When Young's double slit experiment is conducted in water,

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

When the experiment is conducted in water, the wavelength of light changes to $\lambda' = \dfrac{\lambda}{\mu}$, where $\mu$ is the refractive index of water.

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

Since the refractive index of water $\mu > 1$, the new fringe width $\beta'$ is less than the original fringe width $\beta$.

Therefore, the fringe width will decrease.