A Young's double-slit apparatus with slit separation 0.40 mm and screen distance 1.5 m is completely immersed in water of refractive index 4/3. A…

The film adds extra optical path $(n_f-n_w)t$ in the path of the slit it covers, because the same thickness $t$ of water would otherwise be traversed there. The central fringe moves towards the covered slit to the point where the geometric path difference compensates this. In water the geometric path difference at a screen position $y$ is $yd/D$, with optical path difference $n_w\,yd/D$. Equating, $n_w\dfrac{yd}{D}=(n_f-n_w)t$, so $y=\dfrac{(n_f-n_w)\,t\,D}{n_w\,d}$. Substituting, $y=\dfrac{(1.5-1.3333)(7.0\times10^{-6})(1.5)}{(1.3333)(0.40\times10^{-3})}=\dfrac{1.75\times10^{-6}}{5.333\times10^{-4}}=3.28\times10^{-3}\ \mathrm{m}$. The shift is therefore $3.28\ \mathrm{mm}$. Using air-film values $(n_f-1)$ would incorrectly give a much larger shift. Because the setup is in water, the fringe width is $\lambda_0D/(n_wd)$, but the wavelength never enters the shift, which depends only on the optical path introduced by the film.