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Question

A beam of unpolarised light having flux 103 watt falls normally on a polarizer of cross sectional area 3×104m2 . The polarizer rotates with an angular frequency of 31.4 rad/s. The energy of light passing through the polarizer per revolution will be:

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

The correct option is C 104 joule

The time period of revolution is given by, T = 2πω
where ω is the angular frequency.
Now, the power of the source is given by, P=103 Watt
Now, T = 2×3.1431.4 = 0.2 sec
Hence, Energy = 0.5 × power × time
=103×0.2×0.5=104 J


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