Its so stupid my testbook tells me nothing like this and expects me to understand. As others has mentioned you are a hero. its so easy to enjoyable to learn with your videos. thumbs up
THANK YOU, why isn't Huygen's principle taught BEFORE two slit interference in all the texts, you'd think this knowing this beforehand would save you from a lot of misguided connections.
Kim double slit interference involves the interference of two point sources. Single slit interference involves the interference of an infinite amount of point sources, and defines diffraction: they are definitely linked.
@@REVIEWSGALATTAyea exactly, its taught before and tbh it caused SO MUCH confusion, i have been stuck on it for a day 🥲 i was trying to find a connection bw huygens and double slit and obviously got more confused
if huygen's interpretation was taught to me before geometrical optics I'd be an astronomer/photonicist. It's such a leap from too simple -> crazy, to weird -> everything become simple
So let me bring to your attention the elephant in the room.... What are the light waves refracting THROUGH?? The universe is pretty simple when you think about it and don't overcomplicate and keep bombarding and bamboozling teenagers with complex formulas and experiments that are just the same as magic tricks (smoke and mirrors) . Thankfully I managed to see around the slit as it where 30 years after being brainwashed😁 n.b. I have left a few comments under single and double slit experiment TH-cam videos that have questioned the narrative and all of a sudden I am now not allowed to post a new comment, just reply to one?? ..I must be onto something 🙄
Very interesting, especially for a lay man who has been told so many times that only a double slit will show interference, thanks to quantum mechanics... By the way, your pronunciation of Huygens is quite good if I compare it how 't Hooft is sometimes announced on TH-cam. :-)
thank u thank u the first sentence that u said is why I'm looking for this in the internet I was so frustrated that all the books I have were talking about waves interfering with out telling me why they interfere in a single slit in the first place
Omg you are a life saver ❤️... thank you very very much. It solved all of my questions which I posted in double slit experiment video regarding if the wave spreads out does that mean wavelength of that particular wave increases?
I understand your explanation of single slit diffraction of light and the production of small maxima either side of the big central maxima. However, whenever single slit diffraction diffraction is set up in a ripple tank there is no sign of the small maxima either side of a central maxima.
Why are we only considering the points spaced w/2, w/4, w/8 apart? Shouldn't we be considering each pair of points instead since they each would contribute to the intensity? Also, it seems like even if the w/2 cancel because they are half a wavelength apart, the points w/4 apart wouldn't cancel because by definition they must be less than half a wavelength apart so I don't see the logic there at all.
I don't know why but I never seem to understand Kahn Academy. It seems like it should make sense (especially compared to my teachers) but it just goes in one ear and out the other.
I don't know if i have got it right but, the way you cal the different wave lengh as w/2*sinΘ confuses me,because the right part of light from the footprint seem to have different wave distance
Well explained, thank you. Question: if hyugen's principle is true, how can you direct any light at a particular direction, for example a pointer laser? Shouldn't any light source always be a source of spherical wave fronts?
@@johanneskrv No I wouldn't because after certain distances lasers do begin to diffract. Other than what I need to know and the brief information I have read, I am not really in a position to comment whether he is right or wrong and he more than likely has greater understanding of the subject than you do and certainly more than me. However, from what I've studied, nothing disproves and there's plenty of people who have devoted their lives to Physics and even this topic which I am sure would have been disproved by now if he were to be wrong.
@@ht-ve9fe i'm not saying he is incompetent. Just wondering why in some cases the hyugens principle seems to be in effect and in others not. This seems contradictory.
This has bothered me for a very long time. Thank you very much for this video. Great explaination! My very minor criticism is when you say that the wave cancel completely at that point. Later, you explain that there is a slight difference [in the length of the triangle]. At that point you could add that’s why they don’t cancel entirely. But, I wouldn’t redo or add to the video if it changes the duration google posted as it gives you precidence! Again, great job and thank you for bringing up Huygen’s! - Jim Foit
@Ryana Imtiaz He took the "topmost" and "middle" point, only for the sake of explanation. But this applies for any two arbitrary "points", as in reality, there's an infinite number of "points" within the band of light that passes through the slit, where you can't really select "points" that are "topmost" and "middle", as concepts like "points" are just an approximation...
@@D8Football actually, it isn't the "official formula" but more of a presumption made to help find an explanation to why it so, i.e., why there's a dark region between the bright ones.
Doubt: what if we took 9 sources(as in odd sources) instead of 8, wouldn't there be further complications with deciding whether it ends us up with constructive or destructive interference ? Kindly respond.
Brilliant. But puzzled by why the interference pattern is always horizontal. If Huygen's is correct then shouldn't there be a spherical pattern? Or... What if there was just a small hole instead of a slit? What would the pattern be?
How does infinitely many diffracting waves make a ‘full’ wavefront before the slit then? Why does the slit make the infinitely many diffracting waves into an interference pattern not a ‘full’ but smaller wavefront? I guess I understand that the slit can make a interference pattern but why doesn’t it do the same thing for a normal wavefront?
Anubhav Bahadur if you calculate it comes out w/2. There are 8 points hence the distance between two consecutive points is w/8. The points he chose had 4 points in between. Hence, the distance is 4*w/8 = w/2
How can we be so sure that destructive interference will occur at those specific points like if the waves from the uppermost and the middle points interfere destructively at that point then it could also interfere constructively with the waves from the other points .... And why should we take half of the slit and not the full one ??? They are just ripping my mind @part!!!
For a destructive interference to happen, all the waves have to cancel out - the sum of their "heights" has to be 0. That means that for every wave with height y there has to be a wave with height -y at that point we are looking at. If it were otherwise, all other waves would cancel out and that wave would be "left alone" so that point wouldn't be a dark spot, the sum of the heights would be equal to its height = y. So whichever way you add the waves, you should always end up with the sum = 0 for it to be a dark spot. for example lets say you add up separately waves with positive heights at that point in a specific time, and separately the ones with negative heights then. lets say the heights are y,x,c,v,b... and -y,-x,-c,-v,-b... or what ever names you want to call them. The sum of the positives is y+x+c+v+b+..., and the sum of the negatives is -y-x-c-v-b.... and the ULTIMATE sum is (y+x+c+v+b.)+(-y-x-c-v-b...). As you can see whatever the sum of the positives is, the sum of the negatives is always - that, so the sum is always zero for that point. So pairing off the waves is just editing the sum (y+x+c+v+b.)+(-y-x-c-v-b...) to y-y + x-x + c-c + v-v + b-b +... (You must note that the heights of the waves are dependent on time so when I say that a wave's height is positive I mean its positive in a certain moment in time, not always, same for negative). You might ask well how do bright spots appear then. Well for bright spots at a specific moment there are more "positive" waves than "negative" ones, or vice versa, so not all of them cancel out, and thus the sum is not equal to zero and there is a constructive interference. Hope this helped :)
IKstreme Yeah that was a great explanation . But what I am thinking is that why should we apply plain Mathematics to it .... If I apply some logic , what my mind says is that there shouldn't be any destructive interference . Like , for example all the waves arrive at the same time ... You said there will be y for -y and x for -x and so on . But let say the y . Thus y could also be in destructive interference with the x wave or -x or what else you call it . Then , why should we add up the two y's (positive and negative) . Now you might say that x and y can't be added because they are different variables but let's forget about that because we don't have an equation like that . It could be numbers and then numbers could be added in any way . What my concern is that why should we be adding two waves from two distinct points while there are waves from numerous and numerous points . Based on your explanation , I think when we add them all together we won't be getting either a constructive or destructive interference anywhere ...There shouldn't be one of them.. I hope you get my point.
if y and -y are in a destructive interference, and y and -x are also in destructive interference, then so are x and -y, so any way you pair them up the total sum is always zero: (y-y)+(x-x)= 0+0 = 0 (y-x)+(x-y)=0+0 = 0 Even if you add waves that are not in destructive interference the total sum will be 0. lets say y+x=a, then by simple maths (-y) + (-x) = -(x+y) = -a. So even if you do: (y+x)+(-y-x)=a-a=0 you still get zero because ultimately they somehow all cancel out
IKstreme Ok it makes sense but I am still very much confused . I am just thinking that how can there be + for every - when the waves are pointing at a point far away from the central point ?
yes but when you make the middle of the slit half polarized then they should not interference with the other neighbourhood. also the cancellation is must happen at the slit point only. its not really like a lake of waves.. once it cancells its done, that wave wont de-cancell due to phase swap.. thats why when u move the wall back and forward, you dont see alternating blobs that would represent the peaks and destruction points intersecting the wall...
This is easy to understand. But what confuses me, is when two waves that run along each other completely PARALLEL and apparently never intercept each other can interfere with each other.
At the centre of the screen, there is zero path difference between each pair of sources and they interfere constructively . But the resulting phase of each contructive interference isnt necessarily in the same phase isn't it . That means the centre of the sreen which we see much brighter shouldn't be that bright.
0:45 this representation of a light wave is absolutely wrong. Light is not one big wave which can span across space like a wayer wave. Its point fluctuations in the EM field. So multiple packets of light are actually going through the slit at the same time causing the interference
Technically, it should be single slit diffraction AND interference. Since, even in YDSE, the waveform diffracts and spreads through the slit, the key really is the interfetence which occurs.
1 and 8 pair constructively, rest pair up somewhat between constructive and destructive. However the 4 resultant waves have different phases and when they are added up, resultant will still be 0, albeit by applying more complicated mathematics.
It's actually fairly simple when you think about how interference occurs. It's addition that causes constructive and destructive interference, so its almost like doing this: 1 + 1 + 1 + 1 - 1 - 1 - 1 - 1 = 0 I am trying to simplify it with a really dumbed down explanation, but the concept is pretty much the same. His pairing is pretty much doing this: (1 - 1) + (1 - 1) + (1 - 1) + (1 - 1) = 0 There are no changes to the results, it is rearranged in this manner purely for explanation purposes.
It was clear. But why are we choosing two points which are half a slit distance apart. If we choose the top and the bottom point they would still interfere and form destructive pattern at lambda by 2 path difference. I am not understanding why we are choosing two points one in the top and the other one in the middle of the slit. Can anyone pls explain?
Around the same width as the wavelength of light being diffracted. Too large of a width would not create a diffraction pattern, instead only a slit would appear on the screen.
All light is diffracted by any size aperture. Diffraction is wavelength and aperture size dependant but diffraction exists in all cases. The smaller the aperture is with respect to the wavelength, the stronger the diffraction (when you cannot see it, it's more a case of "just because the diffraction is very small it doesn't mean it is not there" :)). In optical systems (things like telescopes) we see something called Airy Disks which are the single aperture diffraction effects (the resultant intensity Khan draws on the right in green), except it is a 2D incarnation. In the example of a telescope, the objective aperture (the "slit" doing the diffracting) can be metres wide, and the wavelenth is half a millionth of a metre; but the diffraction is still there and clearly visible. See... en.wikipedia.org/wiki/Airy_disk
Huygens principle is actually like having infinite slits and getting a uniform pattern because infinite points on wavefront form infinite wavefronts . may be!!
Don't understand why Theta remains the same at all point pairs surely it must change enough to make a difference. When the wavelength of light is in nm, (really tiny) a miniscule change in Theta would move the point from destructive to constructive.
Not sure if this counts but I have a growing light for plants. Unintended experiment. There is a small single slit where light breaks through. I get 4 single strands of light in a tight angle, Red, Purple, Red, Purple. What the heck is going on here? Edit: Ah I get it, now, the edges spits the light. 1 edge, two beams. 2 edges 4 beams. I've never seen this before, I was just mindblown how it could happen without trying. Perfect conditions.
If single slit can produce interference why don't we consider it in the ydse Why we consider slit in the ydse as point sources and works on only interference of that pattern
How come the points on the end of the wavefronts don't spread out as there's no point to one side of them? Also, when the wave meets a slit, how come the points next to the barrier on the reflected wave diffract, for similar reasoning as above?
I too as bothered by that, but then I wrote it down and realized that it does not matter, we do it so that the math and the equation derived is simple. As long as the separation between any two chosen waves is constant(because we want the path difference to be lambda/2 for destructive interference at a given point), everything works out.
It does matter, as the angle theta is not the same, if we keep the path difference the same for point sources on a straight parallel to the screen line.
I never said that the value of theta does not matter. In fact the whole equation is driven by it. I said that the separation between the rays does not matter. For a separation A, there will be a specific theta1 or for a seperation B, there will be a theta2. So, it does not matter if we take w/2 or w/3. We can use both to describe the equation. We choose w/2 because the math becomes easy.
I'm going to be very nitpicky: his full name is Huygens, with an 's' at the end, so it's actually 'Huygens' principle'. 'Huygen's principle' would suggest his name is 'Huygen'. The pronunciation is acceptable: I can't think of an English word with the same 'ui' sound, even the 'RightSpeech' video on TH-cam does'n get Huygens name right... Nice video BTW...
While pairing with each secondary sources wave, I noticed If I take the two end points. This doesn't the same relationship. So I want to ask, why am I taking the sources which have difference of W/2
It's actually complicated. We have to take slit width W tending to 0. We get number of waves.Then we use calculus to find integral of all waves coming from infinitesimally small width. Frensel did the calculations for us. But for students, we assume W/2. We can also take W/4. But it has to be even for destructive and odd for constructive.
@@amber9643 I forgot to add that it's proven experimentally that, For W/2, W/4 and so on, slit produces minima And for W/5, W/7 and so on, slit produces maxima. It actually has an explanation but you have to search it yourself.
there is no interference 1.all the points on the wave front are the source for new wave. 2.circles are drawn a the points as centres 3.the circles are joined tangentially 4.and light travelles
Spandan Bhattacharya we know tat amplitude =1+cos (teta), in forward direction, teta = 0deg hence amplitude will be 2..... in backwards direction, teta= 180deg ; ampli= 1+(-1) = 0....!!!! tats y there are no backwards wavelets..!! :)
![Single Slit Diffraction [AHL] - IB Physics](/img/n.gif)
You made single-slit and double-slit diffraction easy to understand and now I am not panicking over physics anymore, you're the true og, my good sir.
indeed
As representative of the state general of the Dutch royal democracy I hereby pardon your mispronunciation.
Loal
Lol
Lolol what even
lole
Lol 😂
i like how he didn't ask for a like and sub in the end and simply said "you're all good"
Ahmed Ibrahim which deserves both a sub and a like
Because it's Khan academy
That is called nonprofit organization...
Its so stupid my testbook tells me nothing like this and expects me to understand. As others has mentioned you are a hero. its so easy to enjoyable to learn with your videos. thumbs up
Same my book is also doesn't explain this topic well..This vedios explain it in detail
This concerpt never made sense to me until I watched this video,,, thank you a lot for the lesson
THANK YOU, why isn't Huygen's principle taught BEFORE two slit interference in all the texts, you'd think this knowing this beforehand would save you from a lot of misguided connections.
milkwasabadcholce because it doesnt have anything to do with the doubleslit experiment. Different causes..
Kim double slit interference involves the interference of two point sources. Single slit interference involves the interference of an infinite amount of point sources, and defines diffraction: they are definitely linked.
science is all about being hard ;)
in india huygen's principle and wave nature of light is taught first and then we move onto young's double slit experiment.
@@REVIEWSGALATTAyea exactly, its taught before and tbh it caused SO MUCH confusion, i have been stuck on it for a day 🥲 i was trying to find a connection bw huygens and double slit and obviously got more confused
if huygen's interpretation was taught to me before geometrical optics I'd be an astronomer/photonicist. It's such a leap from too simple -> crazy, to weird -> everything become simple
So let me bring to your attention the elephant in the room.... What are the light waves refracting THROUGH?? The universe is pretty simple when you think about it and don't overcomplicate and keep bombarding and bamboozling teenagers with complex formulas and experiments that are just the same as magic tricks (smoke and mirrors) . Thankfully I managed to see around the slit as it where 30 years after being brainwashed😁 n.b. I have left a few comments under single and double slit experiment TH-cam videos that have questioned the narrative and all of a sudden I am now not allowed to post a new comment, just reply to one?? ..I must be onto something 🙄
The ONLY single slit diffraction video that’s actually makes sense. Thank so so so much Khan Academy!! Saved my A level!!
😭😭I wanna cry coz I still don’t understand this
Please refer to " Science Alert, current research in Physics".
4:53 Engineers approximation: 8≈∞
hahaha
Geniss i mean genius
I mean I'm drunk
Can someone please just make a playlist of all videos by David SantoPietro. That would just make life so much more easier.
Twhen my school teacher taught me this, i was unable to understand it, but with your explanation i think i understood it just fine.
THANK YOU,
I have never had such a strong sense that something is wrong
one of the best explanations of single-slit diffraction. good job
Khan Academy....The Savior
Your videos are absolutely brilliant! Excellent job and thank you for making this topic so clear and visual!
I had to watch this 3 times to finally get it
You made this critical topic easy to me. Thank you so much sir. Lives in Bangladesh 🇧🇩
Finally! Thank you for your amazing explanation. Absolute legend.
omg, what a good way to explain it visually, exactly like how i like to learn physics. THANK YOU
Very interesting, especially for a lay man who has been told so many times that only a double slit will show interference, thanks to quantum mechanics...
By the way, your pronunciation of Huygens is quite good if I compare it how 't Hooft is sometimes announced on TH-cam. :-)
Fransamsterdam in your comment quantum mechanics solved the main question Hahn that each point acts as a wave thanks........
thank u thank u the first sentence that u said is why I'm looking for this in the internet I was so frustrated that all the books I have were talking about waves interfering with out telling me why they interfere in a single slit in the first place
thank you .... now i understod that.... that was killin my brain for years :3
This just motivated me alot to learn physics:D
Omg you are a life saver ❤️... thank you very very much. It solved all of my questions which I posted in double slit experiment video regarding if the wave spreads out does that mean wavelength of that particular wave increases?
Nice 😊
Thank you for this precious resources for the students
I understand your explanation of single slit diffraction of light and the production of small maxima either side of the big central maxima. However, whenever single slit diffraction diffraction is set up in a ripple tank there is no sign of the small maxima either side of a central maxima.
Why are we only considering the points spaced w/2, w/4, w/8 apart? Shouldn't we be considering each pair of points instead since they each would contribute to the intensity? Also, it seems like even if the w/2 cancel because they are half a wavelength apart, the points w/4 apart wouldn't cancel because by definition they must be less than half a wavelength apart so I don't see the logic there at all.
Straightforward,pure teaching
I don't know why but I never seem to understand Kahn Academy. It seems like it should make sense (especially compared to my teachers) but it just goes in one ear and out the other.
Thank you so Much!!! The start of the video was exactly answering what I was looking for I had no idea their was so much more to it
Brilliant...thanks a lot🤗
wonderful explanation💥
Terima kasih, materinya sangat bagus
A true lifesaver
I don't know if i have got it right but, the way you cal the different wave lengh as w/2*sinΘ confuses me,because the right part of light from the footprint seem to have different wave distance
For once in my lifetime I was proud of myself for being clever but then "Ah! Not really." 14:14
This helped a lot.
Well explained, thank you. Question: if hyugen's principle is true, how can you direct any light at a particular direction, for example a pointer laser? Shouldn't any light source always be a source of spherical wave fronts?
That are polarized light
I think it's because Laser light is collimated which means the light rays are all parallel so, it spreads very little on it's journey.
@@ht-ve9fe ok so would agree then that hyugen's principle in fact is not true?
@@johanneskrv No I wouldn't because after certain distances lasers do begin to diffract. Other than what I need to know and the brief information I have read, I am not really in a position to comment whether he is right or wrong and he more than likely has greater understanding of the subject than you do and certainly more than me. However, from what I've studied, nothing disproves and there's plenty of people who have devoted their lives to Physics and even this topic which I am sure would have been disproved by now if he were to be wrong.
@@ht-ve9fe i'm not saying he is incompetent. Just wondering why in some cases the hyugens principle seems to be in effect and in others not. This seems contradictory.
This has bothered me for a very long time. Thank you very much for this video. Great explaination! My very minor criticism is when you say that the wave cancel completely at that point. Later, you explain that there is a slight difference [in the length of the triangle]. At that point you could add that’s why they don’t cancel entirely. But, I wouldn’t redo or add to the video if it changes the duration google posted as it gives you precidence! Again, great job and thank you for bringing up Huygen’s! - Jim Foit
2:40 , 3:21 🌟
this is awesome I wish someone could have taught like this to me earlier
This man is speaking the language of gods
Good explanation. Thank You..
Why are we taking the topmost and middle point of the slit to get a dark point? How do we know?
Ya I don`t get it either. He said "if it's destructive at half the slit length" and proceeded to make that scenario the official formula
@Ryana Imtiaz
He took the "topmost" and "middle" point, only for the sake of explanation. But this applies for any two arbitrary "points", as in reality, there's an infinite number of "points" within the band of light that passes through the slit, where you can't really select "points" that are "topmost" and "middle", as concepts like "points" are just an approximation...
@@D8Football actually, it isn't the "official formula" but more of a presumption made to help find an explanation to why it so, i.e., why there's a dark region between the bright ones.
I was so confused with this equation that path difference of minima is getting for constructive. Thank you for video
2:11
is really cool and "UGE"
The diagram is so damn satisfying! 😬
I know this is weird but it is (to me) 🙄
weird*
Omg hyungugons thing is cool
Love ur teaching skills🤘❤️
Doubt: what if we took 9 sources(as in odd sources) instead of 8, wouldn't there be further complications with deciding whether it ends us up with constructive or destructive interference ? Kindly respond.
I don't know if i should be happy or disappointed at myself because I'm not upset about anything lol
How is 5th source w/2. If there are 8 sources than the 4th source should have been w/2.
God bless you Sir
Brilliant. But puzzled by why the interference pattern is always horizontal. If Huygen's is correct then shouldn't there be a spherical pattern?
Or... What if there was just a small hole instead of a slit? What would the pattern be?
If we make a small hole, then the pattern is, indeed, spherical. It's called an Airy disk.
Please refer to " Science Alert, current research in Physics".
How does infinitely many diffracting waves make a ‘full’ wavefront before the slit then? Why does the slit make the infinitely many diffracting waves into an interference pattern not a ‘full’ but smaller wavefront?
I guess I understand that the slit can make a interference pattern but why doesn’t it do the same thing for a normal wavefront?
Great video ! Very informative !
Just one question , how is the difference between the two arbitrary points chosen , w/2 ?
Anubhav Bahadur if you calculate it comes out w/2.
There are 8 points hence the distance between two consecutive points is w/8.
The points he chose had 4 points in between.
Hence, the distance is 4*w/8 = w/2
Actually it's not arbitrary, you set and divide the slit line up in such way that two points superposition cancels and things pair up.
@@papiyaghosh102each two pairs cancel out assuming slit is very small right?
i was wondered that if the screen is a curved arc, what would the interference pattern look like?😮😊
Thanks, man.
I am the best in my physics class! lol
path difference for destructive points is (2n+1)*half lambda,so in this case why is asin theta=n*lambda not asin theta=(2n+1)*lambda
Well then are there no troughs??...u can keep zooming in and all u see are wave fronts??
please make a video on maxima minima and everything please?.
How can we be so sure that destructive interference will occur at those specific points like if the waves from the uppermost and the middle points interfere destructively at that point then it could also interfere constructively with the waves from the other points .... And why should we take half of the slit and not the full one ??? They are just ripping my mind @part!!!
Sheraz Khan experiment
For a destructive interference to happen, all the waves have to cancel out - the sum of their "heights" has to be 0.
That means that for every wave with height y there has to be a wave with height -y at that point we are looking at.
If it were otherwise, all other waves would cancel out and that wave would be "left alone" so that point wouldn't be a dark spot, the sum of the heights would be equal to its height = y.
So whichever way you add the waves, you should always end up with the sum = 0 for it to be a dark spot.
for example lets say you add up separately waves with positive heights at that point in a specific time, and separately the ones with negative heights then. lets say the heights are y,x,c,v,b... and -y,-x,-c,-v,-b... or what ever names you want to call them. The sum of the positives is y+x+c+v+b+..., and the sum of the negatives is -y-x-c-v-b.... and the ULTIMATE sum is (y+x+c+v+b.)+(-y-x-c-v-b...). As you can see whatever the sum of the positives is, the sum of the negatives is always - that, so the sum is always zero for that point. So pairing off the waves is just editing the sum (y+x+c+v+b.)+(-y-x-c-v-b...) to y-y + x-x + c-c + v-v + b-b +...
(You must note that the heights of the waves are dependent on time so when I say that a wave's height is positive I mean its positive in a certain moment in time, not always, same for negative).
You might ask well how do bright spots appear then.
Well for bright spots at a specific moment there are more "positive" waves than "negative" ones, or vice versa, so not all of them cancel out, and thus the sum is not equal to zero and there is a constructive interference.
Hope this helped :)
IKstreme Yeah that was a great explanation . But what I am thinking is that why should we apply plain Mathematics to it .... If I apply some logic , what my mind says is that there shouldn't be any destructive interference . Like , for example all the waves arrive at the same time ... You said there will be y for -y and x for -x and so on . But let say the y . Thus y could also be in destructive interference with the x wave or -x or what else you call it . Then , why should we add up the two y's (positive and negative) . Now you might say that x and y can't be added because they are different variables but let's forget about that because we don't have an equation like that . It could be numbers and then numbers could be added in any way . What my concern is that why should we be adding two waves from two distinct points while there are waves from numerous and numerous points . Based on your explanation , I think when we add them all together we won't be getting either a constructive or destructive interference anywhere ...There shouldn't be one of them.. I hope you get my point.
if y and -y are in a destructive interference, and y and -x are also in destructive interference, then so are x and -y, so any way you pair them up the total sum is always zero:
(y-y)+(x-x)= 0+0 = 0
(y-x)+(x-y)=0+0 = 0
Even if you add waves that are not in destructive interference the total sum will be 0. lets say y+x=a, then by simple maths (-y) + (-x) = -(x+y) = -a.
So even if you do:
(y+x)+(-y-x)=a-a=0
you still get zero because ultimately they somehow all cancel out
IKstreme Ok it makes sense but I am still very much confused . I am just thinking that how can there be + for every - when the waves are pointing at a point far away from the central point ?
oh my god u saved me thank uuuuuuuuuu
yes but when you make the middle of the slit half polarized then they should not interference with the other neighbourhood. also the cancellation is must happen at the slit point only. its not really like a lake of waves.. once it cancells its done, that wave wont de-cancell due to phase swap.. thats why when u move the wall back and forward, you dont see alternating blobs that would represent the peaks and destruction points intersecting the wall...
Im living for your "w" prononcuation, cuz its so cute
If all integers give destructive, what gives constructive besides 0??
When path difference is an even number of half wavelengths you get constructive whereas an odd number of half wavelengths gives destructive.
This is easy to understand. But what confuses me, is when two waves that run along each other completely PARALLEL and apparently never intercept each other can interfere with each other.
Wow thank you!
"Crazy, but true!"
thank you!
At the centre of the screen, there is zero path difference between each pair of sources and they interfere constructively . But the resulting phase of each contructive interference isnt necessarily in the same phase isn't it . That means the centre of the sreen which we see much brighter shouldn't be that bright.
Don't you put a convex lens at the opening of the slit?
That's how it's explained in my textbook
0:45 this representation of a light wave is absolutely wrong. Light is not one big wave which can span across space like a wayer wave. Its point fluctuations in the EM field. So multiple packets of light are actually going through the slit at the same time causing the interference
umm shouldn't this be called single slit diffraction?
Technically, it should be single slit diffraction AND interference. Since, even in YDSE, the waveform diffracts and spreads through the slit, the key really is the interfetence which occurs.
In india it is diffaraction lol
are you dumb? there is no medium change so no diffraction. you must be braindead ffs
@@yaoooy are you confusing with refraction?
@@yaoooy Diffraction happens in same medium. When medium changes, it's refraction.
Actually at 11:03 shouldn't it be m+1/2??
thanx sir
So, a wave front is a full constructive interference ?
Can you tell me in which video you derived dsin(theta)=delta x?
What if I pair up topmost and bottommost point ???
1 and 8 pair constructively, rest pair up somewhat between constructive and destructive. However the 4 resultant waves have different phases and when they are added up, resultant will still be 0, albeit by applying more complicated mathematics.
It's actually fairly simple when you think about how interference occurs.
It's addition that causes constructive and destructive interference, so its almost like doing this:
1 + 1 + 1 + 1 - 1 - 1 - 1 - 1 = 0
I am trying to simplify it with a really dumbed down explanation, but the concept is pretty much the same.
His pairing is pretty much doing this:
(1 - 1) + (1 - 1) + (1 - 1) + (1 - 1) = 0
There are no changes to the results, it is rearranged in this manner purely for explanation purposes.
Leyton Zhang holy shit u just cured my brain cancer gj broo
It was clear. But why are we choosing two points which are half a slit distance apart. If we choose the top and the bottom point they would still interfere and form destructive pattern at lambda by 2 path difference. I am not understanding why we are choosing two points one in the top and the other one in the middle of the slit.
Can anyone pls explain?
how wide does the slit need to be for the effect to be present?
Around the same width as the wavelength of light being diffracted. Too large of a width would not create a diffraction pattern, instead only a slit would appear on the screen.
wendy L Are you sure? Veritasiums video shows a slit at something like 0.1mm slit. How long/wide is a wavelength?
Depends on the type of electromagnetic wave
Abdul Sharapov uhm, but its not 0.1mm...
All light is diffracted by any size aperture. Diffraction is wavelength and aperture size dependant but diffraction exists in all cases. The smaller the aperture is with respect to the wavelength, the stronger the diffraction (when you cannot see it, it's more a case of "just because the diffraction is very small it doesn't mean it is not there" :)).
In optical systems (things like telescopes) we see something called Airy Disks which are the single aperture diffraction effects (the resultant intensity Khan draws on the right in green), except it is a 2D incarnation. In the example of a telescope, the objective aperture (the "slit" doing the diffracting) can be metres wide, and the wavelenth is half a millionth of a metre; but the diffraction is still there and clearly visible.
See... en.wikipedia.org/wiki/Airy_disk
Huygens principle is actually like having infinite slits and getting a uniform pattern because infinite points on wavefront form infinite wavefronts . may be!!
Cbse students hit like.
Which app you use for this demonstrate
Don't understand why Theta remains the same at all point pairs surely it must change enough to make a difference. When the wavelength of light is in nm, (really tiny) a miniscule change in Theta would move the point from destructive to constructive.
Not sure if this counts but I have a growing light for plants. Unintended experiment. There is a small single slit where light breaks through. I get 4 single strands of light in a tight angle, Red, Purple, Red, Purple.
What the heck is going on here?
Edit: Ah I get it, now, the edges spits the light. 1 edge, two beams. 2 edges 4 beams. I've never seen this before, I was just mindblown how it could happen without trying. Perfect conditions.
Thank you sir you really give a very nice and simple explanation 🙂. Sir is it possible to also add mathematics of diffraction also
Cool video
If single slit can produce interference why don't we consider it in the ydse
Why we consider slit in the ydse as point sources and works on only interference of that pattern
How come the points on the end of the wavefronts don't spread out as there's no point to one side of them? Also, when the wave meets a slit, how come the points next to the barrier on the reflected wave diffract, for similar reasoning as above?
nice
It's like single-slit interference is like an infinite-slit, continuous version of the double-slit interference, I guess
why does it have to be the rays that are half the slit length apart being half a wavelength out of phase?
+Kelvin Fan how come its not like (w/3)sine theta = lambda/2
I too as bothered by that, but then I wrote it down and realized that it does not matter, we do it so that the math and the equation derived is simple. As long as the separation between any two chosen waves is constant(because we want the path difference to be lambda/2 for destructive interference at a given point), everything works out.
Parth Upmanyu Ohh. Thanks for taking the time to reply! haha. I will think about it. I think what you're saying makes sense.
It does matter, as the angle theta is not the same, if we keep the path difference the same for point sources on a straight parallel to the screen line.
I never said that the value of theta does not matter. In fact the whole equation is driven by it. I said that the separation between the rays does not matter. For a separation A, there will be a specific theta1 or for a seperation B, there will be a theta2. So, it does not matter if we take w/2 or w/3. We can use both to describe the equation. We choose w/2 because the math becomes easy.
Huygen is "HOW-chen," with 'ch' pronounced as in 'loch."
Walter Lewin said "HHURRHENNS"
I'm going to be very nitpicky: his full name is Huygens, with an 's' at the end, so it's actually 'Huygens' principle'.
'Huygen's principle' would suggest his name is 'Huygen'.
The pronunciation is acceptable: I can't think of an English word with the same 'ui' sound, even the 'RightSpeech' video on TH-cam does'n get Huygens name right...
Nice video BTW...
Lifesaver
While pairing with each secondary sources wave, I noticed If I take the two end points. This doesn't the same relationship. So I want to ask, why am I taking the sources which have difference of W/2
It's actually complicated. We have to take slit width W tending to 0. We get number of waves.Then we use calculus to find integral of all waves coming from infinitesimally small width. Frensel did the calculations for us. But for students, we assume W/2. We can also take W/4.
But it has to be even for destructive and odd for constructive.
@@haemophilusinfluenza9199 so this derivation is to understand it simply. Thank you very much for your response.
@@amber9643 I forgot to add that it's proven experimentally that,
For W/2, W/4 and so on, slit produces minima
And for W/5, W/7 and so on, slit produces maxima.
It actually has an explanation but you have to search it yourself.
@@haemophilusinfluenza9199 ok. I will do some research.
there is no interference
1.all the points on the wave front are the source for new wave.
2.circles are drawn a the points as centres
3.the circles are joined tangentially
4.and light travelles
ajey bhat But the circles that traverse backwards get cancelled out by incoming wavefronts right? I don't see whats wrong with his presentation.
Spandan Bhattacharya we know tat amplitude =1+cos (teta), in forward direction, teta = 0deg hence amplitude will be 2..... in backwards direction, teta= 180deg ; ampli= 1+(-1) = 0....!!!! tats y there are no backwards wavelets..!! :)