Question

# A beam of plane polarized light falls normally on a polarizer (cross-sectional area 3×10−4 m2) which rotates about the axis of the ray with an angular velocity of 31.4 rad/s. Find the energy of light E passing through the polarizer per revolution and the intensity Iavg of the emergent beam if the flux of energy of the incident ray is 10−3 W.

A
E=104 J, Iavg=1.67 W/m2
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B
E=103 J, Iavg=1.97 W/m2
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C
E=104 J, Iavg=1.97 W/m2
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D
E=103 J, Iavg=1.67 W/m2
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Solution

## The correct option is A E=10−4 J, Iavg=1.67 W/m2As, I=I0cos2θ The average value of I over one revolution can be calculated as, Iavg=∫2π0I dθ∫2π0dθ =12π∫2π0I0cos2θ dθ=I02 Now, Intensity=PowerArea ∴I0=powerArea=10−33×10−4 W/m2=103 W/m2 As, Iavg=I02=102×3=53=1.67 W/m2 The energy of light passing through the polarizer per revolution E=Iavg×Area×time period of revolution =53×3×10−4×2πω =5×10−4×2π31.4 ∴E=10−4 J Hence, option (A) is correct.

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