Fresnel Equations Part 2: Deriving the Equations

[@anthonyw4496]

The most clear derivation I have ever heard!! thank you so much!!!

[@akashpatel6620]

Posted 10 years ago but still teaches better than my professor right now. Thank you doing the explanations so clearly.

[@gillesR337]

Splendid video!! Never saw such a clear and concise derivation in my life!

[@Lugotax]

Grazie mille, una delle spiegazioni migliori che abbia sentito o letto. A volte i libri sono molto approfonditi e uno perde la vera essenza del problema, ma grazie alla tua spiegazione si va subito al punto senza dare per scontato tutto ciò che ci sta intorno. Grazie. Cheers from Italy

[@KieranIAm]

Thank you so much! I'm a 3rd year Physics student and I was having such issue with this, as my lecturer was going in on it at 3rd year University terminology and my Layman mind couldn't get my head around it! This was perfect! Very clear! All the best

[@privonfn7833]

professor tried to explain this to us with shitty powerpoint presentation, no one understood it. This probably saved my course of electromagnetism, thank you so much

[@skinemax]

Extremely well done. The most clear derivation I have ever seen/heard.

[@nashs.4206]

You are a godsend! This has got to be the most clear and intuitive derivation of Fresnel equations I've ever seen. Even better than Griffiths!

[@haaardcoooreee]

Im studying Physics, you have no idea how many times you had saved me! Im very very grateful of your work, i really appreciate it. Greetings from Mexico!

[@DeepakJangra-up8ek]

Only video on the internet... helped me one night before exam thanks

[@anton2956]

Thumbs up for "ive rearranged the camera so that costheta2 is on screen" i love your consistency keep it up

[@anasofiamarulanda5548]

i love you, i've got an optics test and I was sooooo confused, now I understand this thanks to you

[@rohanchauhan9924]

I do not know if you are active here anymore. But thank you for creating this video. Helped a lot

[@revgro]

Another worthwhile video. I like the Mother-in-Law clock.

[@ianmurray1144]

I really appreciate that you took into consideration that the permittivity and permeability constants are different in a medium than in a vacuum. Most of the derivations I've seen for the Fresnel Equations assume mu naught and epsilon naught are their vacuum values.

[@moienr4104]

thank you sir , these are some really hard subjects in Fields and waves , but you teach them like they are nothing.

[@globe7396]

This video distinguishes very well the difference between s and p- polarized light. That helped me a lot :)

[@smrutiswarupbishi6418]

The way you are explaining is very good,sir...cant comment...and thumbs up

[@arijitmukherjee5846]

Thank you so much for the video. It really helped me for my University exams. Congratulations from India! Keep up the good work!

[@araya2812]

Thks teacher!! You made tomorrow's Optic experiment lecture be easy!!

[@adithiyer9732]

Thank you so much! This is out of my sllabus but I am really interested in optics, this video really helped.

[@bobsxu2387]

Thanks for the excellent explanation Dr Eagle!

[@negargh4208]

Thank you .you really make it easy for me to understand .

[@ΟριζόντιοςΚατακόρυφος]

I give a vote for the lecture and two on the clock.

[@IntergrateThisFool]

Thank you. Very clear explanation. After my lecture course I was under the impression that Bt was not continuous.

[@bidaubadeadieu]

I believe I've spotted an error! Your F3, the parallel polarization reflectivity, should have n1cos(2)-n2cos(1) in the numerator but instead it has those flipped. That is to say, your F3 must be multiplied by negative one. I wasn't watching closely enough to see where this error was made, but Griffith's electrodynamics and wikipedia seem to agree with me.

[@Harpercent]

HELP HELP HELP HELP When you say the E fields will cross over the boundary ( at 9:20 ) do you mean the resolved component (that is parallel to the boundary) of the electric field will cross over the boundary? Also, why is it Ei + Er = Et ?? If all of the horizontal (parallel to the boundary) components of the Ei field transfer straight through the boundary unaffected, and non is reflected, then Ei = Et ?! Where does Er come into it?! Another reason why that doesn't make sense to me is that you're ADDING Er. I can understand the Et having some component of the Ei because the wave goes from from Ei to Et. I'd think that the components of Ei are split between Er and Et because of it interacting with the boundary and splitting off into two different parts, does that make sense? so that it would be Er + Et = Ei? HELP HELP HELP HELP

[@unixzee]

Yeah I have the same problem. I feel like it should be Et=Ei-Er

[@benjuliebenjulie7414]

Is there a simple explanation of how the dialectric boundary creates the relfected wave?

[@zengrz]

@6:21 Does E_i have to be on the left? Wouldn't it work out regardless if you are consistent with your definition?

[@abdelrahmangamalmahdy]

+zengrz Not necessarily on the left. it could be on the right as long as you're consistent, but the problem was that the electric fields of the incident and reflected waves had to be pointing to the same direction.

[@Kemaaaaa]

Did I miss something? If intensity of reflected light is equal to r^2, why is not intensity of transimited light t^2 ? It cant be because it would be just 64% plus 4% which is not 100%. I for sure missed something, I just dont know what

[@dpedalv]

Lukáš Herman (Kema) Hello, you must take care about that the speed of light in the medium 2 is different from medium 1, so the value of B vector and therefore the Poynting vector , too. Regards.

[@jameskeeble6981]

Should the vertical component have a unit vector of j? and not i?

[@jameskeeble6981]

ah ignore me I've just moved from the waves section, I'm getting confused with Unit Vectors! :O

[@physicsstories6664]

For parallel part it's n2 -n1. For perpendicular part, it's n1-n2. So negative sign which is phase change is not appearing for parallel part when we consider from air to glass.

[@vanshikasharma2503]

Thanks for this amazing explanation

[@engrriz001]

Very nice and interesting explanation but I have one question that what happened to the vertical component of of E field in boundary conditions ? why we don't use the vertical component boundary conditions ?? Thanks :)

[@muhammadnisarqasim636]

Sir how we can find sin theta brewster angle.Please derive this quantity.

[@physicsstories6664]

Sir... What about parallel part.... Negative sign is not appearing in this case

[@the_golden_one4733]

When calculating how the electric field intensity changes from air to glass, why is it that only the perpendicular E field is considered? How about the E field parallel to the paper? so shudnt the 3rd and 4th fresnel equations be used too ?

[@weinihao3632]

The calculation was carried out for the case theta1=0, thus the EM waves were purely perpendicular to the surface. For the general case you are correct.

[@zackbarkley7593]

Thanks! Is there a way to derive the Fresnel equations using a Lagrangian formalism? as we can derive the path of a ray from the shortest time in a medium with varying "n"...shouldn't there (or is there?) a way to do this with polarization?

[@umrankoca3862]

Thank you for this video.

[@marlonsaveri]

Hi, another question: the energy of the wave shouldn't be considered? I.e., the reflection rate doens't depends on the frequency? I tend to think this "4%" would be different for X-Rays, for example.

[@conan1154]

+Marlon S It does depend on frequency, that's a good observation. The dependence on frequency is hidden inside the index of refraction, "n". This index is a function of frequency, which can be calculated for a given material using the Sellmeier equation. Most of the time at introductory levels, n is assumed to be about 1.5 for a typical glass so as not to make everything too complicated immediately. 4% reflection is based on this assumption, but it would change as n changes.

[@marlonsaveri]

Thank you very much for the video. Just two questions: The reflection rate doesn't depends on the thickness? And... Is it also possible to calculate the absorbed part of the light or that's only a empirical result?

[@Nazo_Darkfire]

Very informative and clearly explained. Helped me a lot in preparation for my optics exam. Question in regards to the speed of light in a media though: you had it as n*E/c. You still got the correct derivation at the end, but isn't the original equation of the index of refraction to be n = c/v? If that's the case, shouldn't the B-Field actually be c*E/n? Edit: Whoops, nevermind. I just realized my own mistake! Forgot to invert c/n since we're dividing by v.

[@MrKlein96]

why are only the parallel components used to derive the equations?

[@hussamqasem3356]

Very elegant. Thank you!

[@TheKradok]

I really liked this video and I feel like I understod the concepts but I'm missing some examples where you actually apply the formulas that you derived.

[@laflaca5391]

At 14:25 why can we assume i=r?

[@DrPhysicsA]

Incident and reflected rays are at the same angle to the normal.

[@donaldmoon3527]

are these equations applicable to snell's law ?

[@DrPhysicsA]

Ultimately yes. Maxwell's laws govern everything concerning electromagnetic waves.

[@chloecrusan2206]

Is this derivation for TE or TM?

[@wolfisr]

First things first - your explanations are very clear , thanks for your effort here! just wonder - what can one deduce out of the two unused boundry conditions (Ke1*Ev1 = Ke2*Ev2 and Bv1 = Bv2 ) ? are they redundant - meaning you get the same Fersnel Eq if you choose to use them instead the other two boundry conditions? or maybe they imply for other meaningful relations? BTW, Ke1*Evi = Ke2*Ev2 is quite similar to Snell's law but not exactly. thanks again!

[@bidaubadeadieu]

Mixed bag! One of them is not relevant to the situation, the other gives redundant information. For instance, if we have parallel polarization, then B points out/into the page, and thus there is no component of B that is perpendicular to the boundary, so we couldn't apply the Bperp condition, it gives no info.

[@johnkovolsky8277]

Excelent video

[@osamuyiuwadia7480]

What happens if the incident wave has a magnitude

[@rinakundu6243]

thank you sir. it helped a lot

[@nevermindshort3]

+DrPhysicsA Thank you

[@sumansarkar9866]

Just awesome, i really enjoyed, thank u.

[@debodyutikar7106]

Tune to 14.28 if u wanna hear the sound of a vampire :-P

[@baraskparas9559]

Beautiful circular argument that does not illuminate why light bends in the denser medium. Maxwell equations are wrong for bou dary conditions. Each photon striking the denser medium will be either reflected, scattered ( reflected at an indeterminate angle) or will experience a one off torque in the direction of the denser medium since it has dimensions. Apart from temp, and impurities only density of silica quartz and incident angle play a part with deviation from the straight ahead path being most relevant descriptively.

[@jonaherran]

Thank you. I really appreciate.

[@claudiomorassuti]

Thank you... you're the best!

[@raseenak6096]

Thank you sir...

[@alexleviyev]

Dat ending

[@bbqsq8150]

Nice clock! ;)

[@nickneutrino3405]

Thanks sir

[@ukstudy_tourism_guide]

good step

[@akankshasharma4277]

Thanks

[@fernandoantunezantonio3450]

That's a nice clock

[@KrabTrilogie]

Thank you!

[@jacobvandijk6525]

@ 12:16 The speed of light in "a media", hihi. Yep, the Romans didn't waste too much of their time in Britain.

[@jameskeeble6981]

Subscriber 141,413!

[@jimdogma1537]

I suppose I should have watched this video first before making my so-called prescient comment about the alarm clock :-/

[@iqbalazis6520]

can you replace my prof :(

[@คําสอนหลวงพ่อฤาษีลิงดําวัด-ฬ1ส]

LOLs tha cow!