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# A beam of plane polarized light falls normally on a polarizer of cross sectional area 3×10–4 m2. Flux of energy of incident ray is 10–3 Watt. The polarizer rotates with an angular frequency of 31.4 rad s–1. The energy of light passing through the polarizer per revolution will be -

A
104 Joule
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B
103 Joule
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C
102 Joule
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D
101 Joule
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Solution

## The correct option is A 10−4 JouleUsing Matus law, I=I0cos2θ As here polariser is rotating that means all the values of θ are possible. Iav=12π2π∫0Idθ=12π2π∫0I0cos2θdθ On integrating, we get Iav=I02 where I0=EnergyArea×Time=pA=10−33×10−4=10 Watt3 m2 ⇒Iav=12×103=53 Wattm2 and Time period T=2πω=2×3.1431.4=15 sec So, energy of light passing through the polariser per revolution. =Iav×Area×T=53×3×10−4×15=10−4 J  Suggest Corrections  0      Similar questions  Related Videos   PHYSICS
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