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JEE Mains · Physics · STD 12 - Wave optics
  • JEE Mains 2020
  • Hard
  • Numerical

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 \(3 \times 10^{-4}\, m ^{2}\) ) 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 ........ \( \times 10^{-4} \,J\)

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Answer & Solution

Correct Answer

(C) \(1\)

Step-by-step Solution

Detailed explanation

Intensity, \(I=3.3 Wm ^{-2}\)

Area, \(A=3 \times 10^{-4} m ^{2}\)

Angular speed, \(\omega=31.4 rad / s\)

\(T=\frac{2 \pi}{\omega}\)

\(I = I _{0} \cos ^{2} \omega t\)

\(\because \cos ^{2} \theta =\frac{1}{2},\) in one time period

\(I=\frac{E}{A\cdot t}\)

\(\therefore\) Average energy \(=\frac{I_{0}}{2} \cdot \frac{2 \pi}{\omega} A=\frac{3.3 \times 3.14 \times 3 \times 10^{-4}}{31.4}=9.9 \times 10^{-5} \)

\(\approx 10 \times 10^{-5} \approx 1 \times 10^{-4} J\)

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