In a Young's double-slit experiment, light of wavelength 500 nm in air is used, with slit separation 0.5 mm and screen distance 1.0 m. Point O on the…

The fringe width is $\beta=\lambda D/d=\dfrac{500\times10^{-9}\times1}{0.5\times10^{-3}}=1\ \mathrm{mm}$ and is not changed by the film, so (A) is incorrect. The film adds an extra optical path $(\mu-1)t=0.5\times5\ \mu\mathrm{m}=2.5\ \mu\mathrm{m}=5\lambda$ to the path via $S_1$. At $O$ the path difference is therefore an integral number ($5$) of wavelengths, so $O$ is bright (a fifth-order maximum), not dark; (B) is incorrect. In water the wavelength becomes $\lambda/\mu_w$, so $\beta'=\beta/\mu_w=0.75\ \mathrm{mm}$, not $1.33\ \mathrm{mm}$ ($1.33$ would result from multiplying instead of dividing); (C) is incorrect. The zero optical path difference now occurs where the geometric path difference $S_2P-S_1P=(\mu-1)t$, i.e. at $y=\dfrac{(\mu-1)tD}{d}=\dfrac{2.5\times10^{-6}\times1}{0.5\times10^{-3}}=5\ \mathrm{mm}$ toward the covered slit $S_1$ (equivalent to $5$ fringes). So (D) is correct. Only (D).