Glad it helped - again there is so much going on here but it is a start to try and explain why white light is separated into different colours on reflection - enjoy your physics and thanks for taking the time to comment.
Thanks Will. Pleased to hear it - I am aware that it is a bit of an over-simplification and there are other things going on but the main thing is to understand path differences of the order of a wavelength of light in the reflected waves causing interference effects. Thanks for taking the time to comment.
Thanks so much for the comment - as ever, makes it all worthwhile. I don't feel it is the best explanation of diffraction directly - perhaps I should do a video on diffraction especially.
Bahar - so pleased you found it useful. I had never realised that this short film would help so much! Thanks for leaving your comment and enjoy your physics!
Glad you liked it - I felt I hadn't done too good a job on this as the explanation of what happens and the surface of a DVD are quite complex but, as ever, I enjoyed making it.
Beautifully explained! Sir, I have a question regarding interference. When we perform the derivations for interference in Young's double slit experiment, for constructive interference, we get the path difference as ( n lambda) where lambda is the wavelength. But when we consider diffraction of light at a single slit, then ( n lambda) gives the path difference for destructive interference. In case of Young's experiment, i could visualize the situation by imagining two wave crests adding up just like you said in this video. But I cannot understand how does the reverse happen in diffraction i.e : how can ( n lambda) be the condition for destructive interference. Please help me in visualizing this concept just like you explained diffraction and interference in this video. I am preparing for my exams and I want to understand this concept better.
Good question. So interference is n lambda, where n is the number of whole wavelength path differences - I know you understand this. But, for single slit it should be m lambda, where m is the 'order' of the minima, not the number of wavelengths. Does that make sense?
@@AnthonyFrancisJones What do we precisely mean by the order of the minima? It is written in the notes, that for diffraction due to a single slit, if path difference = lambda, then P will be the point of minimum intensity because of every point in the upper half (of the lens used in the diagram while deriving) , there is a corresponding part in the lower half of the lens for which the path difference between the secondary waves reaching P is lambda/ 2. Thus destructive interference takes place at P and hence P is the point of first secondary minimum. I just can't figure out how to add the waves from the upper half and lower half of the lens. This all is getting me confused. I simply want to figure this out in terms of addition or subtraction of wavelengths like two crests giving us constructive interference and a full wavelength plus half wavelength giving us destructive interference. Can you help me in understanding this concept in terms of addition or subtraction of wavelengths so that I can visualize why destructive interference is taking place? And thank you very very much for replying. Also I want to know the basic difference between Young's experiment of interference and this concept of single slit diffraction. This concept is complicated as well as remarkable!
@@TN-id4yx All good questions. So, and as you will appreciate in the future, all explanations at this stage are simplified. Diffraction can be considered to be a single slit effect - spreading out of a wave going through a single wavelength width slit. Understanding how this forms is best done by looking at how the formula is derived from looking at a wave from the bottom and the top of the slit meeting at any angle other than straight through. Every place on the intensity curve that is a minima is called an order with the central one being the zero order. (I am not sure why a lens was needed in the case you mention) Do please look in another (UK) textbook to help with this. Now you will notice that even though we talk of diffraction only from a single slit, multiple wavefronts meet to produce the diffraction pattern and so it is an interference effect too but we usually do things such that the very narrow slit width makes us only see the central order maxima - light spreading out into a circular wavefront from a plane one. Now with Young's slits we already know that each slit will spread the point light sources out into two circular wavefronts (diffraction) so we consider where the maxima are caused by the interference of these two wave fronts meeting each other. My lesson on interference might help you with a real ripple tank example ruclips.net/video/cXEEjSkkFIw/видео.html Also have a look at Huygen's Construction which will help too. Hope that moves you forward a bit further!
@@AnthonyFrancisJones Thank you very much. I am getting this concept better now. In the NCERT text, the derivation of diffraction from a single slit involves a lens and both the derivations for interference and diffraction were confusing me but I think now I am getting a better grip of the concept. I will definitely watch your video on interference.
@@TN-id4yx If only I could post pictures here I could put up a diagram of simple single slit diffraction. Yes you get diffraction from lenses too but they are very wide apertures. You can of course use the lens to get a parallel beam of light to shine on the slit!
Thanks - glad you liked it. It is a bit of a simplification of what is going on as the surface effect of the DVD on light is a bit more complex but I think it in general captures the idea of a diffraction grating. Hope you enjoy the others too.
@@AnthonyFrancisJones btw sir can i ask you something? The pits in the CD in uniform arrangement cause the light to reflect or diffract? Im sorry for asking if you not understand my english. Im confused a little bit between these two.
@@muhammadridhwan266 Good question and your English is excellent! The pits are made shiny so they reflect the light - the light rays coming off each pit then interfere with each other to give a diffraction pattern pattern you see - I think of it more as an 'interference pattern' So there are two things happening. It is what we would call a reflection grating.
@@AnthonyFrancisJones so this means that the colour rainbow we see is the interference pattern of constructive part? I mean if we change our angle of view then suddenly we see the rainbow colour change place too, thats mean the constructive and destructive plays the role here? Im sorry again for asking. I hope you can understand the question.
@@muhammadridhwan266 yes you are right - I agree that I have oversimplified lots of things here but you basically have it right. Normally you would see maxima for each colour giving the rainbow (spectrum) where each colour's wave is in phase where they overlap. You should also see second order maxima too. Basically think of a DVD surface as being just a diffraction grating with white light landing on it and rather than the light going through slits it reflects off them. Normal diffraction gratings will do this too. Hope that helps.
Any CD or DVD will do this if held up to a bright white light source. I think I was using the light from the window here. The other demonstration is a bright light source (projector) with a slit in front of it to give a narrow beam and a glass prism. It is possible to take measurements but this was just to show the idea rather than to delve into the maths. Hope that helps!
Glad you liked it - not an easy explanation and these things are always more complex than it seems but the idea of a periodic pattern on the disc causing interference is a good starting point. Keep enjoying your physics!
Good question - if the gap is small enough - close to the wavelength of the light - then the answer is yes, but our eye lashes are a bit far apart and not evenly spaced so it is unlikely you will see a nice spectrum of colours with white light nor be able to focus it effectively!
Piyush, not an easy one... So light onto the DVD is white (Red, Blue and Green waves mixed together). When they reflect waves of each colour overlap and therefore add up mathematically. Now for the easy bit... if a crest/peak of a wave meets the crest of a similar wave they add up to make stronger light - however, if a crest overlaps with a trough of another similar wave they will add up to zero and cancel that colour out. So in some places the colours can be seen - constructively interfere and in other places they are cancelled out - destructively interfere. That is a massive simplification but hope it helps. Look up Young's Slits and also Constructive and Destructive Interference - that should help you too. Thanks for asking! Well done in trying to understand this tricky and odd bit of physics!
Good question. I guess as the atmosphere is especially transparent to RGB wavelengths we have evolved to use these wavelengths the most. We can of course feel infrared too. LED 'pixels' produce the colours that we need to be able to see an image by mixing R, G and B in different amounts. I will do a video on that at some stage. Isn't it strange that R + G LEDs produce light that when we see it we perceive it as yellow for example!
Anthony Francis-Jones isn't it possible to run completely with white led? And also when the TV is showing space in some fiction or solar system why the led tv lights black colour and how it emits black colour by which combination? Why doesn't it shut off during those areas to show more dark black and to save power?
@@unicornlion8078 The rainbow is caused by refraction (and some reflection) in water droplets in rain. This is different from the diffraction happening here. Topic for another video I feel!
@@unicornlion8078 Good question. A lot of LED screens get more sensitive control over the colours and brightness by having multiple LEDs. Some used to have one R and one B and two G running at half intensity to get more colour control. The way the eye/brain responds to colour is interesting if an intense light source is next to a very dim on the eye/brain sees the dim one as black or off. This is a really interesting and complex topic. Well done for spotting it.
OK, I guess you realise that this is with electrons so nothing to do with colour..But as you say if similar physics is done with light there is an equation which states where each maxima will be knowing the wavelength (colour) you are interested in. [n Lambda = s sin(theta)]. This is the diffraction grating or multiple slit equation. Sorry I could not write it in a better way - one to google.
Yousef, I am not sure I get your question. I will try and answer but will need a little more detail. This video uses classical physics to explain the colours off a DVD. Yes, a bit simplistic but it does allow for some correct predictions.
thanks a million!...I was overwhelmed by this concept and now just landed
Thanks - glad it was helpful.
Thaaaank you for this explanation. This phenomenon has boggled my mind forever!!!
Glad it helped - again there is so much going on here but it is a start to try and explain why white light is separated into different colours on reflection - enjoy your physics and thanks for taking the time to comment.
It was really helpful to visualise this phenomena with different example that u did
Great - pleased that it helped you understand this interesting phenomenon. Thanks for taking the time to comment and as ever enjoy your physics!
Great demonstration and explanation, I wish you are my physics prof.
That's kind of you. Glad it helped you with the understanding of this tricky effect!
Amazing sir...loved the explanation!
Thanks Arjun. Glad you liked it and the explanation helped. Thanks for commenting - makes it all worthwhile! As ever, enjoy your physics!
@@AnthonyFrancisJones thanks
Very helpful. Thanks, FJ👍🏼👍🏼
Thanks Will. Pleased to hear it - I am aware that it is a bit of an over-simplification and there are other things going on but the main thing is to understand path differences of the order of a wavelength of light in the reflected waves causing interference effects. Thanks for taking the time to comment.
This was sooo helpful more than my text book which just said it "diffraction"
Thanks so much for the comment - as ever, makes it all worthwhile. I don't feel it is the best explanation of diffraction directly - perhaps I should do a video on diffraction especially.
Super helpful!
I wish I had a Teacher like you.
Excellent - glad you found the film useful and, as ever, enjoy your physics!
You are a good explaner. 👍 Great
Thanks, that's very kind and glad you enjoyed it.
Thank you for your help!
Glad it helped! Keep enjoying your physics!
You deserve Nobel prize cuz u made me to understand it
Bahar - so pleased you found it useful. I had never realised that this short film would help so much! Thanks for leaving your comment and enjoy your physics!
Anthony Francis-Jones
And thank you for giving free practical lectures!
@@baharosman1416 Pleasure. Plenty more to come!
Thanks a bunch sir. Hats off for you
Thanks Steven. Glad you found it useful. I want to see what other tubes we have in the lab and do some of those at some stage. Enjoy your physics!
your most welcome sir
Well explained, Well done!
Glad you liked it - I felt I hadn't done too good a job on this as the explanation of what happens and the surface of a DVD are quite complex but, as ever, I enjoyed making it.
thank you sir it waas very helpful for me as i could not answer when somebody asked that what is dispersion of light. Thank you very much😀😀😀😀😀😀😀
Thanks, again, for taking the time to comment. I know I have oversimplified a bit here but I am so pleased it helped you. Keep enjoying your Physics!
Helped me a lot, thanks
Great Danilo. So pleased it helped - not an easy topic but I hope you feel you have a better understanding of it now.
Beautifully explained!
Sir, I have a question regarding interference. When we perform the derivations for interference in Young's double slit experiment, for constructive interference, we get the path difference as ( n lambda) where lambda is the wavelength. But when we consider diffraction of light at a single slit, then ( n lambda) gives the path difference for destructive interference. In case of Young's experiment, i could visualize the situation by imagining two wave crests adding up just like you said in this video. But I cannot understand how does the reverse happen in diffraction i.e : how can ( n lambda) be the condition for destructive interference. Please help me in visualizing this concept just like you explained diffraction and interference in this video. I am preparing for my exams and I want to understand this concept better.
Good question. So interference is n lambda, where n is the number of whole wavelength path differences - I know you understand this. But, for single slit it should be m lambda, where m is the 'order' of the minima, not the number of wavelengths. Does that make sense?
@@AnthonyFrancisJones What do we precisely mean by the order of the minima? It is written in the notes, that for diffraction due to a single slit, if path difference = lambda, then P will be the point of minimum intensity because of every point in the upper half (of the lens used in the diagram while deriving) , there is a corresponding part in the lower half of the lens for which the path difference between the secondary waves reaching P is lambda/ 2. Thus destructive interference takes place at P and hence P is the point of first secondary minimum.
I just can't figure out how to add the waves from the upper half and lower half of the lens. This all is getting me confused. I simply want to figure this out in terms of addition or subtraction of wavelengths like two crests giving us constructive interference and a full wavelength plus half wavelength giving us destructive interference. Can you help me in understanding this concept in terms of addition or subtraction of wavelengths so that I can visualize why destructive interference is taking place? And thank you very very much for replying.
Also I want to know the basic difference between Young's experiment of interference and this concept of single slit diffraction.
This concept is complicated as well as remarkable!
@@TN-id4yx All good questions. So, and as you will appreciate in the future, all explanations at this stage are simplified. Diffraction can be considered to be a single slit effect - spreading out of a wave going through a single wavelength width slit. Understanding how this forms is best done by looking at how the formula is derived from looking at a wave from the bottom and the top of the slit meeting at any angle other than straight through. Every place on the intensity curve that is a minima is called an order with the central one being the zero order. (I am not sure why a lens was needed in the case you mention) Do please look in another (UK) textbook to help with this. Now you will notice that even though we talk of diffraction only from a single slit, multiple wavefronts meet to produce the diffraction pattern and so it is an interference effect too but we usually do things such that the very narrow slit width makes us only see the central order maxima - light spreading out into a circular wavefront from a plane one. Now with Young's slits we already know that each slit will spread the point light sources out into two circular wavefronts (diffraction) so we consider where the maxima are caused by the interference of these two wave fronts meeting each other. My lesson on interference might help you with a real ripple tank example ruclips.net/video/cXEEjSkkFIw/видео.html Also have a look at Huygen's Construction which will help too. Hope that moves you forward a bit further!
@@AnthonyFrancisJones Thank you very much. I am getting this concept better now. In the NCERT text, the derivation of diffraction from a single slit involves a lens and both the derivations for interference and diffraction were confusing me but I think now I am getting a better grip of the concept. I will definitely watch your video on interference.
@@TN-id4yx If only I could post pictures here I could put up a diagram of simple single slit diffraction. Yes you get diffraction from lenses too but they are very wide apertures. You can of course use the lens to get a parallel beam of light to shine on the slit!
Thank you sir it was very helpful
Pleasure. Glad you found it helpful! Good luck with your studies.
Thanks sir for the wonderful video
Thanks - glad you liked it. It is a bit of a simplification of what is going on as the surface effect of the DVD on light is a bit more complex but I think it in general captures the idea of a diffraction grating. Hope you enjoy the others too.
thank you sir . your explanation is really understandable.
Thank for your kind comment - so pleased it was useful. It's good to hear because I was not sure I really made a good job of it!
@@AnthonyFrancisJones btw sir can i ask you something? The pits in the CD in uniform arrangement cause the light to reflect or diffract? Im sorry for asking if you not understand my english. Im confused a little bit between these two.
@@muhammadridhwan266 Good question and your English is excellent! The pits are made shiny so they reflect the light - the light rays coming off each pit then interfere with each other to give a diffraction pattern pattern you see - I think of it more as an 'interference pattern' So there are two things happening. It is what we would call a reflection grating.
@@AnthonyFrancisJones so this means that the colour rainbow we see is the interference pattern of constructive part? I mean if we change our angle of view then suddenly we see the rainbow colour change place too, thats mean the constructive and destructive plays the role here? Im sorry again for asking. I hope you can understand the question.
@@muhammadridhwan266 yes you are right - I agree that I have oversimplified lots of things here but you basically have it right. Normally you would see maxima for each colour giving the rainbow (spectrum) where each colour's wave is in phase where they overlap. You should also see second order maxima too. Basically think of a DVD surface as being just a diffraction grating with white light landing on it and rather than the light going through slits it reflects off them. Normal diffraction gratings will do this too. Hope that helps.
Thanks sir
Pleasure - glad it helped and good luck with your studies.
how did you do it can you share with me , thank you
Any CD or DVD will do this if held up to a bright white light source. I think I was using the light from the window here. The other demonstration is a bright light source (projector) with a slit in front of it to give a narrow beam and a glass prism. It is possible to take measurements but this was just to show the idea rather than to delve into the maths. Hope that helps!
thx this was helpful for my lesson
Thanks - glad it helped - not an easy thing to explain and a bit 'simplified' here but if it gets you started that's great!
Thank u , great job !!!
Glad you liked it - not an easy explanation and these things are always more complex than it seems but the idea of a periodic pattern on the disc causing interference is a good starting point. Keep enjoying your physics!
Plz let me know do light diffract after passing through our eye lashes? Do they behave as slits?
Good question - if the gap is small enough - close to the wavelength of the light - then the answer is yes, but our eye lashes are a bit far apart and not evenly spaced so it is unlikely you will see a nice spectrum of colours with white light nor be able to focus it effectively!
Sir please explain simply and in easy way
Piyush, not an easy one... So light onto the DVD is white (Red, Blue and Green waves mixed together). When they reflect waves of each colour overlap and therefore add up mathematically. Now for the easy bit... if a crest/peak of a wave meets the crest of a similar wave they add up to make stronger light - however, if a crest overlaps with a trough of another similar wave they will add up to zero and cancel that colour out. So in some places the colours can be seen - constructively interfere and in other places they are cancelled out - destructively interfere. That is a massive simplification but hope it helps. Look up Young's Slits and also Constructive and Destructive Interference - that should help you too. Thanks for asking! Well done in trying to understand this tricky and odd bit of physics!
What would be the best way of myself weighting in the 70s. When i get a cd to weight in the 70s then all i have to do is place them on the ground.
Jason, would like to help but I am really sorry but I do not understand your question. Can you explain further please.
Why is our eyes sensitive to RBG colours and why is the pixels of led are in those colours?
Good question. I guess as the atmosphere is especially transparent to RGB wavelengths we have evolved to use these wavelengths the most. We can of course feel infrared too. LED 'pixels' produce the colours that we need to be able to see an image by mixing R, G and B in different amounts. I will do a video on that at some stage. Isn't it strange that R + G LEDs produce light that when we see it we perceive it as yellow for example!
Anthony Francis-Jones isn't it possible to run completely with white led? And also when the TV is showing space in some fiction or solar system why the led tv lights black colour and how it emits black colour by which combination? Why doesn't it shut off during those areas to show more dark black and to save power?
And does this also explain the phenomenon of rainbow?
@@unicornlion8078 The rainbow is caused by refraction (and some reflection) in water droplets in rain. This is different from the diffraction happening here. Topic for another video I feel!
@@unicornlion8078 Good question. A lot of LED screens get more sensitive control over the colours and brightness by having multiple LEDs. Some used to have one R and one B and two G running at half intensity to get more colour control. The way the eye/brain responds to colour is interesting if an intense light source is next to a very dim on the eye/brain sees the dim one as black or off. This is a really interesting and complex topic. Well done for spotting it.
Sir how do I understand which colour is produced when constructive interference takes place? Assume that white light is shone.
OK, I guess you realise that this is with electrons so nothing to do with colour..But as you say if similar physics is done with light there is an equation which states where each maxima will be knowing the wavelength (colour) you are interested in. [n Lambda = s sin(theta)]. This is the diffraction grating or multiple slit equation. Sorry I could not write it in a better way - one to google.
Oh now I understand. Thanks sir
@@joyboy7679 Pleasure.
What about the quantum mechanics that allow light to reflect emitted from a different source?
Yousef, I am not sure I get your question. I will try and answer but will need a little more detail. This video uses classical physics to explain the colours off a DVD. Yes, a bit simplistic but it does allow for some correct predictions.
「あなたの動画はとても良いですし、メッセージがた
Hope you found it useful!