Question

A beam of the plane polarized light having flux ${10}^{-3}$ watt falls normally on a polarizer of a cross sectional area $3\times {10}^{-4}{m}^2$. The polarizer rotates with an angular frequency of $31.4rad/s$. The energy of the light passes through the polarizer per resolution will be_____

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

Answer 1

The correct option is A ${10}^{-4}$ Joule

The time period of revolution is given by, T = $\frac{2\pi }{\omega }$
where $\omega$ is the angular frequency.
Now, the power of the source is given by, $P={10}^{-3}$ Watt
Now, T = $\frac{2\times 3.14}{31.4}$ = 0.2 sec
Hence, Energy $=$ 0.5 $\times$ power $\times$ time
$={10}^{-3}\times 0.2\times 0.5={10}^{-4}$ J