0}{2}$
where $I_0=\frac{\text{ Energy }}{\text{ Area }\times \text{ Time }}=\frac{p}{A}$
$=\frac{10^{-3}}{3\times 10^{-4}}=\frac{10watt}{3m^2}$
$\therefore I_{\alpha v}=\frac{1}{2}\times \frac{10}{3}=\frac{5}{3}$ watt
and Time period $T=\frac{2\pi }{\omega }=\frac{2\times 3.14}{31.4}=\frac{1}{5}sec$
Energy $\Rightarrow 0.5\times$ power $\times$ time
$\Rightarrow 0.2\times 0.5\times 10^{-3}$
$\Rightarrow 10^{-4}J$