Q. A beam of unpolarized light having flux $10^{- 3}$ W falls normally on a polarizer of cross sectional area $3 \times 10^{- 4}\ m^{2}$. The polarizer rotates with an angular frequency of $31.4\ rads^{- 1}$. The energy of light passing through the polarizer per revolution will be
A) $10^{- 4}\ J$
B) $10^{- 3}\ J$
C) $10^{- 2}\ J$
D) $10^{- 1}\ J$
Explanation
(a) Here, $\omega = 31.4\ rads^{- 1}$
$\therefore$ Time period of revolution,
$T = \frac{2\pi}{\omega} = \frac{2 \times 3.14}{31.4} = 0.2s$
Energy transmitted/revolution
$= (IA)T = \left( \frac{I_{0}}{2}A \right)T$
$= \frac{\phi_{0}T}{2} = \frac{10^{- 3} \times 0.2}{31.4} = 10^{- 4}J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.