Thin-Film Interference using Fresnel Equations

/ edmundsj
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Ready for some advanced-level thin-film interference? This video walks through interference in a thin film using Fresnel's equations, and making absolutely no assumptions, I show what conditions must be satisfied for the wave to be completely transmitted (a perfect anti-reflection coating).
This is part of my graduate series on optoelectronics / photonics, and is based primarily on Coldren's book on Lasers as well as graduate-level coursework I have taken in the EECS department at UC Berkeley.
Hope you found this video helpful, please post in the comments below anything I can do to improve future videos, or suggestions you have for future videos.

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[@danielribastandeitnik9550]

There is something weird when you set the amplitude of the total reflected wave to zero. What you normally want in an interference is that the norm squared of the field, which is proportional to the intensity, be zero (or a minimum since sometimes zero is not possible) for the total destructive case. Interference is all about absoluting squaring the fields interfering, because it manifests itself on the crossterms.

[@learnUpready]

Thank you so much for making such videos

[@alexisphillips471]

How can these euqations be related to Bragg reflection in cholesteric liquid crystals? Specifically trying to derive why/how there is 100% reflection efficiency when the material is uniaxially mechanically strained.

[@dhiagarbaya5283]

A question please: If we want to get the reflectance as a function of the wavelength (to see the effect of the ARQ on other wavelengths for example in solar cells) we will calculate the real part of the complex amplitude (divided by the initial one) of the total reflected wave and then take the square. It works. But, should we worry about the variation of indices according to lambda (Cauchy's law)? Or just the variation of delta-phi is to take into consideration (delta-phi = 2pi.(2L)/lambda ) ?

[@SmileyNyan]

How can e^-(j2phi) = -1 at 11:00. I thought that expression always yield a 1. Even if I consider that the phase can be out of phase with phase = -pi, it won’t came out with -1 Edit : Nvm, I can see a phase of -pi/2 can result in -1 but I still dont understand. I think better question to ask is why do you put the e^…. to be equal to 1 and -1?

[@dhiagarbaya5283]

Are the two conditions (on L and on n1) equivalent ?

[@dhiagarbaya5283]

Oh no! They're not. Indeed, The first condition (on the thickness L) assure the destructive interference but the amplitudes of the reflected waves aren't necessarily the same . Consequently, the interference results on an attenuated but non-null wave. We won't get a totally destructive interference (nothing reflected) unless we impose the 2nd condition on n1. Thanks... It was really helpful !

[@serdaraliandrnloglu3220]

you should add in your title that calculations is for s polarized light! and s-pol picks up an additional 180 phase shift upon relfection from air to glass. So I am assuming this is all glass to air reflections is that right?

[@danielribastandeitnik9550]

Notice that he assumes a normal incidence. Therefore, the light is necessarily s-polarized since the polarization of the electric field will be parallel to the boundary by construction. If he had considered the general case with the incident light making an angle theta, than yes, he would have to make a different calculation for the s and p-polarization cases. Nevertheles, I think the normal incidence case is more important.

[@serdaraliandrnloglu3220]

@@danielribastandeitnik9550 I think you meant that at normal incidence, frenel equations are same and air to glass or glass to air transmission doesn't give reflected wave a pi phase shift. I don't get why did you say normal incidence require s-pol.

[@danielribastandeitnik9550]

@@serdaraliandrnloglu3220 The incident EM field is coming orthogonal to the boundary, therefore the wave-vector is normal to the boundary and the electric field must be parallel to the boundary since it must be orthogonal to the wave-vector. The definition of s-polorized lighe is "light whose electric field is normal to the plane of incidence is called s-polarized". Note that for a normal incident EM field the plane of incidence is kind of ill-defined since there are infinite candidate planes. Nevertheless, for a given electric-field, there is one plane of incidence for which it is normal, therefore it is a s-polarized beam.

[@RohitSharma-mi8gt]

how do we account for the interence of reflected waves ?

[@soumenkundu8394]

Why are you only considering "S polarized " reflection and transmission coefficient in the calculation?

[@JordanEdmundsEECS]

You can do it for p-polarized light as well (I believe I do in one of the next videos). S-polarized light is just easier to deal with and fortunately it’s also more common to find in practice.

[@SePidEh2024]

honestly I could not calculate how r01 = -r12 ,

[@huonghuongnuquy7272]

can we apply this in the case of multi structures ?

[@JordanEdmundsEECS]

Yup! But if you try to do it with infinite sums it gets pretty complicated, and so most people use the Transfer Matrix method for this (I have a couple videos on that as well).

[@huonghuongnuquy7272]

@@JordanEdmundsEECS i saw your videos, i have one question. In the matrix method, for the case of the incident light have a certain incident angle, how can i do ? I imagine that i have to use the frenel equations in the case of incident angle instead of the frenel equations in the case of normal angle. But i don't know for the optical length, we have to change something or not. Thank you.