In Young's double slit experiment, two slits S₁ and S₂ 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) from S₁ and S₂ respectively. The central bright fringe spot will shift by ___________ number of fringes. Numerical answer type. Work it out, then open the answer.

Answer: 10. Net extra path = (μ₂ - μ₁)t = (1.55 - 1.51)(0.1×10⁻³) = 4×10⁻⁶ m. Fringes shifted = 4×10⁻⁶4000×10⁻¹⁰ = 10 So the central fringe shifts by 10 fringes.

In Young's double slit experiment, two slits S₁ and S₂ are ' d '… | JEE Main 2023 (30 Jan, Shift 1)

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YDSE: Film, Medium & Two Wavelengths

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 $0.1\text{ mm}$ but refractive index $1.51$ and $1.55$ are introduced in the path of beam $(\lambda = 4000\ \text{\AA}$ ) from $S_1$ and $S_2$ respectively. The central bright fringe spot will shift by ___________ number of fringes.

Numerical answer type. Work it out, then open the answer.

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Answer: 10

Net extra path $= (\mu_2 - \mu_1)t = (1.55 - 1.51)(0.1\times10^{-3}) = 4\times10^{-6}\text{ m}$ .
Fringes shifted $= \dfrac{4\times10^{-6}}{4000\times10^{-10}} = 10$
So the central fringe shifts by 10 fringes.

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