Hello, thanks for the explanation. I have a question: Unlike the other phenomena when we calculate the path difference, in the Fabry Perot interferometer there is the minus (2AB-CD) I would like to know why do we use minus?
I have a question about path difference(3:40). Does it mean that I moved as much as the CD during the time that moved from BA to B'A inside the etalon?
The path difference at 3:39 is 2*d*cos(Theta). However, at 4:00 the path difference is written as d*cos(Theta) which is then put equal to m*lambda...shouldn't it be 2*d*cos(Theta)=m*lambda, in order to have constructive interference?
@@harshitabhuyan8892 Sorry for the late answer, but I believe the pattern is circular because constructive interference can only occur at certain angles \theta between the incident ray and the optical axis (the horizontal line). In a 2 dimensional system, you would therefore obtain a typical "on-off" interference pattern, but in a 3 dimensional one this pattern is rotated around the optical axis. The shape of the lens is also a factor, because only a radially symmetric lens will create such a pattern.
WAAAAAY better than Wikipedia.
Mark another one for *daddy Pitagoras*
2:37 best beat switch ever
please explain why the shape of the fringes are circular.
Probably due to the geometry of the focusing lens.
Thank you so much!!!!❤❤❤❤
Hello, thanks for the explanation. I have a question: Unlike the other phenomena when we calculate the path difference, in the Fabry Perot interferometer there is the minus (2AB-CD) I would like to know why do we use minus?
becoz that is the path the first ray travels extra after getting transmitted from plate2
Very nice sir!
I have understood all the things.
I want to request you to upload a video on origin off circular fringes in michelson Interferometer...
how did the path difference became 2dcostheta
Is there any phase change as light travels from rarer to denser medium ie stokes law??
No , there is no phase change as light travels from one medium to other (refraction) , while the phase changes by 180 during reflection.
Why don't you take into account the value of cd?!!
GRACIAAAAASSS!
@Maximilian Marcel did you find anything interesting?
is the second girl Chinese
I have a question about path difference(3:40). Does it mean that I moved as much as the CD during the time that moved from BA to B'A inside the etalon?
ruclips.net/video/D-JbtOSeX9w/видео.html
The path difference at 3:39 is 2*d*cos(Theta). However, at 4:00 the path difference is written as d*cos(Theta) which is then put equal to m*lambda...shouldn't it be 2*d*cos(Theta)=m*lambda, in order to have constructive interference?
Federico Pevere you are right It is 2dCos(x)
can someone tell me why the pattern is circular?
@@harshitabhuyan8892 Sorry for the late answer, but I believe the pattern is circular because constructive interference can only occur at certain angles \theta between the incident ray and the optical axis (the horizontal line). In a 2 dimensional system, you would therefore obtain a typical "on-off" interference pattern, but in a 3 dimensional one this pattern is rotated around the optical axis. The shape of the lens is also a factor, because only a radially symmetric lens will create such a pattern.
nice and simple intro
very useful .............. thank u
Very useful! Thank you \O.O/
thks!
THANK U !
ruclips.net/video/D-JbtOSeX9w/видео.html
good content
Thank you :)
great, thanks !
Thanks sir
Good for beginners
Thanks
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Omg you talk too fast even 0.5 is faster then a normal human talking ugh
ruclips.net/video/D-JbtOSeX9w/видео.html
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