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Q.

An analyser is inclined to a polariser at an angle of 300 . The intensity of light emerging from the analyser is 1nth of that is incident on the polarizer. Then n is equal to

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a

4

b

43

c

83

d

14

answer is C.

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Detailed Solution

Here ,  θ=300 ,I=I0n

Intensity of polarized light I=I02cos2θ

                                           I0n=I02cos230

                                                 I02.34

                                         I0n=3I08

                                       n=83

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