# Thin film inteference derivation by using wave equations

Currently I'm studying about possible ways to measure the thickness of a thin film through the use of a simple laser beam, by changing the incident angle. I have found a ton of helpful stuff around the web but my main problem relies on my attempt to reach the solution from the scratch.

So, a laser beam strikes the surface and part of it is reflected in the form of R1 in the picture and part of it is transmitted (R2) until it reflects on the second surface of the substrate below the film.

The equation of a wave is $$E=E_0 \cos (\omega t - 2\pi x/\lambda) .$$ Superposition principle states that $$E=E_1+E_2 =E_0 [\cos(\omega t - 2\pi x/\lambda) + \cos(\omega t - 2\pi (x+\Delta x)/\lambda)]$$ where $\Delta x$ can be derived from the geometry of the picture.

If I try to carry on from this point, everything is smooth and fine, BUT one thing that is crucial and is missing from the last equation is that the wavelength of the transmitted wave is not $\lambda$, but $\lambda/n_2$ (we're considering $n_1=1$ here). So the last equation eventually becomes: $$E=E_1+E_2 =E_0 [\cos(\omega t - 2\pi x/\lambda) + \cos(\omega t - 2\pi (x+\Delta x)n_2/\lambda)] .$$ If I carry on with the current situation by using the trigonometric identity $$\cos A+\cos B=2\cos((A+B)/2)\cos((A-B)/2) ,$$ things get pretty ugly and I can't reach the point where my intensity would be proportional to $4 E_0^2 [\cos(\Delta x \pi/\lambda)]^2$ as would happen if my $\lambda$ was the same. I'm really puzzled and I can't find anything related on the net. Any help? Thanks in advance and sorry for my bad english and the horrible presentation.

The interference occurs in the exterior region, where both wavelengths are $\lambda$. The wavelength of the light is only equal to $\lambda/n_2$ within the thin film.
$$\delta = \frac{2\pi}{\cos(\theta_2)} \frac{2d}{\lambda/n_2} = \frac{4\pi d n_2}{\lambda \cos(\theta_2)}$$
Additionally, whether or not there is an additional $180^\circ$ phase shift at due to the reflection at the bottom interface depends on the refractive index of the layer below.