Fresnel Equations at Normal Incidence

/ edmundsj
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Probably the most important equations in optics, the Fresnel equations allow us to figure out how much of an electromagnetic wave is reflected off a surface - and this gives us immeasurable power to design waveguides, mirrors, and all manner of optical devices.
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.

[@Akshay-xo5iy]

A big round of applause for you sir. I have been following Computational Photonics book by Marek S Wartek and had a hard time understanding the topics as no geometry was involved. Now you have made my life simpler.

[@valeriereid2337]

Excellent! Thanks you.

[@alan262]

The internet states that Fresnel is pronounced "Fray nel". You do a nice job. And don't forget Einstein's statement about simplicity.

[@haya4895]

thank you so! do you have a video explaining the antireflection coating?

[@sablebonnet2104]

You're the best

[@amvin234]

are we missing a factor of n_2 for the E_t equation?

[@infinity-and-regards]

Thanks for the video! So a good mirror would have n2 >> n1?

[@JordanEdmundsEECS]

Exactly! ... or an n2

[@panthadipmaji1433]

Why exponential terms will cancel out? I think there is a term k.z. But for different medium k will be different because k=(w/v),where w is the angular frequency of the wave and v is the velocity of the wave in the medium and w is same but v is different for different medium.

[@Saptarshi.Sarkar]

Why can't we have E_i = E_r + E_t as the third equation? Won't the magnitude of the Electric field be conserved?

[@sebastianteran3750]

Pregunta 1 Hello, I need your help plz, good video.. Estimate the approximate output coupling power losses (in percent) from Fresnel reflection at normal incidence in a phosphor/air interface, where the phosphor has an index of 2.3 and air has a refractive index of 1.

[@fmazzelli5264]

Thank you for this video! Suppose we want to calculate R and T for the second interface, external to which is air again, so that, for instance n2=1,5 and n3=1. Can we use the same equations? It appears to me that T becomes greater than 1 and I do not understand how this can be. Thank you very much.

[@thedarkcider4450]

I think it's a mistake, power transmission coefficient is (1-R^2), the part written around minute 13 is wrong and violates power conservation, he needs no multiply T by n2/n1 I guess if u want to state it that way, but i just use T=(1-R^2), this can be proved when you calculate the power of the incident and reflected wave, power = (E x H*)

[@kangningyu6075]

Thank you, I also think there are some mistakes with transmission coefficient. But it is strange that the lecturer can solve the reflection coefficent correctly by these equations, but not correct for transmission.@@thedarkcider4450

[@sl2357]

Thank you! At z = 0, why time dependence cancels out, please? OK, the phase term becomes a constant?

[@JordanEdmundsEECS]

Yes, when we take ratios of amplitudes the e^(iwt) term will cancel out, so we usually ignore it.

[@sl2357]

@@JordanEdmundsEECS Thank you!

[@RohitSharma-mi8gt]

do u mean the corkscrew rule ?

[@varunpatt]

11:48

[@jerrylg8087]

why Ei=nHi? i think it should be (n/c)Ei=Hi

[@jordanedmunds4460]

The n-looking thing is actually the Greek letter eta (the wave impedance), not the refractive index. Good eye!