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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 1n  th 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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