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Question
A beam of unpolarised light having flux 10−3 watt falls normally on a polarizer of cross sectional area 3×10−4m2 . 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 beam of unpolarised light having flux 10−3 watt falls normally on a polarizer of cross sectional area 3×10−4m2 . The polarizer rotates with an angular frequency of 31.4 rad/s. The energy of light passing through the polarizer per revolution will be:
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Solution
The correct option is C 10−4 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=10−3 Watt
Now, T = 2×3.1431.4 = 0.2 sec
Hence, Energy = 0.5
× power × time
=10−3×0.2×0.5=10−4 J
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=10−3 Watt
Now, T = 2×3.1431.4 = 0.2 sec
Hence, Energy = 0.5
× power × time
=10−3×0.2×0.5=10−4 J
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