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

The sheet adds an extra optical path $(\mu_g-\mu_l)t$ in one arm, where $\mu_l=4/3$ is the index of the surrounding liquid, since the same thickness of liquid is replaced. The extra optical path is $(1.5-4/3)\times12\ \mu\mathrm{m}=2.0\ \mu\mathrm{m}$. In the liquid the geometrical path difference at a screen point is $yd/D$, and the corresponding optical path difference is $\mu_l\,yd/D$. The central fringe is where the total optical path difference vanishes, so $\mu_l\,\dfrac{yd}{D}=2.0\ \mu\mathrm{m}$ and $y=\dfrac{2.0\times10^{-6}\times1.0}{(4/3)\times10^{-3}}=1.5\ \mathrm{mm}$, towards the side of the glass sheet. Ignoring the liquid gives $(\mu_g-1)tD/d=6.0\ \mathrm{mm}$; forgetting the factor $\mu_l$ gives $2.0\ \mathrm{mm}$; using half the sheet effect gives $0.75\ \mathrm{mm}$.