In a double-slit experiment the angular width of a fringe is found to be 0.2° on a screen placed 1 m away.
Question:

In a double-slit experiment the angular width of a fringe is found to be 0.2° on a screen placed 1 m 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? Take refractive index of water to be 4/3.

Solution:

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

Wavelength of light used, $\lambda_{1}=600 \mathrm{~nm}$

Angular width of the fringe in air, $\theta_{1}=0.2^{\circ}$

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

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

Refractive index is related to angular width as:

$\mu=\frac{\theta_{1}}{\theta_{2}}$

$\theta_{2}=\frac{3}{4} \theta_{1}$

$=\frac{3}{4} \times 0.2=0.15$

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