A double slit experiment is immersed in a liquid of refractive index 1.33. Separation between the slits is 1.0 mm and he distance between slit and screen is 1.33 m. If slits are illuminated by a parallel beam of light whose wavelength is 6300 Å, then fringe width will be   

0.63 mm 

$ \lambda_l = \frac{\lambda_{air}}{\mu} $

$ \beta = \frac{\lambda_l D}{d} = \frac{\lambda_{air} D}{\mu d} = 0.63mm$