Q 12-10-131JEE MainJEE Main 2020 (4 Sep, Shift 1)Medium
A beam of plane polarized light of large cross-sectional area and uniform intensity of $3.3\ \text{W m}^{-2}$ falls normally on a polarizer (cross-sectional area $3\times10^{-4}\ \text{m}^{2}$), which rotates about its axis with an angular speed of $31.4\ \text{rad s}^{-1}$. The energy of light passing through the polarizer per revolution, is close to:
Answer: (B) $1.0\times10^{-4}\ \text{J}$
Transmitted intensity $I_0\cos^{2}\omega t$ averages to $\dfrac{I_0}{2}$ over a revolution. Time for one revolution $T = \dfrac{2\pi}{31.4} = 0.2$ s.
$$E = \frac{I_0}{2}AT = \frac{3.3}{2}\times3\times10^{-4}\times0.2 \approx 1.0\times10^{-4}\ \text{J}$$
Solution by Sreeraj P, M.Sc Physics