In Young's double slit experiment, the fringe width is found to be $0.4 \mathrm{~mm}$. If the whole apparatus is immersed in a liquid of refractive index $\dfrac{4}{3}$ without changing geometrical arrangement, the new fringe width will be

The fringe width $\beta$ in Young's double slit experiment is given by the formula $\beta = \dfrac{\lambda D}{d}$, where $\lambda$ is the wavelength of light, $D$ is the distance between the slits and the screen, and $d$ is the distance between the two slits.

When the apparatus is immersed in a liquid of refractive index $\mu$, the wavelength of light changes to $\lambda' = \dfrac{\lambda}{\mu}$.

The new fringe width $\beta'$ becomes $\beta' = \dfrac{\lambda' D}{d} = \dfrac{\lambda D}{\mu d} = \dfrac{\beta}{\mu}$.

Given $\beta = 0.4 \text{ mm}$ and $\mu = \dfrac{4}{3}$, we calculate the new fringe width as $\beta' = \dfrac{0.4}{4/3} = 0.4 \times \dfrac{3}{4} = 0.3 \text{ mm}$.

Answer: 0.30 mm