In a Young's double-slit experiment the slits are 1.0 mm apart and the screen is 1.0 m from them. The entire apparatus is immersed in water of…

The sheet adds an optical path $(\mu_g-\mu_w)t$ relative to the water-filled path, where $\mu_g=1.5$, $\mu_w=4/3$, $t=24\ \mu\mathrm{m}$: $(1.5-\tfrac43)(24\ \mu\mathrm{m})=4\ \mu\mathrm{m}$. A point at distance $x$ from the centre has geometrical path difference $xd/D$ in water, i.e. optical path difference $\mu_w xd/D$. The central maximum moves to where this cancels the sheet's contribution: $\mu_w x d/D=(\mu_g-\mu_w)t$. So $x=\dfrac{(\mu_g-\mu_w)tD}{\mu_w d}=\dfrac{4\times10^{-6}\times1}{(4/3)\times10^{-3}}=3\ \mathrm{mm}$. The value $12\ \mathrm{mm}$ is the air result $(\mu_g-1)tD/d$, and $4\ \mathrm{mm}$ forgets to divide by $\mu_w$. This corresponds to $8$ fringe widths of $0.375\ \mathrm{mm}$.