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Question

# Unpolarized light of intensity 32 Wm−2 passes through three polarizers such that transmission axes of the first and second polarizer makes an angle 30∘ with each other and the transmission axis of the last polarizer makes an angle θ with the axis of second polarizer as shown in figure. The intensity of final emerging light will be-

A
32 Wm2
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B
16 Wm2
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C
12 Wm2
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D
3 Wm2
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Solution

## The correct option is D 3 Wm−2 Let, P1, P2 and P3 be the three polarisers, and angle between P1 and P2=30∘ And angle between P2 and P3 is, θ=90∘−30∘=60∘ I0=32 Wm−2 The intensity of light transmitted by P1 is, I1=I02=322=16 Wm−2 According to Malu's law, the intensity of light transmitted is given by, I=I0cos2θ Hence, intensity of light emerging out from the P2 is, I2=I1cos230∘=16⎛⎝√√32⎞⎠2=12 Wm−2 Similarly, intensity of light emerges out from the P2 is, I3=I2cos260∘=12(12)2=3 Wm−2 <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, (D) is the correct answer.

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