In a Young's double-slit set-up in air, slit separation is d=0.50 mm, screen distance D=1.0 m and wavelength λ=600 nm. A thin transparent film of…

In air the fringe width is $\beta=\lambda D/d=1.2\ \mathrm{mm}$. The film adds extra optical path $(\mu-1)t=0.9\ \mu\mathrm{m}=1.5\lambda$, so the central maximum moves towards the covered slit by $1.5\beta=1.8\ \mathrm{mm}$; (A) is true. At $O$ the path difference is $1.5\lambda$, an odd multiple of $\lambda/2$, so $O$ is dark and (B) is false. In water the wavelength is $\lambda/\mu_w=450\ \mathrm{nm}$, so $\beta_w=0.9\ \mathrm{mm}$ and (D) is false. With the film in water, its extra optical path relative to the surrounding water is $(1.5-\tfrac43)(1.8\ \mu\mathrm{m})=0.3\ \mu\mathrm{m}$, which equals $0.3/0.6=0.5$ of the vacuum wavelength, i.e. a shift of $0.5$ fringe widths. The shift is $0.5\times\beta_w=0.5\times0.9=0.45\ \mathrm{mm}$, so (C) is true.