JEE Main 2023 · 31 January · morning · Single correct · +4 / −1

Two polaroide A and B are placed in such a way that the pass-axis of polaroids are perpendicular to each other. Now, another polaroid C is placed between A and B bisecting angle between them. If intensity of unpolarized light is I₀ then intensity of transmitted light after passing through polaroid B will be:

  1. (a) I₀/4
  2. (b) I₀/8
  3. (c) Zero
  4. (d) I₀/2
Answer and solution

Answer (b) I₀/8

Solution

Physics sir · 4 steps

Halve the intensity at the first polaroid, then apply Malus's law twice, once for each 45° turn.

  1. First polaroid A. Unpolarised light loses half its intensity at any polaroid, so after A the intensity is I₀/2. The light is now polarised along A's pass-axis.
  2. Through C. C bisects the 90° angle, so it sits at 45° to A. By Malus's law, I = I₀/2 cos² 45° = I₀/2 × 1/2 = I₀/4.
  3. Through B. B is also at 45° to C, so the intensity is halved once more: I₀/4 × 1/2 = I₀/8.

    I_B = I₀/2 cos² 45° cos² 45° = I₀/8

  4. So. The light coming out of B has intensity I₀/8, which is option (b).

The idea to keep. Unpolarised light always drops to half at the first polaroid. After that, each polaroid multiplies the intensity by cos² of the angle between it and the one just before it.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Wave Optics

Asked about 140 times in JEE Main since 2020. This question is from 2023.