JEE Main 2020 · 8 January · evening · Single correct · +4 / −1

In a double slit experiment, at a certain point on the screen the path difference between the two interfering waves is 1/8th of a wavelength. The ratio of the intensity of light at that point to that at the centre of a bright fringe is :

  1. (a) 0.853
  2. (b) 0.568
  3. (c) 0.672
  4. (d) 0.760
Answer and solution

Answer (a) 0.853

Solution

Physics sir · 3 steps

Turn the path difference into a phase difference, then use I = I₀cos²(φ/2).

  1. Path to phase. A path difference of one wavelength is a phase difference of 2π. So a path difference of λ/8 gives φ = 2π/λ × λ/8 = π/4.
  2. Intensity rule. For two equal waves, the intensity at phase difference φ is I = I₀cos²(φ/2). Here I₀ is the intensity at the centre of a bright fringe, where φ = 0.

    I/I₀ = cos² π/8 = (1 + cos π/4)/2 = (1 + 0.7071)/2

  3. So. I/I₀ = 1.7071/2 ≈ 0.853, which is option (a).

The idea to keep. In YDSE, first convert path difference to phase using φ = 2π/λΔ x. Then I = I₀cos²(φ/2) gives the intensity as a fraction of the bright-fringe maximum.

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 2020.