In a Young's double slit experiment the slit separation is 0.30 mm and the screen is 1.5 m from the slits. A thin transparent film of refractive…

Immersion changes the effective index of the film relative to the surrounding medium: $\mu=\mu_f/\mu_w=1.5/(4/3)=1.125$. The film adds an extra effective path $(\mu-1)t$ (in units of the surrounding medium) to the beam through the covered slit. The central maximum moves towards the covered slit until the geometrical path difference $d\,y/D$ cancels this, giving the standard result $\Delta y=(\mu-1)\dfrac{tD}{d}$. Thus $\Delta y=0.125\times\dfrac{6.0\times10^{-6}\times1.5}{0.30\times10^{-3}}=0.125\times0.030=3.75\times10^{-3}\ \mathrm{m}=3.75\ \mathrm{mm}$. In air the shift would be $(1.5-1)\times6\times10^{-6}\times1.5/0.3\times10^{-3}=15\ \mathrm{mm}$; immersion reduces it, and the wavelength does not enter the shift. Using the bare index $1.5$ instead of the relative index is the typical error.