As shown in the figure, in Young's double slit experiment, a thin plate of thickness t=10 μ m and refractive index μ=1.2 is inserted infront of slit S₁. The experiment is conducted in air (μ=1) and uses a monochromatic light of wavelength λ=500 nm. Due to the insertion of the plate, central maxima is shifted by a distance of x β₀ . β₀ is the fringe-width befor the insertion of the plate. The value of the x is _____________.

Solution

Physics sir · 4 steps

The plate adds an extra path (μ-1)t; divide it by λ to get the shift in fringe widths.

1. Extra optical path. Inside the plate, light covers an optical path μ t instead of t. So the ray from S₁ picks up an extra path of (μ - 1)t compared with the ray from S₂.
2. New central maximum. The central maximum sits where the total path difference is zero. Because the extra path is added to the S₁ ray, the geometric path from S₂ must be longer to balance it. That point lies on the S₁ side, above P, at a distance (μ-1)tD/d from P.

   x = shift/β₀ = ((μ-1)tD/d)/(λ D/d) = (μ-1)t/λ

3. Put in numbers. (μ-1)t = 0.2 × 10 × 10⁻⁶ m = 2 × 10⁻⁶ m. Dividing by λ = 5 × 10⁻⁷ m gives x = 4.
4. So. The central maximum shifts by 4β₀ towards S₁, so x = 4.

The common slip: 24. Using μ t instead of (μ-1)t for the extra path gives 1.2 × 10⁻⁵/(5 × 10⁻⁷) = 24. The plate only adds the part of the path beyond what air would give.

The idea to keep. A thin plate shifts the whole pattern towards its own slit by (μ-1)t/λ fringes. The fringe width itself does not change, so D and d never matter.

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