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.
"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?
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.
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
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
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 ?
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.
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.
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?
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
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
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 ?
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.
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.
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?
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?
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
i was asking about condition and theory proof of bright fringes.........like u hv shown for dark fringes in this video.......................please reply
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!
hello doctor schuster your videos have been really helpful but there is something i am still not sure about.....see thing is in our physics textbook the equation for a dark fringe id defined as (d)sin(theta)=(m+1/2)*lamda and but when you do it you juts put the half and not add the m....can you please clear that for me coz that added m is very confusing
Refilwe Senosha The m is an arbitrary integer, so adding it allows you to describe the infinite set of (in this case) dark fringes. Without the m, I must be referring to just one fringe.
Your enthusiasm is contagious. Keep enjoying life. Nicely done!
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...
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
The popcorn was good
7:44 golden moment
Wow your videos are unbelievably better than the crap videos they give me at my university. Thank you so much
I love your enthusiasm!
Love the enthusiasm. He puts fun in Physics more than Sheldon Cooper. :D
Thank you so very freaking much. All of your videos are epic.
"thats a dark fringe yo!"
I LOVE THIS PLZ DONT STAHHHHHPPP EVER
You are great .... You make physics very very interesting . ThanX
Keep going
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! = )
Thanks! Happy to help.
Thanks for deriving the equations- I've found that to be key for understanding physics!
Well, my brain is now non existent!
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.
awww I like the cute little Newton doll at the beginning I want it. hah
REALLLYYYYY HELPFULLLL , THANK YOU !!
Thank you so much Doc!! You’re amazing
You are amazing!:D
Mr. Schuster: Are you taking notes?
Me: 👀
Also me skipping back to take notes: 😕 🤔
"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 =)
u r a beaut teacher doc!!
Excellent video. Really helped me a lot. Thanks so much!
love your videos!
The way he said “goodbye”
this is hilarious, thanks for the laughs :))
6:30 the example made me laugh, thank you. I was not having a good day but this has brightened me :))
It's too chaotic for me.. i feel more confused then i was before 😅
I can't believe Benson went back to school to get a physics degree
I wish my class were this fun...Thank youfor this!!
Thanks! I should really look these things up before I start, right?!?
Thank you so much! even though Im not good at English, I can understand from your drawings. really good and easy explanation.
OH MY GOD THANK YOU SO MUCH SIR!
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?
im here sipping my lemonade and getting hyped as hell
I loved this lecture.
single slit diffraction made me want to to cry
thank you for this!
Geez... but I seriously am sitting here eating popcorn and not taking notes!
I am eating popcorn right now
Same!
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!
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.
thank you so muchhhhh !
Great vid! Keep it up!
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
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!
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
Thanks for the help. I wish I was as interested as you in physics. I never do this when I study 12:12.
When you "separate" the slit into 4, do the rays converge to a single point on the screen?!? And it represents a single dark spot?
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...
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 ?
Thank you for this video! :) Subscribing
Welcome, friend.
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.
Very good.
btw, whole no. include zero @15:16
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 see what you're doing and I like it.
But he did stick needles in his eyes, so that's cool hahaha
Doc Schuster Damn ur lucky
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?
How are the bright fringes defined in the single slit diffraction?
I never enjoyed physics that much that I did today
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
you are now my most favorite person
How do rays of light moving parallel to each other, and starting at different origins, ever meet on a screen and interfere? 8:05
i like your funny style.. Nice work
"Come on dude, fix your bowtie."
how do you know that the second ray that is interfering is in the middle of the hole?
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
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 ?
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.
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!
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.
So essentially single slit diffraction proves that parallel line do intersect eventually...woah
You remind me of Jake Peralta from Brooklyn 99, fun listening to this
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?
oh my gooooooddd single slit diffraction T_____T
Nothing weird just that light is not a particle. It bends on the walls of slit.
thanks sir it really really helped
Why divide the slit into powers of two? Why can't we split in into 3 parts or nine parts etc?
i love the title
Mistake - whole no. Starts with 0,1,3.... Not 1,2,3.....
You only think that 'cuz you can't see my face, though!
Yay!
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?
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
You sound like Amir from Jake & Amir - College Humor
amazing.......thnks u really are my teacher.......:) love from me
What causes the bright fringes in between the integers of m
i was asking about condition and theory proof of bright fringes.........like u hv shown for dark fringes in this video.......................please reply
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!
How do the rays interfere when they're parallel to each other? (At 8:04) I'm probably just being stupid but I just don't get it :)
is this how cinema theaters work?
Why is when the distance between two rays w/4, the difference in wavelength is still 1/2? Shouldn't it be 1/4?
double slit exp gives different intensities at different points??? (the initial part of the video)
And that's supposed to be Natural number set!!
it's probably not my first time, but i still wanna say thank you!!!!!
hello doctor schuster your videos have been really helpful but there is something i am still not sure about.....see thing is in our physics textbook the equation for a dark fringe id defined as (d)sin(theta)=(m+1/2)*lamda and but when you do it you juts put the half and not add the m....can you please clear that for me coz that added m is very confusing
Refilwe Senosha The m is an arbitrary integer, so adding it allows you to describe the infinite set of (in this case) dark fringes. Without the m, I must be referring to just one fringe.
Is your textbook describing double or single slit interference?
its double slit interference
but i do understand now
wooh...its like studing from a frnd..!
thanks doc
Come on Newton fix your bow tie XD