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A beam of plane polarized light falls normally on a polarizer of cross sectional area 3×104 m2. Flux of energy of incident ray is 103 Watt. The polarizer rotates with an angular frequency of 31.4 rad s1. 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 104 Joule
Using Matus law, I=I0cos2θ
As here polariser is rotating that means all the values of θ are possible.
Iav=12π2π0Idθ=12π2π0I0cos2θdθ
On integrating, we get Iav=I02
where
I0=EnergyArea×Time=pA=1033×104=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=53×3×104×15=104 J

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