But it is also slightly untrue. This means that if you go to infinity you cannot tell whether there is odd or even number of wavelet sources. The real reason for diffraction pattern being based on these equations is because average distance of all sources to a point on the screen is equal to the distance of a single point in the very center of the slit to the point on the screen. If you have an access to Crawford's "Waves" there is full explanation.
@@Great_Himalayan_ASMR cuz I remember not understanding diffraction no matter how much I read it and I loved wave optics.... After I found the meaning of diffraction, I can't help but see beauty in it
For everyone wondering this result, this is just an approximation for maxima you need to solve (tan beta= beta) where beta is pi b by lamda into sin theta
Yes I'm pretty sure you found this using the resultant wave eqn and a resultant intensity equation . In which case do know of any way to approximate to this result from that equation . Minima part was pretty clear but I can't quite figure out a way to come to this result via that eqn . And in this case is the sin theta = theta assumption the one due to which this ambiguity arises ?
This video does clarify my doubts on how can constructive interference occur when we divided the slit in odd number sections thanks for your work sir, but does all the tiny section left on the slit contribute to constructive interference or just some portion of it?(eg , (1/3)a ) , if it does, how can we prove it?
The formula derived in this video is not exact. It is an approximation. There is no simple expression for single-slit maxima. One must numerically solve a transcendental equation for the intensity profile to find the maxima. The numerical solution is close but not exactly what is shown in this video.
0:23 from where we get the condition for 1st minima. should we memorize it? is this the education system where we learn to memorize without understanding? This is unexpected from khan acedemy👎
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this level of clarity is nowhere found on youtube. Thank you for solving the doubts we didn't even have but needed.
omg vikrant sirr your content is great
But it is also slightly untrue. This means that if you go to infinity you cannot tell whether there is odd or even number of wavelet sources. The real reason for diffraction pattern being based on these equations is because average distance of all sources to a point on the screen is equal to the distance of a single point in the very center of the slit to the point on the screen. If you have an access to Crawford's "Waves" there is full explanation.
im crying,,,this is beautiful
why!
@@Great_Himalayan_ASMR cuz I remember not understanding diffraction no matter how much I read it and I loved wave optics.... After I found the meaning of diffraction, I can't help but see beauty in it
@@joshnirohina318 share ur contact with me
@@Great_Himalayan_ASMRbro what😂
how
Khan academy always saves lives❤❤
For everyone wondering this result, this is just an approximation for maxima you need to solve (tan beta= beta) where beta is pi b by lamda into sin theta
You are absolutely correct by differentiating the i res we get that equation but it is impossible to locate those points.
Yes I'm pretty sure you found this using the resultant wave eqn and a resultant intensity equation . In which case do know of any way to approximate to this result from that equation . Minima part was pretty clear but I can't quite figure out a way to come to this result via that eqn . And in this case is the sin theta = theta assumption the one due to which this ambiguity arises ?
You earned a big thanks brooo😊
Saved my life fr
ok
This video does clarify my doubts on how can constructive interference occur when we divided the slit in odd number sections thanks for your work sir, but does all the tiny section left on the slit contribute to constructive interference or just some portion of it?(eg , (1/3)a ) , if it does, how can we prove it?
you have shown a maxima at 3lambda/2 but shouldn't destructive interference lead to a minima instead of maxima
thats cuz there's still one part that does not destructively interfere and hence does not get cancelled and so contributes to the maxima
but why that one part@@joshnirohina318
@@joshnirohina318dude tell me the same about lembda, why lemda on minima, path difference.
How about if n = 0, so a.sin(theta) = lamda/2 --> is this the maxima at the center?
Yes but n is a natural no. and starts from 1 so it cannot be 0 :)
Just wanna say to WOW
Thank you so much sir
The formula derived in this video is not exact. It is an approximation. There is no simple expression for single-slit maxima. One must numerically solve a transcendental equation for the intensity profile to find the maxima. The numerical solution is close but not exactly what is shown in this video.
Isn't "n lembda" the condition for constructive interference?
thats for ydse...this is diffraction...u may ask why and thats exactly wat this vid explains
i love you
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Love u too bro
0:23 from where we get the condition for 1st minima. should we memorize it? is this the education system where we learn to memorize without understanding?
This is unexpected from khan acedemy👎
All you have to do is watch previous videos..👎👎
Please refer to previous videos
1:54 how path diff is 3lamda/2
Please try to speak in hindi