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 Aᵒ) from S₁ and S₂ respectively. The central bright fringe spot will shift by ___________ number of fringes.

Solution

Physics sir · 3 steps

Both slabs add extra path, so only the difference of the two extra paths shifts the fringes.

1. Extra path per slab. A slab of thickness t and refractive index μ adds an extra optical path of (μ - 1)t to its beam. Each beam here gets its own slab, so you compare the two extra paths.
2. Net path difference. The two extra paths differ by (1.55 - 1)t - (1.51 - 1)t = (1.55 - 1.51)t = 0.04t. With t = 0.1 mm = 1 × 10⁻⁴ m, this is 4 × 10⁻⁶ m.

   n = ((μ₂ - μ₁) t)/λ = (0.04 × 1 × 10⁻⁴)/(4 × 10⁻⁷) = 10

3. So. Each fringe width corresponds to a path difference of one λ = 4 × 10⁻⁷ m. The central bright fringe shifts by 10 fringes, towards S₂, the side of the denser slab.

The common slip: 137.5. Using only the slab in front of S₂, (1.55 - 1)t/λ, gives 137.5, not 10. The slab in front of S₁ cancels most of that path.

The idea to keep. A slab shifts the fringe pattern by (μ - 1)t/λ fringes towards its own side. With slabs on both slits, use (μ₂ - μ₁)t/λ.

Answer from the NTA final answer key. Solution by PrepWiser.