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Why do we only consider interference of the first two reflected waves when studying thin-film interference (see attached diagram)? Is there a rigorous treatment that considers the infinite number of reflections and refractions that can occur at two interfaces?

I thought that the third and fourth reflected waves and so on would be negligible, but that doesn't seem like the case, based on the Fresnel equations.

Wikipedia diagram

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  • $\begingroup$ Can you picture what you consider "third and fourth reflected waves" here in the same diagram or in own pencil schematics? $\endgroup$ Mar 23 at 5:12

4 Answers 4

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You are quite correct that in general we need to consider multiple reflections. For example a Fabry-Pérot etalon relies upon multiple reflections to function properly.

However in simple treatments of thin films we are often interested in the thicknesses and/or angles that lead to fully constructive or fully destructive interference. That is, when the path difference is an integral or half integral number of wavelengths. In this case the path difference is integral or half integral for all the reflected rays, so we need only consider two rays because the result we get for two rays applies to all the rays.

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    $\begingroup$ That sounds reasonable, but I'm not entirely convinced yet. Suppose $n_1<n_2$. Then, the phase difference between the first and second reflected wave is $\Delta\phi = k_2 (AB + AC) - k_1 (AD)$, where I've canceled out the $\pi$ phase shifts that each wave encounters. But between the first and third reflected wave, the phase difference is $2\Delta\phi + \pi$, since the third wave was reflected off the ground one more time. Wouldn't this cause consecutive reflected waves to alternate between constructive and destructive interference? $\endgroup$
    – antoine
    Mar 23 at 21:52
  • $\begingroup$ Just wanted to follow up on my previous comment ^. Thanks! $\endgroup$
    – antoine
    Mar 25 at 5:12
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As has been pointed out by @JohnRennie the Fabry–Pérot interferometer is a device which relies on significant multiple reflections between two parallel surface.

However you will note that the condition for a maximum, $2\,n_2\,d\,\cos (\theta_2) = m\,\lambda$ using the symbols defined in the OP's diagram, is not affected by the number of reflections and so it is easier to consider as few as possible.
Also only about $4\%$ of near normal incidence light is reflected from a glass surface and so the intensity of the multiple reflections drops off quite rapidly as the number of reflections increases.

However, a Fabry–Pérot etalon with reflecting surfaces such that multiple reflections are significant is often configured in an interferometer so that the interference of transmitted light is observed.

enter image description here

Multiple reflections reduce the width and increase the intensity of the principal maxima and hence increase the resolution (ability to differentiate between two wavelengths which are close together) of the device as shown below with increasing reflectance $R$.

enter image description here

You may have met the idea that increasing the number of significant reflections results in a reduction in the width of the principal maxima when discussing multiple slit interference and considering the difference between two slits and many slits (diffraction grating)?

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Thin film interference is typically used to create anti-reflection coatings. In this situation (unless the angles allow total internal reflection), there are not multiple reflections. The number of reflections is equal to the number of interfaces.

Also, the ray actually interferes with itself. At each interface, a portion of the ray is transmitted and a portion is reflected (I'm ignoring losses like absorption and diffuse reflectance). The ratio in intensity between reflected and transmitted portions shows the percentages. The intensity of the ray is equal to the sum of its transmitted and reflected portions.

Suppose that light traveling through air encounters a material with index of refraction $n_m$ The transmission of light through the material can be described as

$$I_0 = I_{R} + I_{T}$$ $$I_{T} = I_0 - I_{R}$$ where $I_{R}$ and $I_{T}$ are the reflected and transmitted intensities.

Suppose now, a different material of given thickness and different index of refraction is deposited on top of the original material (anti-reflection coating AR), then the transmission of light through this material can be described as:

$$ I_0 = I_{AR(reflected)} + I_{AR(transmitted)}$$ and then that light is reflected from and transmitted through the material/AR boundary $$I_{AR(transmitted)} = I^{'}_{AR(reflected)} + I^{'}_T $$ Finally

$$I_0 = I_{AR(reflected)} + I^{'}_{AR(reflected)} + I^{'}_T$$

Clearly if the two reflected components could cancel each other out$$I_{AR(reflected)} + I^{'}_{AR(reflected)} = 0$$ then 100% transmission would be possible $$I^{'}_T = I_0$$

This is what AR coating design takes advantage of and how the destructive interference that cancels these reflected waves increases transmission.

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One such rigorous treatment of the multiple reflections is the transfer matrix method, well described on its wikipedia page, or in this article by Steven J. Byrnes regarding a python implementation.

Citing Wikipedia :

The transfer-matrix method is based on the fact that, according to Maxwell's equations, there are simple continuity conditions for the electric field across boundaries from one medium to the next. If the field is known at the beginning of a layer, the field at the end of the layer can be derived from a simple matrix operation. A stack of layers can then be represented as a system matrix, which is the product of the individual layer matrices. The final step of the method involves converting the system matrix back into reflection and transmission coefficients.

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