# Young's double slit experiment with a thin film covering one slit

In the Young’s double slit experiment, one of the slits is covered by a thin film of thickness $d$ and refraction index $n$. I want to determine the phase difference of the waves coming from each slit:

$\psi_1=\psi_0cos(\omega t-kr_1+\phi_1)=\psi_0cos(\Lambda_1)$

$\psi_2=\psi_0cos(\omega t-kr_2+\phi_2)=\psi_0cos(\Lambda_2)$

$I=4I_1cos^2(\delta/2)$

$\delta=\Lambda_2-\Lambda_1=\frac{2\pi}{\lambda}\Delta r+(\phi_2-\phi_1)$

But, having a thin film of refraction index n in front of the second slit, I don’t know how to determine this phase difference. I have searched about this and I found that it would be given by

$\delta=\frac{2\pi}{\lambda}[r_2-(r_1+t(n-1))]$

Where does this formula come from?

The thin film causes light to travel more slowly through it: $$c/n$$ instead of $$c$$. The time it takes to get through the film is $$t = \frac{d}{c/n} = \frac{dn}{c}$$. In that same time, light through the other slit has traveled a distance $$x = ct = dn$$. The difference in the distances traveled ($$d$$ vs. $$x$$) in terms of the wavelength gives you the initial phase difference (where $$2\pi$$ represents a full wavelength): $$\delta = \frac{2\pi}{\lambda}(dn - d) = \frac{2\pi}{\lambda}d(n - 1)$$ Once you have this, the rest of the phase difference comes from the difference in distances from the two slits to the image plane ($$r_1$$ and $$r_2$$ in the last equation). In your last equation, the letter $$t$$ is the thickness of the film, rather than $$d$$ in your initial problem.