In a Young's double-slit experiment the slits are 1.0 mm apart and the screen is 1.0 m away. The whole apparatus is immersed in a liquid of…

The sheet adds an extra optical path $(\mu_s-\mu_\ell)t$ in the beam through it, where the surroundings have index $\mu_\ell=4/3$. The central fringe moves to the side where the optical path is otherwise shorter, i.e. towards the covered slit, until the geometric path difference in the liquid compensates: $\mu_\ell\,\dfrac{d\,y}{D}=(\mu_s-\mu_\ell)t$. So $y=\dfrac{D}{d}\left(\dfrac{\mu_s}{\mu_\ell}-1\right)t=\dfrac{1}{10^{-3}}\left(\dfrac{9}{8}-1\right)6\times10^{-6}=0.75\ \mathrm{mm}$. With the sheet in air ($\mu_\ell=1$) the shift would be $3\ \mathrm{mm}$.