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A beam of unpolarized light having flux 10-3 W falls normally on a polarizer of cross-sectional area 3×10-4 m2. The polarizer rotates with an angular frequency of 31.4 rads-1. The energy of light passing through the polarizer per revolution will be

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a

10−4J

b

10−3J

c

10−2J

d

10−1J

answer is A.

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Detailed Solution

Here, ω=31.4rads−1∴Time period of revolution, T=2πω=2×3.1431.4=0.2sEnergy transmitted/revolution, =(IA)T=I02AT=ϕ0T2=10−3×0.22=10−4J

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