The Fabry Perot Interferometer

[@CognitiveEvolution]

Thank you. I have watched several videos on the FB interferometer, but this is the first one that clearly explained that and how all the light is transmitted out the mirror opposite the entrance mirror.

[@family-accountemail9111]

Same here the explanation here have me the eureka moment of understanding that's like a drug ! I want more!

[@nkeiruu7866]

This is an awesome explanation in plain English.Thank you so much from the USA

[@nohaivce2614]

awesome, super well explained and introduced thru a practical POV as well! thank you!

[@Bert0ld0]

Thank you! You are very good at explaining things.

[@khatrnafar]

Thank you for this clear and concise explanation.

[@prateekjawalkar4112]

Thks q soo much sir love and respect from India

[@amersaad8939]

Thanks you professor for this wonderful course

[@karthikknarayanan776]

Great explanation, thanks for making this!

[@dhruvg550]

That's such a beautiful concept!

[@shahnoorali1904]

Excellent Explanation

[@JJZB9]

So helpful, thank you!

[@benjaminsteinberg8975]

Thank you for this excellent presentation. I'm looking forward to your next one, is there a way I could follow you on youtube?

[@jonathanpilgram1119]

MARTIN!!! Goed filmpje.

[@safdarsafdari7289]

two questions: does from where input light enters light reflected end transmitted from left mirror has 2 times intensity? in confocal interferometer light transmitted from right mirror doesn't diverge?

[@superbialm]

Thank you so much!

[@MaratBikmullin]

Спасибо!

[@samantahira8692]

What is nanoparticles based on Etalon?

[@FiduzP88]

Question: why do you have to slightly focus inside the confocal FP cavity? Is it possible to input just a narrow collimated beam? Thanks!

[@cornekoks5211]

Yes, that is possible! The amount of light you couple into the FP is however proportional to the mode overlap between the incoming beam and the cavity eigenmode. A slightly focussed beam has a larger mode overlap than a collimated beam. :)

[@sudipnayak1188]

Thanks a lot

[@lambda4931]

Very informative

[@sanampallegowtham7571]

excellent but why dont you have more views

[@ElPrestigo]

If 100% of the signal is transmitted, then 0% is reflected. However when I try to do the calculations I end up with the wrong answer. How would the calculation for reflection look like? Sadly I cannot show my calculations, because if I do the comment gets removed.

[@ElPrestigo]

Only if I take the reflection from outside to inside the cavity (r) different from the reflection that occurs inside the cavity (r') with r = -r', then it does add up to zero. Is that true and why?

[@ElPrestigo]

@@DW-iq8lt I would say that it is precisely what 100% transmission means. Simply put, everything that goes in on one end comes back out the other end. And this is also what the professor in the video is showing and several times stating. Why do you think otherwise and how would you then define 100% transmission?

[@DW-iq8lt]

My mistake. You were right about the transmitted percentage. And at resonance, reflected intensity is zero.

[@jaredjones6570]

The magic of interference :)

[@therealkeshavsehgal]

Good one Thanks

[@SaiPraneethMunjeti]

and that's the magic of interference

[@hanaizdihar4368]

Cant read the green writing , its so well explained but cant read

[@therealkeshavsehgal]

Fyi I'm just a beginner so it's a little high for me

[@anujKumar-vs2cy]

Sir bsc2 mai physics ke important question explain kr dijiye

[@therealkeshavsehgal]

😐😐😐