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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?

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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.

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  • $\begingroup$ Ok, thanks, I see. But then we are supposing that the light from the slit is perpendicular to the surface of the film, don’t we? And then this would be only acceptable for the case the light from the first slit goes perpendicularly to the plane of the screen... $\endgroup$
    – Quaerendo
    May 18, 2018 at 9:34
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    $\begingroup$ @Quaerendo That's why they specify a "thin" film, so the angle and added distance don't matter. In practice, these films would be microns thick. $\endgroup$
    – Mark H
    May 18, 2018 at 9:56
  • $\begingroup$ @Mark H For thin films the angle of incidence does matter. $\endgroup$
    – my2cts
    Apr 19, 2020 at 9:51

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