In Young's double slit experiment, two slits and are distance apart and the separation from slits to screen is (as shown in figure). [2023]

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Solution

Q.
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. [2023]

Ans.
(10)

$Path difference at P be Δ x$

$Δ x = (μ_{2} - μ_{1}) t = (1.55 - 1.51) 0.1 mm = 0.04 \times 10^{- 4}$

$Δ x = 4 \times 10^{- 6} = 4 μ m$

$y = \frac{Δ x D}{d} = 4 \times 10^{- 6} \frac{D}{d}$

$[y is the distance of central maxima from geometric center]$

$Fringe width = \frac{λ D}{d} = 4 \times 10^{- 6} m \frac{D}{d} = 4 μ m \frac{D}{d}$

$∴$ $Central bright fringe spot will shift by' x'$

$Number of shift = \frac{y}{β} = \frac{4 \times 10^{- 6} D / d}{4 \times 10^{- 7} D / d} = 10$

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