Wave Optics

JEE Main level, single correct, medium

179 of 191

Question 179 of 191

A polarizer - analyser set is adjusted such that the intensity of light coming out of the analyser is just 3636% of the original intensity. Assuming that the polarizer - analyser set does not absorb any light, the angle by which the analyser needs to be rotated further, to reduce the output intensity to zero, is (sin⁡−1(35)=37∘)\left({\sin }^{-1}\left(\dfrac{3}{5}\right)=37^{\circ}\right)

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Answer: (B) 37o{37}^{o}

I=I0cos2θI={I}_{0}co{s}^{2}\theta
0.36I0=I0cos⁡2θ0.36{I}_{0}={I}_{0}{\cos }^{2}\theta
cos⁡(θ)=35\cos \left(\theta \right)=\dfrac{3}{5}
⇒θ=53∘\Rightarrow \theta =53^{\circ}

angle rotated should be =90o−53o=37o={90}^{o}-{53}^{o}={37}^{o}

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