Fraunhofer Diffraction Explained

/ edmundsj
If you want to see more of these videos, or would like to say thanks for this one, the best way you can do that is by becoming a patron - see the link above :). And a huge thank you to all my existing patrons - you make these videos possible.
In this video, I describe the process of Fraunhofer diffraction (also known as far-field diffraction) in terms of the Fourier Transform and Fourier Optics. I go over the assumptions that underlie Fraunhoffer diffraction (both the paraxial approximation and the small-aperture approximation), and give the mathematical form that it takes.
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.

[@adrianoseresi3525]

When I see j being used as the imaginary unit I get visibly ill. But thanks for the video.

[@sanjaythorat9914]

@Jordan Edmunds, This is really a nice video, but there seems to be something missing after @10:24 and @12:54. Can you please correct it? I think it's too important to be missed. I will be grateful if you could correct it.

[@Hubieee]

Thanks alot for the video. Sadly there are some gaps in between that leave out probably simple stuff but in the end after 1 or 2 minutes are skipped it just makes the overall very good video a little fishy. I assume for the last part kx you too the paraxial approximation where the sin(theta) = kx / k and tan(theta) = xs / d, and sin ~ tan so kx/k ~ xs/d so kx = k*xs/d.

[@zacharythatcher7328]

It took me rewatching several times to figure out why the aperture equation comes out of nowhere at the end, so here is the explanation. The aperture equation is assumed to be a decimal representation of the percentage of light that can go through. 1 being 100% and .5 being 50%, as an example. It essentially represents the density (intensity) of point sources at the plane of diffraction, because it is a proportional multiplier on the point source at the infinitely small point in the integral.

[@sungbeom8796]

omg this explanation sould've been part of the lecture thanks!

[@jacobvandijk6525]

Jonathan can't teach. He is just showing off. Or too dumb to realize what he is doing.

[@bendaknee8640]

Great video, my only suggestion would be to try and be a bit more organized? It's still comprehendible nonetheless.

[@JordanEdmundsEECS]

Omigod yes you are so right it’s not even funny. These few Fourier optics videos are some of my worst in that regard.

[@jacobvandijk6525]

A bit??? Hahaha!

[@nenntmichbond]

The older songs were better

[@yogitshankar6348]

The portion where you explain why is it called Fourier optics has been edited out from the video

[@codynelson2633]

Great videos Jordan. However, I think you should consider updating this one specifically for better continuity. Being the first video in the playlist, I wasn’t looking forward to any of the others. However, I was pleased when they were not choppy. Updating this one will entice people to continue the playlist. As you can see the views here are an entire order of magnitude greater than the rest.

[@victorkislovsky40]

This is a really great video. If I am not mistaken, starting from 10:24, there should be a factor 1/2 in the power of exp outside of the diffraction integral.

[@chenlecong9938]

not really,evened out with the 1/2.but I’m lost anyway

[@victorkislovsky40]

@@chenlecong9938 Why are you lost? Can I help you?

[@chenlecong9938]

@@victorkislovsky40 well,at 3:00 he sad that the sperical wave is proportional to the e^jkr/r.But that's really ambiguous,did the meant by the,say,electric field of the that sperical wave,or the radius?I don't get it,there ain't any physical significance of saying a "sperical wave is proportional " to something,i supoose…But after all,where does that propotionality relation even come from?Is it experimental observation?Or theoretical calculation?Where can I get reference from?

[@victorkislovsky40]

@@chenlecong9938 Let's remove the ambiguity. The monochromatic (single wavelength!) and with singular polarization Electrical Fields E is proportional to e^jkr/r. This is actually a solution to Maxwell equations that can be reduced to Nabla^2(E)=delta_function(r=0) in 3D space (not 2D!) with a pointwise source placed at r=0. This solution is, by definition, also a Grin function. Additional assumptions: no free electrical charge or currents, only vacuum. More questions?

[@sagibruck6788]

there is a problem the expponent inside is exp(-0.5*jk/d(x_s-x)^2)-you took the x_s and took it out'you should multiply in 0.5 but tjats not my problem'my problem is that exp(-0.5jk/d*(-2x_s*x) is exp(jk*x_s*x/d) for foriour transform you need a minous in the exponent, your lesson is good but i just dont understand till the end

[@eklhaft4531]

It's kinda scarry and beautiful at the same time that this video might make a difference between me getting and not getting my masters degree 😂 Anyway very nicely explained. Thanks 👍

[@adlib8096]

Lost me in the first few secs but had to watch the whole thing anyway😂. Happy to see that those integrals i had to learn in school seem to have some use somewhere👍🏻😱😂😂😂

[@gokulkrishnan486]

Great video, thanks. But after 12:23, you have directly included g(x) in the equation. If you can explain that how you included g(x) it would be helpful. Thanks

[@MegaChilisauce]

there is a seperate video about it (i guess that´s the reason for jumping the explanation): kzbin.info/www/bejne/a2qXpWV-esiKpZI

[@4m0nr3]

Thanks, The ending is confusing due to reverse movie clips that are not in order it seems... also an equation editor would help.

[@JordanEdmundsEECS]

Yup, this is certainly not my finest work in therms of organization. I have started making actual written outlines and following them to improve that. Thanks for the feedback!

[@XiulianShan]

good vedio except some critical knowledge missing in the end

[@sungbeom8796]

I had several challenges but the comments below totally helped. You got great subscribers :)

[@Diana-he5rl]

It feels like the fragment on 10:25 - 10:40 is falling from the general story line. Can you explain the connection in this sudden jump between trigonometrical derivation of kx, Fresnel integral and further assumption you've made afterwards. I really struggle to see the connection.

[@AirborneLRRP]

Yeah, something must have gotten screwed up on the editing

[@ViceroyoftheDiptera]

There are ads every 3-4 minutes. Really brings down the enjoyment / learning aspect.

[@JordanEdmundsEECS]

Yeah, I agree, I don't love them either. Removing them is my next goal on Patreon: www.patreon.com/edmundsj

[@ViceroyoftheDiptera]

@@JordanEdmundsEECS i appreciate that you are trying to make some money, but there is a balance and you are putting people off (such as myself) from subscribing due to the number of ads. This is something you have some amount of control over. If I were you, I'd lower the ad count, as it is in fact counterproductive right now.

[@Deniz-le9xp]

Great video, my professor really left me hanging because he didn't bother to explain how he came up with the fourier part. Thanks for helping

[@alexandermuller8858]

Interesting. I never thought about that. The intensity distribution of the double slit experiment I did in school is nothing but the Fourier transform of the sources. This is somehow equivalent to the far field of a dipol antenna. We approximate the distance and the phase of the greens function with frauenhofer, and the resulting integral is also a Fourier transform in space.

[@onizhang2748]

Thanks for your videos! Your drawings + explanations are super helpful

[@TheSakr289]

What would be the analytical calculations (via aperture function) for double slit diffraction?

[@asmaa.ali6]

Professional!... thanks

[@sammyapsel1443]

hey, the video kinda jumped at the end. can you briefly explain those last minutes?

[@ANJA-mj1to]

Brilliant application of Fourier theory to Fraunhofer diffaction problem and interference phenomena genereally. As you simplify Fraunhofer diffaction we can assume all apertures bounding the transparent part of the surface as rectangulat and of lenght unity propendicular to the plane of the diagram.

[@sourabhsharma7538]

Which is sort of fronshofer diffraction... 😂 such a comical definition is needed to start up the topic!

[@grandaurore]

comparing 12:52 and 12:55 , why the sign of factory change?

[@ubbowiersema5300]

what is the meaning if capital R in what comes out of the slit? why can it later be written as small r?

[@Neamhain97]

why is it e^{-jkr}? is j the unit vector? where is i?

[@chriscen2774]

Thank you for providing us this impressive and high-quality video.

[@sebas12]

Hi Jordan, nice video!. Could you share which software and tablet are you using? Planning to do similar videos on maths

[@GiftJordawe-gz1xr]

Great but it's too complicated,😊

[@ksviety]

I only wanted to listen to the music, not this...

[@mp3lwgm]

If the structure is truly a slit, then you don’t want points, but here rather cylinders with differential cross sections. The field from each goes ~ 1/r^(1/2), and not ~1/r.

[@JordanEdmundsEECS]

Yup. You are correct, the goal here was to build the framework that would be used in 2D. Should have clarified this, thanks!

[@nesslange1833]

Thanks for sharing your insights. My only questions is: when you're Fourier Transforming a function of time, you get a function of Frequency. Here we're Transforming a function of position getting a function of what? Wave Vector?

[@gabrielsommer2136]

Yes! The spatial FT is a function of the wavevector!

[@henrytoussant9385]

So at the end of the video, the kx is just the spatial frequency of the Fourier component of that direction?

[@JordanEdmundsEECS]

Precisely.

[@CohenSu]

Amazing! so, the slit, acts like a fourier transform of the optic function and then acts on the wave function. so are other optics, CTF funtions.

[@ANJA-mj1to]

If I can give some comment to help. You can look "optical" path as (n) x (geometrical path) , the passage of light through a distance (x) in a medium of refractive index (n) introduces an extra path in the air or vacuum. So you can get variation of phase insted of transmission across the aperture, so the function is complex. In prism (angle) and the aperture with (a), the thiclnedd of the prism at its base is (a) tan(angle) and so if parallel wavefront coming from minus infinity and passed through the prism, the phase that we can choose to be zero at the centre, at the apex and the base of the prism are 0 and add you can write basic function. As well by describing the Huygens walvet you can get The Fourier transform - integrate and multiply the amplitude distribution. I see yours comment where you see the identical diffaction pattern.

[@derasaderasa7058]

👍🏻👍🏻👍🏻👍🏻

[@cmdrbobert9862]

you lost me once you started talking about what (k*d) "could be". It sounded like a phase difference?

[@JordanEdmundsEECS]

Mmm yes, this. set of videos was not my clearets. k*d is indeed the accumulated phase.

[@cmdrbobert9862]

@@JordanEdmundsEECS Excellent, thank you so much for the reply.

[@auwaluidi5374]

Sir, please I don't understand how you arrive at: Kx= K Xs ÷ d

[@nesslange1833]

look carefully: tan(θ) = xs / d sin(θ) = kx / k for small angels tan(θ) ≈ sin(θ) hence kx = k xs / d

[@altugcanturk7939]

Thank you for the well explained video but ı m kinda new in that so could u explain why the exp term has negative power instead of positive?

[@JordanEdmundsEECS]

Great question, this is something that confuses even experienced people. It has to do with the convention you use for a rightward-traveling plane wave. You can say it is exp(jwt - kz), as in these videos, and is pretty standard in E&M and EE, or you can say it is exp(jkz - jwt), as is standard in much of physics. They are both physically equivalent, but when you drop the time term, the kz terms have opposite signs. It doesn’t matter which you use, you just have to be consistent. In optics, one convention will lead you to take the FT of the field, the other will require the inverse FT.

[@comment8767]

grrrr8

[@eastofthegreenline3324]

Great video series! Is there a good lab set-up to demonstrate the --|--|-- sin x transform pair?

[@JordanEdmundsEECS]

Hmm probably multi-slit diffraction, perhaps something like this: physicscourses.colorado.edu/phys2020/phys2020LabMan2000/2020labhtml/Lab5html/lab5.html

[@eastofthegreenline3324]

@@JordanEdmundsEECS First, thank you again for these very helpful videos.. After some foraging I found a similar answer in Hecht at p. 513 which is essentially your answer. There does not appear to be anything better nor--after staring at a lot of patterns--would I have expected it. Thanks again!

[@nagarajprasad3111]

which software do u use to create such videos

[@JordanEdmundsEECS]

I use OBS studio with my iPad connected via Duet Pro and use Sketchbook to draw stuff.

[@optiondrone5468]

Excellent work on explaining Jordan, keep up the good work

[@jakobmessinger8435]

simp

[@jacobvandijk6525]

What a chaotic mess!

[@iamavoidtrippergutterslush666]

How about the idea that different entities and spiritual bodies are real and can only be perceived within different spectrums of light that we humans can't even receive?..... This ties into CERN and parapsychology as well. Or do you just think thats just preposterous??..... Curious about this hypothesis. Ty. \,,/

[@iamavoidtrippergutterslush666]

Receive = Perceive You know.... ;)

[@iamavoidtrippergutterslush666]

And... Have you ever listened to the band Fraunhofer Diffraction? If not, check out their track called, "Forever. ". Really really awesome stuff serious. 3;) \,,/

[@adlib8096]

Quantum mechs goed alot further than that idea alone