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

`therefore (sin theta_2)/(sin theta_1) = "y"_2/"y"_1`   ...(1)

Now, `y prop (lambdaD)/d`

For given d and D,

y ∝ λ

∴ `y_2/y_1 = lambda_2/lambda_1`    ...(2)

Now, `n_w = lambda_1/lambda_2`    ...(3)

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

`(sin theta_2)/(sin theta_1) = lambda_2/lambda_1 = 1/n_w`

∴ `sin theta_2 = sin theta_1/n_w`

`sin theta_2 = (sin 0.2)/1.33`

`sin theta_2 = 0.0035/1.33`

sinθ₂ =  0.0026

θ₂ = sin⁻¹ 0.0026 

θ₂ = 9'

θ₂ = 0.15°

This is the required angular fringe separation.