Question

A double-slit arrangement produces interference fringes for sodium light (λ =589 nm) that are 0.20° apart. What is the angular fringe separation if the entire arrangement is immersed in water (n = 1.33)?

Answer 1

Given : θ₁ = 0.20°, n_w = 1.33

In the first approximation,

D sin θ₁ = y₁ and D sin θ₂ = y₂

sin θ₂/sin θ₁ = y₂/y₁ …..(1)

Now, y ∝ λD/d

For given d and D,

y ∝ λ

∴ y₂/y₁ = λ₂/λ₁ …..(2)

Now, n_w = λ₁/λ₂ …..(3)

From Eqs. (1), (2) and (3), we get,

sin θ₂/sin θ₁ = λ₂/λ₁ = 1/n_w

∴ sin θ₂ = sin θ₁/n_w = \(\frac{sin0.2}{1.33}=\frac{0.0035}{1.33}\) = 0.0026

θ₂ = sin^-10.0026 = 9' = 0.15°

This is the required angular fringe separation