Diffraction grating | Light waves | Physics | Khan Academy

[@SAKA701]

Grate video, must say.

[@chrisbuckle6930]

Aly Azeemi great*

[@SAKA701]

Chris Buckle Ik dude, it was a pun :p diffraction grate-ing. YES IM THE LORD OF ALL BAD PUNS.

[@divyasingh9291]

Aly Azeemi ikr😂😂

[@SAKA701]

TOP COMMENT WOWOWOOW, FeelsGoodMan :D

[@the_golden_one4733]

Dont grate on my nerves

[@wooluke7144]

Wow. Very Well Done. All Capitals Letters.

[@LanNguyen-iu3oe]

Thank you so much! your videos on diffraction and the slit experiment really helped me fully conceptually understand the process.

[@malezacaminante9577]

This videos are just amazing!

[@thesofakillers]

amazing teaching

[@nelsonbaek3494]

Why can't you use the same logic to find maxima in single slits.

[@bboy14]

Khan academy needs to keep you on board, your explanations and commentary style is a treat!

[@Cooltaha]

such a clear and concise voice

[@fluffzxx9615]

I don’t get why in the normal double slit it would be smudgy in the first place, why?

[@albertvandrejer5003]

@@fluffzxx9615 because for a large numer of slits, the condition for constructive interference of all the superimposed waves is only given at very specific positions. For a double slit what you see is just the interference pattern of two waves, which will have areas of constructive interference, destructive interference and everything in between.

[@williamstephenjones3863]

Agreed

[@kushalchakrabarti240]

4:03 .. how about a third hole? This is where it gets interesting... LMAO

[@expectopetronum3851]

😂😂😂😂😂

[@abdallababikir4473]

Brother,😂

[@AbdulHamid-uw2oe]

Mouth..vagina...&...... The thid....

[@mr.j_krr_80]

10:55 "in between you'll get *darkness* " me: I know... *turns off light, sits at the corner and falls into depression*

[@yashgaikwad7516]

Mr. J_Krr_ :(

[@MJ-hq1ie]

A whole mood

[@splinter1817]

sed lyf

[@ponypony4695]

If only my teacher at university could explain this in the way you did... thanks, now it looks much more easier to understand ;)

[@LASTCHANCE-mi8zy]

says who

[@ufo717212]

I just tried to reorganize your points in my own words as the following. At a magical spot, each light travel one wavelength further(or less) than its adjacent light. Therefore all lights results in constructive interference at this spot which is bright. At a non-magical spot, each light has a slight phase difference with its adjacent light. Since there could be hundreds of different lights from hundreds of holes, each light can always be paired with another light with a phase difference of about half wavelength. Thus each pair results in destructive interference. Overall they result in darkness at the spot.

[@kissmyussr1993]

Who are you so wise in the ways of science?

[@vandnashukla3598]

This is exactly how I tried understanding it.i took approximately half of the total number of holes and then each point in between the initial and final hole would pair up destructively with another.

[@Ahmed-vs1ui]

How does that magical spot even exist mathematically

"And in between these bright spots you will get darkness...which is grate."

[@fluffzxx9615]

I don’t get why in the normal double slit it would be smudgy in the first place, why?

[@epickirby101]

Fluffz Xx cause the waves won’t completely cancel out in between, partial destructive or constructive interference

[@smtusharibnesalam8688]

Where's the love button?

[@cortwill4085]

OH... ME... GOSH!! IS THIS... ARE YOU... KHAN ACADEMY KIDS?!?!?!?!?!?!?!?!? I LOVE Khan Academy Kids so much! l couldn't even help myself but to get a game of you guys on my iPad! I like all the characters: Peck, Ollo, Kodi, Raya, and Sandy! But I think... Sandy's my favorite! Khan Academy Kids is all l play! Like, I literally don't want to play anything else on my iPad! CONGRATS!🎉 You get 5 stars!⭐⭐⭐⭐⭐ I wanted to give you 5 stars in the game, but it required a PIN that only Mum and Dad knew. And they won't give me permission to give you 5 stars. Here you go!:⭐⭐⭐⭐⭐

[@simonhwang5624]

Right at 3:00, you say "draw a right angle", but if it's a right angle there's no way the two lines (minus the "extra part") are equal, because geometry: hypotenuse of right triangle must be longer than sides.

[@adityamathur5972]

Same bro, even I didn't get that

[@nononnomonohjghdgdshrsrhsjgd]

@@adityamathur5972 i was scrolling through the comments to find that question and some explanation

[@divyansh_kashyap2901]

​@@nononnomonohjghdgdshrsrhsjgdbecause the angle between both hypotenuse and one side is very very very small...approx 0⁰.. Cos0⁰=1=(base/hypotenuse)

[@Kolinnor]

I think he meant to draw an isoceles triangle.

[@luigiprz5319]

Hi, to keep angles, so green lines, the same, the purple lines should be parallel. We cannot approximate angles because the difference distance is much bigger than wave length. Correct me if I'm not right. It looks the same when draw is not precise but geometrically not. What if first hole will be perpendicular to bright spot and the x hole will be at the border of light point :) but you got good intentions it's the plus

[@Integrelle]

2:12 "Why? Well, lets talk about why." LOL, reminded me of the GMM quote, "Lets talk about that".

[@shashanksadafule]

How is path difference constant for consecutive waves. The theta is different for both of them

[@brianummel5602]

I agree. His explanation is wrong.

[@zeenathunnisa3571]

Gosh!! Such an awesome and clear explanation. Jazakallah Khair.

[@sarahholland5980]

you just explained a full lecture in 15 minutes, and beautifully and simply. bless you.

[@Postermaestro]

This was very well explained. You guys at Khan are so good at making it easy to understand, which makes the learning process so much more fun and time-efficient!

[@Lyoko-wy3si]

I barely understood the idea of diffraction grating until I watched this! Thanks for the help

[@SohamChakraborty42069]

Beautiful explanation. One part I don't understand though. The angle theta is the angle between the perpendicular line joining the middle of the distance between the pair of slits and the screen and another line that joins the middle point and the point of interest on the screen. In diffraction grating, we are choosing different pairs of slits each time, so the middle point changes whereas the point on the screen remains fixed. How is the angle the same, then?

[@haristufail2121]

Beautifully explained Sir.You should get a Noble prize.

[@FDGod99]

-_-"

[@FDGod99]

+Onkar Apte -_-"

[@gracepancalstatela4422]

I agree👍.

[@LogicalCod]

Nobel

[@harrisabdullah2278]

If there was a nobel prize for teaching

[@trailertrashtactics]

This guy is like a young sassy Sal Haha. Love it thank you for the videos.

[@annaszinyei]

You did such a great job explaining! So clear and easy!

[@smallbowl7143]

Thanks, but I'm slightly confused If d (distance between 2 slits remains constant), then at any time no more than 2 waves can be in phase. Furthermore in the maths used to find the equation dsin(theta) = n lambda, you say that the wavelength is the change in x and make a right angled triangle. As it is a right angled triangle the angle opposite the right angle must be the hypotenuse and therefore the two lines must not be the same length and the waves are still out of phase. If I didn't explain clearly then sorry but that's why I'm still confused

[@derdudernan]

Small Bowl I think I understand your question, so let me take a shot.. d*sin(theta) represents the extra distance light has to travel from one opening compared to another. The reason light from several different openings can be in phase is because that distance for those openings will be an integer number of wavelengths. So if the light travels say 2 or 3 or 4 or... extra wavelengths in distance compared to another opening they will still hit the screen in phase. Also, yes in the right triangle you talk about the hypotenuse is a little bit longer then the longest leg, it is assumed the rays are parallel when when leaving the slits, so it is an approximation. But since the distance between the slits is so small compared to the distance to the screen the angle is really small and so assuming they have the same length gives a good approximation, even on the scale of a nanometer. Hope I understood the question, hope that helps. Still attempting to understand it fully myself...

[@smallbowl7143]

derdudernan Since asking the question I studied more physics and came across small angle approximations. In very small angles it is assumed that sin(x) = x = tan(x) This is the reason it works as the hypotenuse is effectively the same length as the adjacent.

[@gausiyakhan6478]

Because of two coherent sources .in huygens theory every point are source .

[@4darekord]

crystal clear explanation man

[@RedBar3D]

Actually drawing them parallell seems to help understanding a whole lot

[@manswind3417]

Quite a marvellous experiment and of great utility, as the instructor points out and I rightfully admit. However, as opposed to YDSE, wouldn't this experiment be a nightmare to actually execute? We are dealing in lambdas of the order of below microns here, so in the process of exacting our Ds across the wall, even if we miss the mark by a nanometer in making the hole we might mess up the whole pattern right? In YDSE we didn't have such a concern since there were only 2 holes so a slight error would only cause a slight shift in the interference pattern, but for diffraction grating wouldn't the error add up due to all holes and effectively yet a dark spot instead of a bright one all because of physical imperfections?

[@tylerlitch3992]

What screen recording software do you use? I would like to know a good one that allows me to pause recording.

[@rafaelrazo18]

i love you

[@MrRak18]

It helped me understand x- ray diffraction.. thank you.

[@jinks6887]

really well explained thanks a lot ..love from india

[@thebeerwaisnetwork8024]

I don't like this explanation. You can say that since each wave differs slightly from each other with non integer values, they cancel each other slightly each time. And if you continue the process "infinitely" many times the wave cancel each other completely. You can think of this as blasting a sin wave with positive-negative paired particles with each additional propagating wave. Each time you blast away some points in your sin wave with the same concept of a pairing argument. This is why you see a blur with only 2 slits since you only blasted your sin wave once. If you blast it "infinitely" many times all the points on your sin wave should be nullified.

[@thebeerwaisnetwork8024]

You can only have constructive interference at whole integer values of x. Any other value will result in some type of destructive interference.

[@smtusharibnesalam8688]

Its Great. Mind blowing. Top class video. Thank you very much for teaching this

[@animeBHARATIYA]

The explanation is very clear, thank you

[@lordasjad4896]

"It's great actually "

[@craetydonutkey1348]

For the deviated spot, i think if you just draw out all the possible waves on the same graph you could tell they are interfered destructively by their adjacent waves at each intersection. In other words, I think any space between the integer wavelength lambda will always be dark, resulting in discrete bright spots on the screen.

[@mariamnaeem6963]

Wow!! Thank you for posting such a wonderful explanation! I was so confused that how wavelength is determined and you have cleared this concept briliantly!

[@adityamathur5972]

THAT CAN'T BE A RIGHT ANGLE PLEASE FIX IT.

[@xiaohanjiang4960]

I am not convinced as to why this will be completely destructive.

[@Hugh.Manatee]

Wait, you claim that the angles are equal, but they can't be. You could have an infinte number of holes going down, but you can't keep adding the same angle infinitely often without crossing behind the projection screen. Am I missing somthing, or is this explanation incomplete?

[@Ottotroph]

+AdenineMonkey At infinity you would not be behind the screen but nearly parallel to it.

[@Hugh.Manatee]

Nick Brylski And the angle between two adjacent lines at infinity would have to be 0. But Khan says the angles are the same for each set of adjacent lines. That's what I didn't understand.

[@xaviervangorp4862]

this is an approximation.For the first couple holes closer to the source (that account for most of the intensity) the angle difference is very small. But you are right to say that you can't go to infinity, these angles all the way down will basically be zero as you go to infinity.

[@alivira1558]

Never understood what the concept was before but this was like magic.. well done and thank you so very much.

[@mastersanjay7642]

Amazing explanation and content hat tip

[@chaitanyavarma1747]

"Double slits are cool" Awesome 😂😂😂

[@MoatterAqeel]

Sie waves k saare chapters ki ik playlist banaden please... !!!! :( ! waves, reflection,refraction,interference , diffraction,polarization of waves, inferno meter, Newton's ring, Diffraction grating, < spectrometer..... Electromagnetic wave Equation,,, Normal and anamolous dispersion, coherence, lasers, .... please create this for waves and arrange your waves lecture in playlist

[@carolwang8647]

啊啊啊啊啊啊，下周就考物理了。希望能考好，好想去UCL

[@sotirissarros3522]

many thanks man. it's clear that you really want to explain the subject and you do it perfectly!! well done!!

[@benjaminzurich6571]

What I still don't understand is: here we only consider interference between single waves of different slits. But we should not forget that every single slit also has it's own diffraction pattern. So what if we consider an angle \theta were we have destructive interference between waves of differents slits, but that the waves of the single slits separately constructively interfere at this angle ? Then we would indeed not become such a discrete pattern. This would in fact be possible if d*sin(theta) = q*lambda (multiple slit, q in Q\Z for destructive interference) and simultaneously for example a/2*sin(theta) = n*lambda (single slit, n in Z for constructive interference) where a is the width of the single slit. Since we are looking for solution in (0, pi/2] because of the symmetry of the problem, and that sin(x) is bijective on that interval, we have arcsin(lambda*q/d) = arcsin(2*n*lambda/a) and so 2*n/a = q/d. This gives a solution for each n in IN.

[@AsmaaAlmurshedi]

2024 and this video still working great 😃👍 thx for the explanation you're doing amazing!

[@workingemail1016]

Very well explained. Thank you.

[@PS-on7jm]

the problem with me is just that when you calculate the difference in tranvelled distance, you assume that either that the rays are parallel, which they are not or you assume that the triange is a isosceles triangle. But is is not one. ok with 2 slits you could say they are almost isosceles triangle if you have big distance, but when you have many slits then you cannot assume this anymore. or you distance has to be very big

[@RahulGupta-pf4yt]

Your style of teaching is remarkable.

[@adxmalad4675]

You tried your best I am satisfied with theory part but stil left with how...want to see that in reality...or I need to go more basic, anyway thankyou

[@tomisz9208]

two pathes are equeal when they can be considered as the legs of an isosceles triangle, so the path length differance is not d*sin theta (what you show here is a right-angled triangle, you basically claim that its hypotenuse and longer leg are having the same length). it also means that no more than two waves are in phase, if d remains constant

[@gausiyakhan6478]

Wave front draw perpendicular in between two waves or Ray then we got d sin theta .it's not a triangle .and next it's just because of two coherent sources .we take d in mm to obtain fringes .

[@pranavkrizz1553]

14:00 isn't d sin(theta) = m lambda the condition for minima

[@Raytolife]

The video was great but the speaker is trying too hard to copy an American accent. Man it would have been okay with your Indian accent too or you can have used hindi too. It doesn't matter what language you speak but I found this accent copying really irritating.

[@rezaapply1450]

You assume that the rays coming out of holes are parallel but it's not true! This assumption is not true. The phenomenon however can be explained more easily.

[@lazyz8969]

Great Explanation

[@CalvinUlbricht]

very well explained video, thanks a lot, really helped me understand it better.

[@kitanomarumilitoni9293]

I understood that very well. Great job sir thank u very much!

[@BlueRangeTV]

Thanks, the Bulgarian student books don't explain this well at all! Finally understood it

[@miyu6720]

khan academy saving my grades I FREAKING LOVE YOUUUU

[@chrisjieyanglee5360]

thanks bro you've helped me alot!!

[@brettgallagher5306]

From his explanation it looks like all points should be cancelled out by destructive interference so do you still get the bright dots?

[@fluffzxx9615]

I don’t get why in the normal double slit it would be smudgy in the first place, why?

[@haniahazem2662]

is d the distance between two slits or holes/cm? the math makes no sense if its distance between slits... unless im doing something completely wrong

[@5ty717]

~4.49 … Theta is not the same bro… so why not say almost the same and be correct rather than wrong.

[@iqbalmahmudmoon3965]

I dont get the thing of the cycle. What is your assumption on this wave cycle?

[@Aditya-f8t5z]

Thank you for making this video. 🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏

[@nyeellis]

I can't explain myself very well but hopefully you understand what I am saying. Wouldn't there be slight variations in the intensity around the maxima because there wont be completely destructive interference as that would need there to be a certain number of waves interfering constructively and an equal number interfering destructively. If there is not exactly an equal number of pairs due to the number of slits then surely you will get small subsidiary maxima?

[@sanyufan8691]

Sir please tell me why we use grating lens in numerical aperture but in Newtown rings we use glass plate

[@wooyoungwang8021]

does that mean that the size of the dot correlates to the wave's amplitude?

[@TimeGrowing]

Hey man you just killed it's been one month on school teaching me,but I didn't understand. And you, just in 14m you make it easier than drinking water. Thank you so much.

[@qualquan]

basically correct but shaky should have drawn the 1.1 wavelength difference's angular extension to the 5th hole so as to get (1.1)x5 =5.5 wavelengths difference. This particular ray with 5.5 wavelength difference would cancel the 1st ray. Similarly the 6th ray would cancel the 2nd and the 7th cancelling the 3rd and so on thus allowing cancellation of "smaller angle" deviation by use of multiple ds (d=inter hole distance). In the 2 slit situation only a single d is available so only larger angle cancellation available leading to smudgy or wjder dots.

[@TheQuotation0]

Sometimes i wonder where is your marker 😅😅 unable to find ,its too small ( i don't know its solution ) but easy nice one

[@tom_something]

This is super clear, thank you. One thing I was wondering as I looked at this. Shorter wavelengths should, then, encounter these positive interference peaks at smaller angles. And I was trying to reconcile that because I know "blue light diffracts more than red light". But I looked it up, and in a diffraction grating, blue dots would be _closer_ together. I had assumed a diffraction grating would work the same as a prism or the sky. But nope. So the observation of dot spacing with regard to wavelength through a diffraction grating is totally consistent with the model you've outlined, which is really satisfying.

[@zihengwang7048]

Sorry but it cannot be a right angle. It is WRONG.

[@snehlapsiwala6949]

An electron beam is accelerated from rest through a potential difference of 200v. (1) calculate the associated wavelength. (2) this beam is passed through a diffraction grating of spacing 3 A. At what angel of deviation from the incident direction will be the first maximum observed.

[@Emily-qw4sy]

Hi, don't know if you'll see this, but just wondering why does light intensity increase when the number of slits increases? Is it simply because there's more light? Or is it because there's a smaller gap so more diffraction? (but then wouldn't intensity decrease if there's more diffraction...this is why I'm confused :( ) thank you!

[@justus7800]

Can you explain why the light is converging? Its my understanding that visible light can be expressed as photon particles, and due to duality exhibit both properties of mass and EM radiation, because of this Im trying to understand why the photons converge (wouldn't momentum continue them straight)? Are they being spun as they slide by the holes which is in turn curving their trajectory? Ive been trying to understand exactly how a visible wave propagates from source to Z (any EM wave for that matter). Ive been assuming its like a spherical compression wave, like audio, except EM, but cant understand how then the holes are able to split the light into its individual WVs or how polarized lenses or UV filtering ect actually blocks certain WVs but not others? Also if its like an EM compression wave, how do photons fit in? I thought I read somewhere that they are spheres of charge that spin and we record the negatively charged side and thus the traditional sin wave form expression? If you did a video on that I would be greatly appreciative. Also, it seems all visible light operates at the same frequency just different WV? Im trying to conceptulize what the different variables of light actually represent.

[@akirasuzuki9269]

Wait, but if we draw these lines perpendicular, even if the path differences are in the form n*λ, the paths themselves aren't equal - they are Hypotenuse and catheter of right triangle.. Path differences can be obtained by drawing lines to form isosceles triangles, but then they aren't n*λ. How's that possible?

[@tayyabhameed9559]

Bro gr8 video, but its not "diffraction grating".its the explanation of "youngs double slit expwriment". I ended up watching whole video tho.

[@markren626]

This super amazing explanation made me think of the animation of 3b1b’s fourier transformation animation, the little dots moving on the sine wave and add up their values is just like the process of wrapping a metal wire around a circle and find its center of mass!

[@RealNovgorod]

Nice visualization... A smartass could now come and ask why then the spots aren't "infinitely" thin, or at least ~N times thinner than the spot from a double slit, if the grating has N lines. But that would be a whole new lecture.

[@inku.1593]

You are amazing 👑👑. Even I don't speak English every time I see you explaining something I feel happy inside of me . Becouse you make studing physics fun .

[@al1383]

But why is any point of the lights “wave point” dimmer or brighter than any other point of the lights wave? Is the top of the wave similar to on, or bright, and the bottom off? Or is it energy, no energy, energy, no energy....?

[@KhaschuluuMunkhbayar]

If rays are parallel each other, you can do with right angle to calculate difference. But those rays isn't parallel. It think your calculation method is wrong, isn't it?

[@adityamathur5972]

Exactly

[@5Youssif]

would be awesome if youd go back and redo sals other physics vids, david. Anyone agree? Love sal but this is david's playing field

[@AndimatterUniverse]

You really should have emphasized that this only works if you assume the screen to be far away from the grating. Because in your animation at around 5:00 its clearly visible that the difference between the second and third ray is not 1 lambda. It only works if you assume the rays to be parallel, wich they are if the screen is far away.

[@ebtsoby]

this is what confuses me, we assume the light is parallel, yet we use triangles

[@ChanakyaVivekananda]

Mannnnnnnn This is such a good explanation!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

[@smitpat1408]

I've never seen it before But I suggest you see some interesting videos on SALEH THEORY-com about behaviors of photon.

[@eduarddez4416]

At the last part when you showed us that if you slightly deviate from the maximum point you said that there will be destructive interference and it will cancel each other out , but won't there also be constructive interference for example after the 10'th slit ?

[@kasamearth1]

Bragg's law states that the intensity of reflected beam from the crystal lattice at a certain angle is maximum when the path difference between the two reflected beam from different planes is an integral multiple of wavelength of X-ray.

[@rithikradhan367]

is it possible to get the result that between the bright spots there is complete darkness using integration, like some arbitrary point where delta x is (1+dx ) lambda

[@nielsn5884]

So i thought that i had understood it, but when i tried with a delta x of 0.2 lambda and i got a wave hitting at the top of the wave but no matter how many waves hits the wall with a difference of 0.2 wavelengths would hit at the bottom to get destructive interference with the one at the top? Can anyone explain this? :)

[@youmnaallam3514]

Its so helpful , one thing is not so clear I have rest or follow to the equation in my book it says "delta y = lamda ×R over d " I dont get how is R related ? Where R is the distance between the double slits & the observation screen

[@lobellaconstanzo4974]

I don't quite understand the difference between diffraction grating and single slit interference. Is it just due to the difference in slit width/ spacing of points on a wave front? Hypothetically if we made really small distances between the gratings would it just effectively demonstrate single slit interference? And why is it that diffraction grating gives the same formula for constructive interference as Young's double slit while single slit uses the same formula but for destructive? This is so hard to understand :(

[@muditjha7235]

I'd say that the thing to keep in mind is that in YDSE and diffraction grating, each opening is treated as a single source while in single slit, we consider many sources as per Huygens Principle. In the diffraction grating, the maxima are more pronounced and sharper because at the maxima, the waves arriving from each slit has a path difference of lambda. This is not the case with single slit.

[@marianbuljovcic]

4:24 Can someone please explain how are the second and third beam apart by the same angle? I don’t see how this is true when the further you go down, the angle between the beams gets sharper (/the angle at which light hits the surface is more obtuse, and because the beams aren’t t parallel, the claim that the distance is longer by exactly one wavelength is false.)

[@pritipaul9353]

It's almost same maybe😅