Question:

In Young’s double slit experiment performed using a monochromatic light of wavelength λ, when a glass plate (μ = 1.5) of thickness xλ is introduced in the path of the one of the interfering beams, the intensity at the position where the central maximum occurred previously remains unchanged. The value of x will be:

Updated On: Mar 19, 2025
  • 3
  • 2
  • 1.5
  • 0.5
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The Correct Option is B

Solution and Explanation

The correct answer is (B) : 2

Path difference at O = (μ-1)t
If the intensity at O remains (maximum) unchanged, path difference must be n 2.
 ⇒ (μ-1)t = nλ 
(1.51) x 2 = nλ 
⇒ x = 2n x = 2n 
For n = 1, x = 2

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Concepts Used:

Young’s Double Slit Experiment

  • Considering two waves interfering at point P, having different distances. Consider a monochromatic light source ‘S’ kept at a relevant distance from two slits namely S1 and S2. S is at equal distance from S1 and S2. SO, we can assume that S1 and S2 are two coherent sources derived from S.
  • The light passes through these slits and falls on the screen that is kept at the distance D from both the slits S1 and S2. It is considered that d is the separation between both the slits. The S1 is opened, S2 is closed and the screen opposite to the S1 is closed, but the screen opposite to S2 is illuminating.
  • Thus, an interference pattern takes place when both the slits S1 and S2 are open. When the slit separation ‘d ‘and the screen distance D are kept unchanged, to reach point P the light waves from slits S1 and S2 must travel at different distances. It implies that there is a path difference in the Young double-slit experiment between the two slits S1 and S2.

Read More: Young’s Double Slit Experiment