Question

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

• A. $E={10}^{-4}\text{J},I_{avg}=1.67{\text{W/m}}^2$
• B. $E={10}^{-3}\text{J},I_{avg}=1.97{\text{W/m}}^2$
• C. $E={10}^{-4}\text{J},I_{avg}=1.97{\text{W/m}}^2$
• D. $E={10}^{-3}\text{J},I_{avg}=1.67{\text{W/m}}^2$

Answer 1

The correct option is A $E={10}^{-4}\text{J},I_{avg}=1.67{\text{W/m}}^2$

As,

$I=I_0{\cos }^2\theta$

The average value of $I$ over one revolution can be calculated as,

$I_{avg}=\frac{{\int }_0^{2\pi }Id\theta }{{\int }_0^{2\pi }d\theta }$

$=\frac{1}{2\pi }{\int }_0^{2\pi }{I}_0{\cos }^2\theta d\theta =\frac{I_0}{2}$

Now, $\text{Intensity}=\frac{\text{Power}}{\text{Area}}$

$\therefore I_0=\frac{\text{power}}{\text{Area}}=\frac{{10}^{-3}}{3\times {10}^{-4}}{\text{W/m}}^2=\frac{10}{3}{\text{W/m}}^2$

As, $I_{avg}=\frac{I_0}{2}=\frac{10}{2\times 3}=\frac{5}{3}=1.67{\text{W/m}}^2$

The energy of light passing through the polarizer per revolution

$E=I_{avg}\times \text{Area}\times \text{time period of revolution}$

$=\frac{5}{3}\times 3\times {10}^{-4}\times \frac{2\pi }{\omega }$

$=5\times {10}^{-4}\times \frac{2\pi }{31.4}$

$\therefore E={10}^{-4}\text{J}$

Hence, option $\left(\text{A}\right)$ is correct.