In a double-slit experiment,

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.