Question

A beam of plane polarized light of large cross-sectional area and uniform intensity of $3.3{\mathrm{Wm}}^{-2}$falls normally on a polarizer (cross sectional area $3\times {10}^{-4}{\mathrm{m}}^2$) which rotates about its axis with an angular speed of $31.4\mathrm{rad}/\mathrm{s}$. The energy of light passing through the polarizer per revolution is close to:

• A. $1.0\times {10}^{-4}\mathrm{J}$
• B. $1.0\times {10}^{-5}\mathrm{J}$
• C. $5.0\times {10}^{-4}\mathrm{J}$
• D. $1.5\times {10}^{-4}\mathrm{J}$

Answer 1

The correct option is A

$1.0\times {10}^{-4}\mathrm{J}$

Step 1: Given data

The intensity of plane polarized light falling on the polarizer, ${\mathrm{I}}_0=3.3{\mathrm{Wm}}^{-2}$

Cross sectional area, $\mathrm{A}=3\times {10}^{-4}{\mathrm{m}}^2$

Angular speed of polarizer, $\mathrm{\omega }=3.14\mathrm{rad}/\mathrm{s}=10\mathrm{\pi }\mathrm{rad}/\mathrm{s}$

Therefore, $\mathrm{\theta }=\mathrm{\omega t}$

$$\begin{gathered} \Rightarrow 2\mathrm{\pi }=10\mathrm{\pi }\times \mathrm{t}\left[\mathrm{\theta }=2\mathrm{\pi }\mathrm{for}\mathrm{one}\mathrm{revolution}\right] \\ \Rightarrow \mathrm{t}=\frac{1}{5}\mathrm{s} \end{gathered}$$

Step 2: Calculating the energy of a light passing through the polarizer

The relationship between plane polarized light of intensity ${\mathrm{I}}_0$ falling on the polarizer and the intensity $\mathrm{I}$ of light coming out of the polarizer is given by:

$\mathrm{I}={\mathrm{I}}_0{\cos }^2\left(\mathrm{\theta }\right)$

The energy of a light passing through the polarizer per revolution is given by:

$$\begin{gathered} \mathrm{E}={\int }_0^{\frac{1}{5}}\mathrm{IA}d\mathrm{t} \\ \Rightarrow \mathrm{E}={\int }_0^{\frac{1}{5}}\left({\mathrm{I}}_0{\cos }^2\left(\mathrm{\omega t}\right)\right)\mathrm{A}d\mathrm{t} \\ \Rightarrow \mathrm{E}={\mathrm{I}}_0\mathrm{A}{\int }_0^{\frac{1}{5}}{\cos }^2\left(\mathrm{\omega t}\right)d\mathrm{t} \\ \Rightarrow \mathrm{E}={\mathrm{I}}_0\mathrm{A}{\int }_0^{\frac{1}{5}}\left[\frac{1+\cos \left(2\mathrm{\omega t}\right)}{2}\right]d\mathrm{t} \\ \Rightarrow \mathrm{E}=\frac{{\mathrm{I}}_0\mathrm{A}}{2}\left[{\left(\mathrm{t}\right)}_0^{\frac{1}{5}}+{\left(\frac{\sin \left(2\mathrm{\omega t}\right)}{2\mathrm{\omega }}\right)}_0^{\frac{1}{5}}\right] \\ \Rightarrow \mathrm{E}=\frac{{\mathrm{I}}_0\mathrm{A}}{2}\left[\frac{1}{5}+\frac{\left(\sin 2\left(10\mathrm{\pi }\right){\displaystyle \frac{1}{5}}\right)}{2\left(10\mathrm{\pi }\right)}\right] \\ \Rightarrow \mathrm{E}=\frac{{\mathrm{I}}_0\mathrm{A}}{2}\left[\frac{1}{5}+\frac{\sin \left(4\mathrm{\pi }\right)}{20\mathrm{\pi }}\right] \\ \Rightarrow \mathrm{E}=\frac{{\mathrm{I}}_0\mathrm{A}}{10} \\ \Rightarrow \mathrm{E}=\frac{3.3\times 3\times {10}^{-4}}{10} \\ \Rightarrow \mathrm{E}=0.99\times {10}^{-4} \\ \Rightarrow \mathrm{E}=1\times {10}^{-4}\mathrm{J} \end{gathered}$$

Therefore the energy of a light passing through the polarizer per revolution is equal to $1\times {10}^{-4}\mathrm{J}$

Hence, option (A) is the correct answer.