A beam of plane polarised light of large cross-sectional area and uniform intensity of 3.3 Wm-2 falls normally on a polariser (cross sectional area 3 × 10-4 m2) which rotates about its axis with an angular speed of 31.4 rad/s. The energy of light passing through the polariser per revolution, is close to: (a) 1.0 × 10-5 J (b) 1.0 × 10-4 J (c) 1.5 × 10-4 J (d) 5.0 × 10-4 J

Answer: (b). As the polariser rotates, I = I₀cos ²ω t; over one revolution cos ² = 1/2, so average transmitted intensity = I₀/2. Time per revolution T = (2π)/(ω) = (6.28)/(31.4) = 0.2 s E = I₀/2 A T = 3.3/2× 3×10⁻⁴× 0.2 ≈ 9.9×10⁻⁵ J So the answer is (b) 1.0×10⁻⁴ J.

A beam of plane polarised light of large cross-sectional area and… | JEE Main 2020 (4 Sep, Shift 1)

Q49JEE Main 2020 · 4 SEP · SHIFT 1Difficult

Polarisation

A beam of plane polarised light of large

cross-sectional area and uniform intensity

of 3.3 Wm-2 falls normally on a

polariser (cross sectional area 3 $\times$ 10-4 m2) which

rotates about its axis with an angular speed

of 31.4 rad/s. The energy of light passing through

the polariser per revolution, is close to:

Show answer and solution

Answer: (b)

As the polariser rotates, $I = I_0\cos^2\omega t$ ; over one revolution $\langle\cos^2\rangle = \frac{1}{2}$ , so average transmitted intensity $= I_0/2$ .
Time per revolution $T = \frac{2\pi}{\omega} = \frac{6.28}{31.4} = 0.2\ \mathrm{s}$
$E = \frac{I_0}{2} A T = \frac{3.3}{2}\times 3\times10^{-4}\times 0.2 \approx 9.9\times10^{-5}\ \mathrm{J}$
So the answer is (b) $1.0\times10^{-4}\ \mathrm{J}$ .

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Polarisation question from JEE Main 2020 (4 Sep, Shift 1)

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