Question

In a double slit experiment the angular width of a fringe is found to be 0.2° on a screen placed 1 metre away. The wavelength of light used is 600 nm.What will be the angular width of the fringe if the entire experimental apparatus is immersed in water refractive index of water is 4 / 3

Answer 1

Given that,

Distance of the screen from the slits, D = 1 m

Wavelength of light used, λ1 =600 nm

Angular width of the fringe in air, θ1 = 0.2°

Angular width of the fringe in water = θ,

Refractive index of water, µ = 4/3

Refractive index is related to angular width as: µ = θ1/ θ2

θ2 = 3/4 θ1 = 3/4 x 0.2 = 0.15

Therefore, the angular width of the fringe in water will reduce to 0.15°.

Answer 2

Answer:-

We know that:-

→ Distance, D = 1m

→ Wavelength, $\lambda_{1}$ = 600 nm

Angular width of the fringe in air $\theta_{1}$ = 0.2°.

Angular width of the fringe in water = $\theta_{2}$

Refractive index of water, $\mu = \frac{4}{3}$

$\mu = \frac{\theta_1}{\theta_2}$is the relation between the refractive index and the angular width.

$\theta_2 = \frac{3}{4}\theta_1 \: \frac{3}{4} \times 0.2 = 0.15$

Therefore, 0.15° is the reduction in the angular width of the fringe in water.