In a double-slit experiment, 0.2° is found to be the angular width of a fringe 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 $\frac{4}{3}$.
Distance of the screen from the slits, $D=1 \mathrm{~m}$
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, $\mu=\frac{4}{3}$
$\mu=\frac{\theta_{1}}{\theta_{2}}$ 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.