Question 68 of 191

In Young’s double slit experiment, two slits ${S}_{1}$ and ${S}_{2}$ are $d$ distance apart and the separation from slits to screen is $D$ (as shown in figure). Now if two transparent slabs of equal thickness $0.1mm$ but refractive index $1.51$ and $1.55$ are introduced in the path of beam $\left(\lambda =4000Å\right)$ from ${S}_{1}$ and ${S}_{2}$ respectively. The central bright fringe spot will shift by ________ number of fringes.

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Answer: 10

The change in path difference due to the two slabs at $P$ is $\Delta x=\left({\mu }_{2}-{\mu }_{1}\right)t$
$=(1.55-1.51)\times 0.1$$=0.04\times {10}^{-4}$
$\Rightarrow \Delta x=4\times {10}^{-6}=4\mu m$
Now, the distance of central maxima from geometrical center is $y=\dfrac{\Delta xD}{d}=4\times {10}^{-6}\dfrac{D}{d}$
Now, fringe width, $n\beta =4\times {10}^{-6}\dfrac{D}{d}$
Or, $\dfrac{n\lambda D}{d}=4\times {10}^{-6}\dfrac{D}{d}$
Number of shift, $n=\dfrac{4\times {10}^{-6}}{4\times {10}^{-7}}=10$.