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Question

An analyser is inclined to a polarizer at an angle of 30o. the intensity of light emerging from the analyser is 1nth of that is incident on the polarizer. Then n is equal to

A
4
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B
43
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C
83
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D
14
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Solution

The correct option is C 83
Given,
θ=300
Lets consider I0= Intensity of incident light
Intensity of light passing though the analyser is given by
Ia=I0n. . . . . .(1)
By malus law,
I=Icos2θ
I0n=I02cos2300
1n=12×(32)2=38
n=83
The correct option is C.

1467938_1096034_ans_d2894b4de05d4e138903347ffb86b0c0.png

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