In young's double slit experiment, the fringe width is 12mm . If...

For a given light wavelength corresponding a medium of refractive index $\mu$

${\lambda}_{med} = \dfrac{{\lambda}_{vacumm}}{\mu}$.

And we know that fringe width $\beta = \dfrac{\lambdaD}{d}$

Therefore, ${\beta}_{med} = \dfrac{{\beta}_{vacuum}}{\mu} = \dfrac{12}{\dfrac{4}{3}} = 9mm$