Question

A double-slit arrangement produces interference fringes for sodium light (λ = 589 nm) that are 0.48° apart. What is the angular fringe separation if the entire arrangement is immersed in a liquid that has index of refraction n = 1.25?

Answer 1

Answer:

0.38°

Explanation:

$\theta$ = Angle

m = Number

d = Distance

n = Refractive index of liquid = 1.25

a denotes air

l denotes liquid

In the case of double split interferance we have the relation

$m\lambda=dsin\theta$

For air

$m\lambda_a=dsin\theta_a$

For liquid

$m\lambda_l=dsin\theta_l$

Dividing the two equations

$\frac{m\lambda_a}{m\lambda_l}=\frac{dsin\theta_a}{dsin\theta_l}\\\Rightarrow \frac{\lambda_a}{\lambda_l}=\frac{sin\theta_a}{sin\theta_l}$

Wavelength ratio = $n$

$n=\frac{sin\theta_a}{sin\theta_l}\\\Rightarrow \frac{sin0.48}{1.25}=sin\theta_l\\\Rightarrow \theta_l=sin^{-1}\frac{sin0.48}{1.25}\\\Rightarrow \theta_l=0.38^{\circ}$

The angular separation is 0.38°

Answer 2

Answer:

0.384°

Explanation:

λ = 589 nm

θ = 0.48°

n = 1.25

When the arrangemnet is immeresed in the liquid, then the wavelength of light is chnaged.

let the new wavelength is λ'

λ' = λ/n  589 / 1.25 = 471.2 nm

λo, the new fringe separation  is θ'

So. θ' / θ = λ' / λ

θ' / 0.48 = 471.2 / 589

θ' = 0.384°

Thus, the new fringe separation is 0.384°.