This was perfect! The explanations were totally clear, absolutely nothing I did not understand. And the excitement was fantastic, too. I feel very well prepared for tomorrows period - thank you! = )
doc, this is the first video of yours that im watchin, and man , i'll tell ya. this video needs more views. your teaching is a reflection of the passion i have for physics. when the teacher is as excited as the kid, then ...well, its a party :D cheers.
Well, if one slit is two, then each slit is W/2 wide. Also, those two slits are W/2 apart from each other. So, yes, width is also separation, but neither is equal to the width of the real, physical slit width.
"It's like you're in a conversation with yourself, and get interfered by a text that you sent your self" Love it! Thanks for bringing the humor to physics =)
I love your enthusiasm when teaching. Really kept me listening with having to struggle to concentrate. I just have a question though, what's the point of treating the single slit as multiple slits? Is it just to get a better equation to use when calculating bright fringe width?
Hey man! Incredible video, first one of yours I've watched as I've been desperately searching for solid info on single and double slit light wave experiments. Tis people such as your self who have inspired me to go on to want to do much the same thing and teach physics at high school or university. The only things I don't seem to understand with all of this is; 1. If Huygen's principle says there's infinite points along a wave front from which 'secondary wave-lets' can exist, then why isn't there simply infinite interference? I don't see how the interference pattern can exist from this viewpoint. (I think someone asked this earlier, but I thought you may know now?). 2. At about point 8.20 in the video where by you talk about these two points from which light rays come out from, you say they're both projected with the same angle theta, but then interfere with each other a relatively large distance away. How would this work if they're projected on the same angle, and are therefore parallel? Unless by them being half a phase out means they're pathways change and meet later on? Cheers :)
1) Very puzzling concept! Unless there is some impediment (a wall or slit, perhaps), there IS infinite interference. The slit allows only some of the new wavelets to exist, which is the whole reason that light is seen at all above and below our slit. You'll have to also agree that the slit is a very large number of very small slits all sitting on top of each other. That allows me the treatment I've made. 2) The rays are of course not perfectly parallel, but are VERY NEARLY parallel since the screen is, as you say, a long way away. That distance allows them to be [almost] parallel and finally to meet. Of course, parallel rays would only meet if the screen were infinitely far from the slit, but it would take too long to put it there. (and then, how would you get it back?!?)
Doc Schuster I see. I guess trying to fully understand how things such as this work is pretty difficult as were only working with models, not reality. Although with Huygen's principle, if spherical waves propagate from all points along the wavefront etc etc, then wouldn't an interference pattern be able to exist on the LHS of the slit, as well as the RHS? It would make sense that there would be to much disturbance behind the slit with incoming waves, but if just one wave were to be sent, then once the wave hits the slit, the wavelets would propagate in all directions from all points along the wave, and so create an interference pattern on both sides of the slit? I understand its a 'forward' moving wave and all, but its almost as though semi-spherical waves propagate from each point, just on the RHS of the point of origin. This could then be seen to make more sense for an interference pattern only occurring on the RHS of the slit? Its all pretty nuts
Throughout the last century, it was great importance to know if the photon's motion is like a wave or like a particle's motion. Saleh Theory give a coherent answer to this question on SALEH THEORY's Video: A Revolution in Light Theory
Diffraction is prominent when wavelength of light is large as compared to the object (small ball for example).In the slit experiment we say that if slit is small then there will be more prominent diffraction ,isn't the distance between the slits acts as a object here ?
Hi Doctor, I have another question for you I suppose that you're dividing the slit into any number of slits, as many as you want, because of Huygens' principle. But you're only taking rays that are at a distance equal to the width divided by a natural number (w/n) to calculate dark fringes in their intersections (interference), at infinite. So you take two rays separated w/2 to calculate the first dark fringe; two rays separated w/4 to calculate the second dark fringe; and so on. The problem I find is: if you just move a little closer one ray to the other after having calculated the first dark fringe, then these two new rays will interfere destructivly just a little higher in the screen, producing a new dark fringe a little higher (the angle theta will not be very much increased). That would produce a totally dark screen, or maybe totally bright. Where is my mistake? It's hard to explain without a picture, and I know it may be hard for you to understand it too, but I hope you will. Thank you very much
if we assume that maxima are found at odd half integers of lambda, for example ø = 3Lambda/2a you can create that maxima by splitting a slit into three slits, slits 1, 2, and 3. so all the waves from 1 interfere destructively with the waves in 2, and only 3 contributes to the maxima at that point. if you have 5 slits, 1 kills 3, 2 kills 4, and only 5 contributes to the maxima, thus ø =5lambda/2a. does that make any sense?
But what about the interference of rays from all the other positions on the two halves of the slit, that are not at a distance of W/2? I don't get it. You can form infinite pairs of rays from the two halves, but we just consider the ones who are at a distance of exactly W/2 (which are also an infinite number of pairs don't get me wrong). What is going on here? What am I getting wrong?
Do you still get interference when the wavelength is exactly the same length as the slit (W)? Huygens explanation states each source will produce wavelets that interact, but if there is only room for one 'wavelet' then how does interference occur? Seems to work with the maths also as if Wsinx=landa then sinX=1 when W=landa, which puts the first dark fringe at 90 degrees which is saying there wont be a dark fringe, just a light fringe gradually decreasing? Thanks for any help and for the video
Thank you so much sir really u r incredible. I would have been more fortunate if you were my physics lecturer. Well here r few topics pls let me know if u can help me in these.....diffraction intensity equationdiffraction at circular aperture (Newton rings)absent spectra in diffraction
+md ajaaz Check my video on Poisson! The other topics appear to various degrees in my diffraction videos, but they don't have their own videos. Thanks for watching and learning. Be sure to work problems.
i have a question .if the wavelength of light is very small,then even a very small distance matters right.then how can we assume parallel rays when we know there will be some extra path difference right and it could be comparable to lights wavelength.
w or w/3 or w/5 are all fine......take w/3.....divide slit into thirds....call points between slits s1 and s2 A B and C.....if path difference between s1 and A is lamda/2 then all points between s1 and B destructively interfere (s1 cancels A, points between s1 and A destroy successive points between A and B....leaving one -third of points, those between B and C to all more or less combine to give a subsidiary maximum at that angle. Geometry says that w/3(Sin theta) = lambda/2...so first subsidiary max occurs at w Sin theta = 3lambda/2. Similar arguments work for w/5 etc etc
Why couldn't we have done the same calculations for the bright spot? Or, let me guess, there are different ranges of bright rather than the one completely dead spot (dark) so we need more complex calculations to calculate it's position?
seems when there is destructive interference we l get a dark spot and in constructive one bright spots with less intensity,so bright fringes,how are dark fringes ? are they having less darkness or less brightness.
In my derivation, I can't have six or ten slits, etc. My simple argument never considers that a single slit be seen as three slits. I guess you'd have to draw TWO red dots on it and see what happens. Good luck!
i have some question when we divide the split into 4 split the wavelength should be h/4 not h/2 ???????? and my sequond question how a sigle wave is interfer with it self i can't imagine that ? do you have some video where i cant watch it ? i saw your Huygen's Principle but i don't get it 3) when do we have the case of 2 split and when we do have 4 split i just cant get it if the first wave interfere with the wave at W/2 and at the same moment it interfere with The wave at w/4 and give us 2 second dark postion ?
Dear Doc: The doublé slit system has its own interference pattern, but each slit also has it´s own interference pattern. So, in the doublé slit experiment we have 3 interferences mixed right? (that is, slit 1 interefence + slit 2 interference + slits 1 and 2 interference). What is the final interference pattern for the doublé slit takeing in count the 3 interferences mixed together?
+Alfpajarito Wow, yes. I have spent some time looking at these patterns and forming them on my retinas, so I can assure you that the double-slit pattern strongly dominates when there are two slits. However, as the two slits each get narrower, the single-slit behavior becomes noticeable. Ultimately, the single-slit diffraction pattern is what causes diffraction-limited optics.
+Doc Schuster Thank you very much for your fast reply. I´m ahppy and surpised you was able to understoond my question because my poor english. Regarding your answer: I was trying to get interference patterns with a green laser I own, both single and doublé slit and I wasn´t able to notice the diference between them. Both patterns seems to be the same intensity, don´t konow may be I´m doing something wrong... But if both kind of patterns has the same intensity why one will dominate over the other?... What I´m missing???
+Alfpajarito Focus on the central peak - is it twice as broad as every other peak or not. The broadening of the central peak is the only distinction between the two.
Your enthusiasm is contagious. Keep enjoying life. Nicely done!
you are an incredible teacher
voice
color
explanation
and most importantly, fluency, it amazes me that u did that in one single take
Peruvian drummer That's really nice of you. I got pretty lucky on that one!
You mean in one single “phase” haha, sorry, sorry, ill stop, ill stop
I LOVE YOUR EVIL LAUGH, thank you so much for this video. I do HL IB Physics so this is great!
What? That's my HAPPY laugh. You should hear my evil laugh, though...
Wow your videos are unbelievably better than the crap videos they give me at my university. Thank you so much
Mr. Schuster: Are you taking notes?
Me: 👀
Also me skipping back to take notes: 😕 🤔
Thank you so very freaking much. All of your videos are epic.
Love the enthusiasm. He puts fun in Physics more than Sheldon Cooper. :D
The popcorn was good
This was perfect! The explanations were totally clear, absolutely nothing I did not understand. And the excitement was fantastic, too.
I feel very well prepared for tomorrows period - thank you! = )
I love your enthusiasm!
Thanks for deriving the equations- I've found that to be key for understanding physics!
7:44 golden moment
Thanks! Happy to help.
doc, this is the first video of yours that im watchin, and man , i'll tell ya. this video needs more views. your teaching is a reflection of the passion i have for physics. when the teacher is as excited as the kid, then ...well, its a party :D cheers.
Yay! Parties! I'm thrilled to hear that you're exited, too.
Excellent video. Really helped me a lot. Thanks so much!
You are great .... You make physics very very interesting . ThanX
Keep going
Well, if one slit is two, then each slit is W/2 wide. Also, those two slits are W/2 apart from each other. So, yes, width is also separation, but neither is equal to the width of the real, physical slit width.
awww I like the cute little Newton doll at the beginning I want it. hah
Thank you so much Doc!! You’re amazing
"thats a dark fringe yo!"
I LOVE THIS PLZ DONT STAHHHHHPPP EVER
6:30 the example made me laugh, thank you. I was not having a good day but this has brightened me :))
u r a beaut teacher doc!!
Please come and teach at my college. I love you, wished my professors had the same enthusiasm like you.
I'd love to, but I probably shouldn't. Thanks for the invitation, though!
I wish my class were this fun...Thank youfor this!!
I loved this lecture.
Well, my brain is now non existent!
you are now my most favorite person
Thank you so much! even though Im not good at English, I can understand from your drawings. really good and easy explanation.
REALLLYYYYY HELPFULLLL , THANK YOU !!
"It's like you're in a conversation with yourself, and get interfered by a text that you sent your self" Love it! Thanks for bringing the humor to physics =)
You are amazing!:D
The way he said “goodbye”
Thanks! I should really look these things up before I start, right?!?
this is hilarious, thanks for the laughs :))
OH MY GOD THANK YOU SO MUCH SIR!
I see what you're doing and I like it.
love how you put fun into your teaching.....i like the "fix you bow tie newton" line.....killed me
Wow, thanks! I like you, too!
I love your enthusiasm when teaching. Really kept me listening with having to struggle to concentrate. I just have a question though, what's the point of treating the single slit as multiple slits? Is it just to get a better equation to use when calculating bright fringe width?
love your videos!
i like your funny style.. Nice work
It's too chaotic for me.. i feel more confused then i was before 😅
ARE YOU NINJA
im here sipping my lemonade and getting hyped as hell
Thank you for this video! :) Subscribing
Welcome, friend.
thank you for this!
Great vid! Keep it up!
I can't believe Benson went back to school to get a physics degree
Hey man! Incredible video, first one of yours I've watched as I've been desperately searching for solid info on single and double slit light wave experiments. Tis people such as your self who have inspired me to go on to want to do much the same thing and teach physics at high school or university.
The only things I don't seem to understand with all of this is;
1. If Huygen's principle says there's infinite points along a wave front from which 'secondary wave-lets' can exist, then why isn't there simply infinite interference? I don't see how the interference pattern can exist from this viewpoint. (I think someone asked this earlier, but I thought you may know now?).
2. At about point 8.20 in the video where by you talk about these two points from which light rays come out from, you say they're both projected with the same angle theta, but then interfere with each other a relatively large distance away. How would this work if they're projected on the same angle, and are therefore parallel? Unless by them being half a phase out means they're pathways change and meet later on?
Cheers :)
1) Very puzzling concept! Unless there is some impediment (a wall or slit, perhaps), there IS infinite interference. The slit allows only some of the new wavelets to exist, which is the whole reason that light is seen at all above and below our slit. You'll have to also agree that the slit is a very large number of very small slits all sitting on top of each other. That allows me the treatment I've made.
2) The rays are of course not perfectly parallel, but are VERY NEARLY parallel since the screen is, as you say, a long way away. That distance allows them to be [almost] parallel and finally to meet. Of course, parallel rays would only meet if the screen were infinitely far from the slit, but it would take too long to put it there. (and then, how would you get it back?!?)
Doc Schuster I see. I guess trying to fully understand how things such as this work is pretty difficult as were only working with models, not reality. Although with Huygen's principle, if spherical waves propagate from all points along the wavefront etc etc, then wouldn't an interference pattern be able to exist on the LHS of the slit, as well as the RHS? It would make sense that there would be to much disturbance behind the slit with incoming waves, but if just one wave were to be sent, then once the wave hits the slit, the wavelets would propagate in all directions from all points along the wave, and so create an interference pattern on both sides of the slit? I understand its a 'forward' moving wave and all, but its almost as though semi-spherical waves propagate from each point, just on the RHS of the point of origin. This could then be seen to make more sense for an interference pattern only occurring on the RHS of the slit? Its all pretty nuts
Geez... but I seriously am sitting here eating popcorn and not taking notes!
thank you so muchhhhh !
I am eating popcorn right now
Same!
Thanks for the help. I wish I was as interested as you in physics. I never do this when I study 12:12.
single slit diffraction made me want to to cry
Very good.
btw, whole no. include zero @15:16
Why does is split in half and not in another quantity, such as 3 or 5?
+Julia Zorthian Try the maff of that split and see what it looks like. I think it would work!
+not anyone I've been wondering why the whole night! still no answer...
Throughout the last century, it was great importance to know if the photon's motion is like a wave or like a particle's motion.
Saleh Theory give a coherent answer to this question on SALEH THEORY's Video: A Revolution in Light Theory
it's probably not my first time, but i still wanna say thank you!!!!!
You remind me of Jake Peralta from Brooklyn 99, fun listening to this
Diffraction is prominent when wavelength of light is large as compared to the object (small ball for example).In the slit experiment we say that if slit is small then there will be more prominent diffraction ,isn't the distance between the slits acts as a object here ?
Awesome another tutorial..
Hi Doctor, I have another question for you
I suppose that you're dividing the slit into any number of slits, as many as you want, because of Huygens' principle. But you're only taking rays that are at a distance equal to the width divided by a natural number (w/n) to calculate dark fringes in their intersections (interference), at infinite. So you take two rays separated w/2 to calculate the first dark fringe; two rays separated w/4 to calculate the second dark fringe; and so on. The problem I find is: if you just move a little closer one ray to the other after having calculated the first dark fringe, then these two new rays will interfere destructivly just a little higher in the screen, producing a new dark fringe a little higher (the angle theta will not be very much increased). That would produce a totally dark screen, or maybe totally bright. Where is my mistake?
It's hard to explain without a picture, and I know it may be hard for you to understand it too, but I hope you will. Thank you very much
thanks sir it really really helped
if we assume that maxima are found at odd half integers of lambda, for example ø = 3Lambda/2a
you can create that maxima by splitting a slit into three slits, slits 1, 2, and 3. so all the waves from 1 interfere destructively with the waves in 2, and only 3 contributes to the maxima at that point. if you have 5 slits, 1 kills 3, 2 kills 4, and only 5 contributes to the maxima, thus ø =5lambda/2a. does that make any sense?
But what about the interference of rays from all the other positions on the two halves of the slit, that are not at a distance of W/2? I don't get it.
You can form infinite pairs of rays from the two halves, but we just consider the ones who are at a distance of exactly W/2 (which are also an infinite number of pairs don't get me wrong). What is going on here? What am I getting wrong?
i love the title
Sir, are you a family member of Walter Lewin......? You teach really great..Thanx for the video.
I am flattered by the question! Thank you!
Rajath Krishna R XD same!
Do you still get interference when the wavelength is exactly the same length as the slit (W)?
Huygens explanation states each source will produce wavelets that interact, but if there is only room for one 'wavelet' then how does interference occur?
Seems to work with the maths also as if Wsinx=landa then sinX=1 when W=landa, which puts the first dark fringe at 90 degrees which is saying there wont be a dark fringe, just a light fringe gradually decreasing?
Thanks for any help and for the video
I never enjoyed physics that much that I did today
wooh...its like studing from a frnd..!
thanks doc
You make me love PHYSICSS...ur'e the COOOOLEST!!
Thank you so much sir really u r incredible. I would have been more fortunate if you were my physics lecturer. Well here r few topics pls let me know if u can help me in these.....diffraction intensity equationdiffraction at circular aperture (Newton rings)absent spectra in diffraction
+md ajaaz Check my video on Poisson! The other topics appear to various degrees in my diffraction videos, but they don't have their own videos. Thanks for watching and learning. Be sure to work problems.
Great !!!!!!!!! you gave me many details
oh my gooooooddd single slit diffraction T_____T
i have a question .if the wavelength of light is very small,then even a very small distance matters right.then how can we assume parallel rays when we know there will be some extra path difference right and it could be comparable to lights wavelength.
How are the bright fringes defined in the single slit diffraction?
why W/2 but not W or W/3 or whatever it is?
w or w/3 or w/5 are all fine......take w/3.....divide slit into thirds....call points between slits s1 and s2 A B and C.....if path difference between s1 and A is lamda/2 then all points between s1 and B destructively interfere (s1 cancels A, points between s1 and A destroy successive points between A and B....leaving one -third of points, those between B and C to all more or less combine to give a subsidiary maximum at that angle. Geometry says that w/3(Sin theta) = lambda/2...so first subsidiary max occurs at w Sin theta = 3lambda/2. Similar arguments work for w/5 etc etc
Why couldn't we have done the same calculations for the bright spot? Or, let me guess, there are different ranges of bright rather than the one completely dead spot (dark) so we need more complex calculations to calculate it's position?
seems when there is destructive interference we l get a dark spot and in constructive one bright spots with less intensity,so bright fringes,how are dark fringes ? are they having less darkness or less brightness.
great dude .........great explanation.😋😊😇
And that's supposed to be Natural number set!!
im actually taking notes, also eaitng popcorn
In my derivation, I can't have six or ten slits, etc. My simple argument never considers that a single slit be seen as three slits. I guess you'd have to draw TWO red dots on it and see what happens. Good luck!
great vid
i have some question
when we divide the split into 4 split the wavelength should be h/4 not h/2 ????????
and my sequond question how a sigle wave is interfer with it self i can't imagine that ? do you have some video where i cant watch it ? i saw your Huygen's Principle but i don't get it
3) when do we have the case of 2 split and when we do have 4 split i just cant get it if the first wave interfere with the wave at W/2 and at the same moment it interfere with The wave at w/4 and give us 2 second dark postion ?
very nice lecture
Doc Schuster 0 is part of whole no at 15:21 sec
It's okay because it will still give you a dark screen xD
But he did stick needles in his eyes, so that's cool hahaha
Doc Schuster Damn ur lucky
What causes the bright fringes in between the integers of m
amazing.......thnks u really are my teacher.......:) love from me
Why does the slit have to be divided into powers of 2? Don't any multiples of 2 work as well?
What about the bright points
Why divide the slit into powers of two? Why can't we split in into 3 parts or nine parts etc?
Dear Doc: The doublé slit system has its own interference pattern, but each slit also has it´s own interference pattern. So, in the doublé slit experiment we have 3 interferences mixed right? (that is, slit 1 interefence + slit 2 interference + slits 1 and 2 interference). What is the final interference pattern for the doublé slit takeing in count the 3 interferences mixed together?
+Alfpajarito Wow, yes. I have spent some time looking at these patterns and forming them on my retinas, so I can assure you that the double-slit pattern strongly dominates when there are two slits. However, as the two slits each get narrower, the single-slit behavior becomes noticeable. Ultimately, the single-slit diffraction pattern is what causes diffraction-limited optics.
+Doc Schuster Thank you very much for your fast reply. I´m ahppy and surpised you was able to understoond my question because my poor english. Regarding your answer: I was trying to get interference patterns with a green laser I own, both single and doublé slit and I wasn´t able to notice the diference between them. Both patterns seems to be the same intensity, don´t konow may be I´m doing something wrong... But if both kind of patterns has the same intensity why one will dominate over the other?... What I´m missing???
+Alfpajarito Focus on the central peak - is it twice as broad as every other peak or not. The broadening of the central peak is the only distinction between the two.
Come on Newton fix your bow tie XD
Where do the dots come from though?
Why is when the distance between two rays w/4, the difference in wavelength is still 1/2? Shouldn't it be 1/4?
is this how cinema theaters work?
You only think that 'cuz you can't see my face, though!
"Come on dude, fix your bowtie."