Question

A beam of unpolarized light having flux 10^-3 watt falls normally on a polarizer of cross sectional area 3? 10^-4.The Polarizer rotates with an angular frequency of 31.4 rad s^-1 .The energy of light passing through the polarizer per revolution will be

Answer 1

Dear Student
$$\begin{gathered} P={10}^{-3}W \\ \omega =31.4rad·s^{-1}=10\pi rad·s^{-1} \\ A=3\times {10}^{-4}{m}^2 \\ {I}_{av}=\frac{I_0}{2}=\frac{Power}{2\times Area} \\ T=\frac{2\pi }{\omega }=\frac{2\pi }{10\pi }=\frac{1}{5}sec \end{gathered}$$
The energy of light passing through the polariser per revolution,
$$\begin{gathered} E=I_{av}\times Area\times T=\frac{P}{2}\times T=\frac{{10}^{-3}}{2}\times \frac{1}{5}={10}^{-4}J \\ E={10}^{-4}J \end{gathered}$$

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