 # A beam of plane polarised light of large cross-sectional area Question:

A beam of plane polarised light of large cross-sectional area and uniform intensity of $3.3 \mathrm{Wm}^{-2}$ falls normally on a polariser (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 polariser per revolution, is close to:

1. (1) $1.0 \times 10^{-5} \mathrm{~J}$

2. (2) $1.0 \times 10^{-4} \mathrm{~J}$

3. (3) $1.5 \times 10^{-4} \mathrm{~J}$

4. (4) $5.0 \times 10^{-4} \mathrm{~J}$

Correct Option: , 4

Solution:

(4) Given:

Intensity, $I_{0}=3.3 \mathrm{Wm}^{-2}$

Area, $\quad A=3 \times 10^{-4} \mathrm{~m}^{2}$

Angular speed, $\omega=31.4 \mathrm{rad} / \mathrm{s}$

Average energy $=I_{0} A\left\langle\cos ^{2} \theta\right\rangle$

$\because<\cos ^{2} \theta>=\frac{1}{2}$ per revolution

$\therefore$ Average energy $=\frac{(3.3)\left(3 \times 10^{-4}\right)}{2} \simeq 5 \times 10^{-4} \mathrm{~J}$  