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 . 1 m

Question

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 $0.1 mm$ but refractive index $1.51$ and $1.55$ are introduced in the path of beam $(λ = 4000 Å)$ from $S_{1}$ and $S_{2}$ respectively. The central bright fringe spot will shift by ________ number of fringes.

TestHub · Accepted Answer

The change in path difference due to the two slabs at $P$ is $Δ x = (μ_{2} - μ_{1}) t$

$= (1.55 - 1.51) \times 0.1$ $= 0.04 \times 10^{- 4}$

$\Rightarrow Δ x = 4 \times 10^{- 6} = 4 μm$

Now, the distance of central maxima from geometrical center is $y = \frac{Δ x D}{d} = 4 \times 10^{- 6} \frac{D}{d}$

Now, fringe width, $n β = 4 \times 10^{- 6} \frac{D}{d}$

Or, $\frac{n λ D}{d} = 4 \times 10^{- 6} \frac{D}{d}$

Number of shift, $n = \frac{4 \times 10^{- 6}}{4 \times 10^{- 7}} = 10$ .

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