A Young's double-slit apparatus is completely immersed in a liquid of refractive index 4/3. The source emits light of vacuum wavelength 600 nm. A…

Only the optical path difference matters. The sheet replaces a thickness $t$ of liquid, so the extra optical path in the covered arm is $(\mu_s-\mu_l)t=(1.5-4/3)(7.2\ \mu\mathrm{m})=1.2\ \mu\mathrm{m}$. A shift of one fringe corresponds to a path difference of one vacuum wavelength, because the condition for brightness is $\Delta=n\lambda_0$ in terms of optical path. The number of fringes shifted is $1.2\ \mu\mathrm{m}/0.6\ \mu\mathrm{m}=2$. Since the covered path becomes longer, the point of zero net path difference lies on the side of the covered slit, so the pattern moves towards it. Ignoring the liquid gives $(1.5-1)t/\lambda=6$, and using $\mu_s/\mu_l-1$ gives $1.5$; both are wrong.