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

0.384°

Step-by-step 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°.

Answer 2

0.38°

Step-by-step 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

`(m\lambda_a)/(m\lambda_l)=(dsin\theta_a)/(dsin\theta_l)\\\Rightarrow (\lambda_a)/(\lambda_l)=(sin\theta_a)/(sin\theta_l)`

Wavelength ratio =
`n`

`n=(sin\theta_a)/(sin\theta_l)\\\Rightarrow (sin0.48)/(1.25)=sin\theta_l\\\Rightarrow \theta_l=sin^(-1)(sin0.48)/(1.25)\\\Rightarrow \theta_l=0.38^(\circ)`

The angular separation is 0.38°