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

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°.