In a Young's double-slit experiment the whole apparatus is immersed in a liquid of refractive index 1.25. A thin transparent film of refractive index…

The geometric paths from the slits are equal, but the film replaces a liquid layer of the same thickness $t$ in one path. The extra optical path is $\Delta=(\mu_f-\mu_l)t=(1.5-1.25)(12\ \mu\mathrm{m})=3\ \mu\mathrm{m}$. Since one fringe corresponds to a change of optical path of one vacuum wavelength $\lambda_0=600\ \mathrm{nm}$ (the fringe width $\beta=\lambda_0D/(\mu_l d)$ arises from the geometric path difference $d\sin\theta$ multiplied by $\mu_l$), the shift in number of fringes is $N=\Delta/\lambda_0=3000/600=5$. The central maximum moves by $5$ fringe widths towards the slit covered by the film. Using $(\mu_f-1)t/\lambda$ in air units gives the wrong value $10$.