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