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Step-by-Step Solution
Step 1: Understand the Scenario
In Young's double-slit experiment, each slit sends light waves to the observation screen, creating an interference pattern. When a transparent slab of thickness $t$ and refractive index $\mu$ is placed in front of one slit, it changes the optical path of that beam compared to the other.
Step 2: Determine the Extra Path Difference
The thin slab adds an extra path difference $\Delta x$ because the light travels through a medium of refractive index $\mu$ instead of air. This is given by:
$ \Delta x = (\mu - 1)\, t $
where $t$ is the thickness of the slab.
Step 3: Relate to Fringe Shift
The resulting shift $y$ in the interference fringes on the screen depends on how the path difference translates to a spatial displacement. Using the geometry of the setup:
$ y = \frac{D}{d} \times \Delta x $
Here, $d$ is the slit separation, and $D$ is the distance between the slits and the screen.
Step 4: Substitute the Extra Path Difference
Substituting $\Delta x = (\mu - 1)\, t$ into the expression for $y$, we get:
$ y = \frac{D}{d} \times (\mu - 1)\, t $
Step 5: Final Expression for Fringe Displacement
Hence, the fringe pattern shifts by:
$ \displaystyle \text{Fringe Displacement} = \frac{D}{d} (\mu - 1) \, t $
This matches the correct answer given in the options.
