In Young's double-slit experiment, the fringe width with the light of wavelength $6000 \mathrm{\AA }$ is found to be $4.0 \mathrm{mm}$. What will be the fringe width using light of wavelength $6000 \mathrm{\AA }$ if the entire apparatus is immersed in a transparent liquid of refractive index $\frac{4 }{3}$?

Wavelength in air is ${\lambda }_a=6000 \mathrm{\AA }$. Let its speed in air be $v_a$. When the apparatus is immersed in a liquid, the frequency of the wave remains unchanged but its wavelength and speed both will change. Let ${\lambda }_l$ be the wavelength and $v_l$ be the speed in the liquid. Since  $\mathrm{v}=\mathrm{f\lambda }$, we have:
$$\begin{gathered} f=\frac{v_a}{{\lambda }_a}=\frac{v_l}{{\lambda }_i} \\ \Rightarrow \frac{{\lambda }_l}{{\lambda }_a}=\frac{v_l}{v_a} = \frac{1}{\mathrm{n}} \end{gathered}$$

where n is refractive index of the liquid.

$$\begin{gathered} {\beta }_l = \frac{{\lambda }_{l}D}{d} \\ {\beta }_a = \frac{{\lambda }_{a}D}{d} \\ \Rightarrow \frac{{\beta }_l}{{\beta }_a} = \frac{{\lambda }_l}{{\lambda }_a} = \frac{1}{n} \\ \Rightarrow \frac{{\beta }_l}{{\beta }_a} = \frac{1}{{\displaystyle \raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}} \\ \Rightarrow {\beta }_l = {\beta }_a\frac{3}{4} = 4 \times \frac{3}{4} = 3 \mathrm{mm} \end{gathered}$$