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 $31.4\, rad / sec$. 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.
$I_{\alpha v}=\frac{1}{2 \pi} \int\limits_{0}^{2 \pi} I d \theta=\frac{1}{2 \pi} \int\limits_{0}^{2 \pi} I_{0} \cos ^{2} \theta d \theta$
On integration we get $I_{a v}=\frac{I_{0}}{2}$
where $I_{0}=\frac{\text { Energy }}{\text { Area } \times \text { Time }}=\frac{p}{A}$
$=\frac{10^{-3}}{3 \times 10^{-4}}=\frac{10 watt }{3 m ^{2}}$
$\therefore I_{\alpha v}=\frac{1}{2} \times \frac{10}{3}=\frac{5}{3}$ 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
$\Rightarrow 0.2 \times 0.5 \times 10^{-3}$
$\Rightarrow 10^{-4} J$