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

$\lambda t=\frac{\lambda_{\text {air }}}{\mu \mathrm{l}}$

$\beta=\frac{\lambda \mathrm{D}}{\mathrm{d}}$, in liquid  $\beta=\frac{\lambda_1 \mathrm{D}}{\mathrm{d}}=\frac{\lambda_{\text {air }} \mathrm{D}}{\mu \mathrm{ld}}$

$=\frac{6300 \times 10^{-10} \times 1.33}{1.33 \times 10^{-3}}$ = 0.63 mm