Question:

In Youngs double slit experiment, a mica slit of thickness t and refractive index $ \mu $ is introduced in the ray from the first source $ {{S}_{1.}} $ . By how much distance the fringes pattern will be displaced?

Updated On: Jan 18, 2023
  • $ \frac{D}{d}(\mu -1) $
  • $ \frac{d}{(\mu -1)D} $
  • $ \frac{D}{d}(\mu -1)t $
  • $ \frac{d}{D}(\mu -1)t $
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The Correct Option is C

Solution and Explanation

Path difference $ {{y}_{p}}=\frac{D}{d} $ Now additional path difference is $ (\mu -1)t $ . Therefore displacement will be $ \Delta y=\frac{D}{d}(\mu -1)\,t $
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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