How is the interference pattern affected if Young's experiment was performed in still water than in air?

Since the refractive index of water is greater than that of the air, the speed of the light used in the experiment will be less in water than in air. 
Since the frequency of light is the same in water and in air, it follows from the relation $\mathrm{\lambda }=\frac{\mathrm{v}}{\nu }$ that the wavelength $\lambda$ in water is less than that in air. Since fringe width $\mathrm{\beta }\propto \mathrm{\lambda }$, the value of $\mathrm{\beta }$ will decrease. Hence, fringes will be narrower.