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JEE Main 2020 (7 Jan, Shift 1) · Polarisation

Polarisation question from JEE Main 2020 (7 Jan, Shift 1)

Q47JEE Main 2020 · 7 JAN · SHIFT 1Medium

Polarisation

A polarizer - analyser set is adjusted such that the intensity of light coming out of the analyser is just 10% of the original intensity. Assuming that the polarizer - analyser set does not absorb any light, the angle by which the analyser need to be rotated further to reduce the output intensity to be zero, is:

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Answer: (c)

  1. Malus's law: I=I0cos⁡2θI = I_0\cos^2\theta; here cos⁡2θ=0.1⇒cos⁡θ=0.316⇒θ=71.6∘\cos^2\theta = 0.1 \Rightarrow \cos\theta = 0.316 \Rightarrow \theta = 71.6^\circ.
  2. Output is zero when the axes are crossed, θ=90∘\theta = 90^\circ.
  3. Further rotation =90∘−71.6∘=18.4∘= 90^\circ - 71.6^\circ = 18.4^\circ
  4. So the answer is (c) 18.4∘18.4^\circ.
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