Verified Solution • 2025–26 Session moderate

QJEEM20Pha9e4: JEE Main 2020 Physics - Class 12 JEEM20Pha9e4 Solution: A beam of plane polarised light of large cross-sectional area…

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 :

Multiple Choice Options

  • A 1.0 ×\times 10-5 J
  • B 1.0 ×\times 10-4 J
  • C 1.5 ×\times 10-4 J
  • D 5.0 ×\times 10-4 J

Step-by-Step Verified Solution

Intensity, I = 3.3 Wm–3

Area, A = 3 × 10–4 m2

Angular speed, ω\omega = 31.4 rad/s

I=I0cos⁡2(ωt)I = {I_0}{\cos ^2}(\omega t)

⇒Iav=I02\Rightarrow {I_{av}} = {{{I_0}} \over 2}

∵\because ⟨cos⁡2θ⟩\left\langle {{{\cos }^2}\theta } \right\rangle = 12{1 \over 2} , in one time period

∴\therefore E=I02×A×(Δt)E = {{{I_0}} \over 2} \times A \times (\Delta t)

and Δt=2πω=2×3.1431.4=15s\Delta t = {{2\pi } \over \omega } = {{2 \times 3.14} \over {31.4}} = {1 \over 5}s

∴\therefore E=3.32×3×10−4×15=1×10−4 JE = {{3.3} \over 2} \times 3 \times {10^{ - 4}} \times {1 \over 5} = 1 \times {10^{ - 4}}\,J

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