8.03 - Lect 20 - Interference, Huygen's Principle, Thin Films, Double Slit Demo

today we're going to talk about of electromagnetic radiation and I will start as a warm-up with the famous historical experiment which was first done by young in 1801 by that time the issue whether or not light was waves or whether those particles were still unresolved Lewton always wanted light to be particles about the Dutch physicist parents wanted them to be waves and the issue was unresolved let us agree that if light are particles and you have a screen that say has two openings and you would throw particles through there like tomatoes tomatoes are particles and you collect them here then those tomatoes that don't get stuck on the screen but that make it would form a pile here of tomatoes and they move form a pile there of tomatoes that's very typical for particles but the situation is different when we deal with waves because the moment that you have wastes coming in say for instance we have plane waves coming in like this then the wave can go through both openings simultaneously and that changes the picture quite dramatically it was no of course already in the 17th century that if you have water waves going through a small opening so here are water waves moving in like so and here is also water that what you see coming out there are circular waves the waves then look like this and they're propagating out in a circular fashion and if the velocity of the waves here is the same as the velocity there then the wave lengths here would be the same as the wave length there but if the velocities are different of course you will see a change a difference in the wavelength so happens my countrymen suggest the in the 17th century your idea which is now known as hyphens principle that we can think of this in a very different way thousands by the way is difficult to pronounce any one of you who knows how to say house ins correctly is also Dutch because you miss the iron in your language and you missed it in your language and the combination of our and is a complete killer for you guys you know you don't get extra 803 course credit if you can come to my office and say how Athens but it will certainly put you in the Dutch category so in the 17th century then how Athens came with an idea which was later amended by Finnell in the 19th century and it is now known as the Huygens I will pronounce it your way against fernell principle which works as follows if we have a plain monochromatic wave and it's incident on a screen with an aperture so this would be an aperture an opening here you would have two apertures two openings then the holden's Fornell principle states the following all points in the aperture plane may be thought of as secondary point sources of spherical waves and the point sources replaced the real source which is flooding the screen and the screen itself is a perfect absorber of the radiation falling upon it and so I will repeat the horrigan Fornell principle which I will need today all points in the aperture plane may be thought of as secondary point sources of spherical waves in the case that we had water this is a two-dimensional surface they would be circles but when you deal with light you can think of them as 3-dimensional that means spherical waves now the Holden's Fornell principle is very powerful though there is a wide range of opinion to its scientific merit and in a very famous book the principle of electrodynamics by Melvin Schwartz I read the following and I quote verbatim from his book he says Huygens principle tells us to consider each point on a wavefront as a new source of radiation and add to radiation from all the new sources together physically this makes no sense at all light does not emit light only accelerating charges emit light thus we will begin by throwing out Huygens principle completely later we'll will see that it actually does give the right answer for the wrong reasons so I will proceed my lecture today to get the right answer but perhaps for the wrong reason suppose now we have a electromagnetic wave I'm particularly thinking of light and I have here an opening and I have here an opening and plane waves are coming in from the left moving with this speed of light and that this opening ba and that this opening the B and here the center of the two is oh this could be a circular opening it could also be a slit most of the experiments I will do today I will slits which are perpendicular to the blackboard so they are very narrow openings imagine now here a point P and so the wave from point A will reach that point P but away from source B will also reach that point B and so the electric vectors there are going to be added of course vectorially imagine now that B P minus a P imagine that that will one half wavelength then what you get is that the mountain of the e vector here will coincide with the valley of the e vector from the other one and so you will get the situation that light plus light will give darkness we call that destructive interference the way from here and the way from there would be hundred eighty degrees out of phase and if you take all the points for which the difference is one-half lambda that is a hyperbola will surface it's like a bowl it's not only in the blackboard but it also comes out of the blackboard it's like a bowl remember from your high school days that if the Sun of the distance AP and BP is a constant then you get an ellipse but if the Sun if the difference is a constant you get a hyperbola so you will get a hyperbola will surface and on everywhere on that surface B minus AP would then be one-half lambda and then you would get the amazing consequence that light plus light would give darkness destructive interference tomatoes don't do that one tomato on top of another tomato does not get no tomato that is distinctly different from particles now of course there were also hyperbola little surfaces for which there is constructive interference in other words if I make this difference n times lambda whereby n is either zero or plus or minus one or plus or minus two then I would get surfaces for which we get constructive interference then the phase difference between the two waves is either zero or 2pi or 4pi so we have surfaces with constructive interference and then we have surfaces with destructive interference and that's going to be at the heart of what I want to discuss with you now so I will start a new drawing I think we don't need this anymore and I will now have a screen with two narrow openings they could be slits as I said before so here is one opening and here is the other opening and let the separation between them be D and the upper one you can call source number one and the other one force number two suppose now that we are looking at this and the distance which is very far away very very far away and where you are located in space seen from this point is an angle theta so the way from this source will reach you and since that point is very far away these lines here are of course parallel to each other very close to parallel because it's very far away the wave that comes from this one has to travel over the larger distance then the wave that comes from this one and that difference can easily be expressed algebraically if this angle is 90 degrees that difference is this much that is how much further this wave has to travel than this one and then they vectorially add very far away at the location where you happen to be and this difference here is D sine theta that's immediately obvious if this angle is Theta this angle is also theta so the path difference is d sine theta and that translates into a phase angle difference between the e vectors theta first of all design theta which is the path difference this is how many wavelengths you can fit on that path difference so I divided by lambda and for each wavelengths that I can fit on there the phase difference is going to be 2 pi and so the phase angle the difference between the way from here and there is therefore 2 pi times D divided by lambda times sine theta and so now I can set the condition for constructive interference to constructive interference what I want is I want Delta to be multiple times 2pi so again I already wrote that down earlier but I will do that again and can be zero it can be plus or minus one it can be close or minus 2 etc if any 0 by the way you are right at this plane which comes perpendicular out of the blackboard clearly when you're on this plane when n is 0 the distance to any point on the plane from source number one and source number two is obviously the same you can also write down that D sine theta and I'll give it a little n which makes connection with this and it's nothing to do with index of refraction that D sine theta is now a multiple times one wavelength that's the same the same thing so this statement here is the same as this statement there's no difference between them and now we can write down the condition for destructive interference so now you want that Delta is PI 3 PI 5 PI so now you have Delta equals to n minus 1 times pi and if you want to write it down in terms of the D sine theta then you get that D sine theta of n which is that end that you see there is now 2 and minus 1/2 times lambda and so this is at the heart of the idea of the interference in this case of two of these sources two slits or two circular openings let me check that we have that right D sine theta the n makes connection with the N equals n lambda Delta is a multiple 2pi destructive interference we get 2 and minus 1 times pi yes I can live with that now I want to be quantitative because I'm going to do a demonstration very much with the numbers that you're going to see now on this blackboard I will give this back to you although you may not need it so we now have the two openings all the way on the left here and they are so close together so this is that screen they closed they're so close together that you can't even separate them anymore so this is the point where the two openings are and we are going to look on the screen which is far away so here is the screen which is going to be that screen and we want to see how the constructive interference that means light shows up on the screen and how the destructive interference that means darkness shows up on that screen and so here are these two openings so light comes in from the left and that this distance BL and so I want to know if I look under an angle theta as seen from the slits I want to know what kind of pattern I will see let's call this x equals zero and we call this X positive x and that's negative x you can also say that the sine of theta if you want that is approximately X divided by L but that's only true for small angles of theta and with this kind of interference we sometimes have angles of theta which are not small at all so I want to warn you that this is not something you can use universally but today you will see that we can use it so I start now with red light lambda is 600 nanometers so that's the light I will use so I use monochromatic light and we will have a distance to the screen which is about five meters the setup is here this is about five meters and we have wrought light from a red laser which is a little larger than 600 nanometers but I just want some round numbers and the separation here D between the two slits is one quarter of a millimeter now you may think that that is rather small well you will see that that's what you need you need very small values of D to even see this phenomenon so now let's evaluate what n is let's evaluate what Delta is let's evaluate what the sign is of theta of n and then we will evaluate what X of n is and we will do that for constructive interference so numbers that you're going to see here is for constructive and of course if you want to you can do the same exercise for destructive interference I need constructive interference so I have to turn to Delta is that and I make them n times 2pi and I start with n equals zero well that's easy any zero Delta is zero the sine theta n is zero and X of n is zero that's where you will see maximum light constructive interference obviously that's the plane that goes right through the middle of those two openings and the distance to each one of those sources everywhere on that plane is of course the same and so you expect here to see light maximum now I go to plus or minus one and I calculate now what Delta is then I get plus or minus two pi and so the sine is now two point four times ten to the minus three you can check that because you know what e d sine-theta is n times leather and I told you what lambda is V sine theta is now n times lambda and we take N equals 1 and you know what lambda is you know what DS so we can calculate the angle sine theta and that translates into an x value of about plus or minus plus n minus 1 point 2 centimeters so now I know what the sine of theta is and I use the small angle approximation which is more than adequate and I know now where on this screen I get again maxima and that is then here one point two centimeters on this side and one point two centimeters on the other side because you get of course plus as well as minus and then I can continue for N equals one N equals two I will just do plus or minus ten so then you get here plus or minus 20 pi so then you get here 24 times 10 to the minus 3 and so you get plus or minus 12 centimeters it's almost linear because the sine of theta for these small angles is still the same as theta in radians now before I show the demonstration I want a little bit more information on the on the shape of the dark and like areas that I'm going to see so what I plot here now is I will lower this and I will raise it later because I want to work above my head so that you can see what I'm doing so I'm plotting here the sine of theta and keep in mind that is linearly proportional with X but I always prefer to plot the sine of theta and so here is sir 0 let here be lambda divided by D that is where my first maximum will come me you can see that and then here at the same distance I have lab 2 lambda divided by D and only and the other side I have minus lambda divided by D and all of these are Maxima constructive interference and the destructive interference clearly Falls smack in the middle I did not calculate them and so the destructive interferences are here and so you get light curves light intensity which is like this light intensity remember is always the pointing vector so this is light intensity and that is in watts per square meter now the first thing that I want you to notice and you will see that today in various demonstrations is that the location of the Maxima depends on the wavelength so that means if red light 650 nanometers if that falls here as a maximum and it falls here is a maximum and it has here a maximum and it has there a maximum that blue light which would have a wavelength say I pick 400 nanometers would not have its maximum at the same location therefore not the minima either because look at the relationship and d sine-theta is n times lambda and so the angles of theta are different and so you would get here roughly it's about two thirds down the way yeah you will get here roughly the maximum for blue you would get here the maximum for blue notice that for N equals zero the red and the blue have the maximum at the same location but you would get the first maximum the blue here you will get the second maximum here and you would get 2/3 maximum almost exactly at the location where you get the second maximum for red the reason being that 2 times 650 is very close to 3 times 400 that's the reason so each lives a life of its own the light intensity that you see is a cosine square function of Delta divided by two and this is so fundamental that first I thought I will ask you to derive it but then I decided to spend one minute on it and the write it for you because it's extremely important that you we use that later in our predictions so let us assume that source number one and the other one I call source number two for the source number one he zero one has an e an e one sorry not easy Row one with E one has an amplitude is zero one when it reaches the screen and then of course you have cosine Omega T and this is the frequency of the light that's the electromagnetic radiation e 2 that is the other source as easy Rho 2 times cosine Omega T and here is that crucial Delta which is the difference between the two in phase that's why you get Maxima and that's why you get minimal that's the whole idea of the interference pattern now if you are very far away from the two openings then these effectors are of course very closely the same remember is proportional to 1 over R but if the distance R is very closely the same then we can assume to good approximation that the e vector the amplitudes are the same so that means the total effective that you get the sum of the two vectors or sum of the two is then going to be 2 e0 I simply call it a zero now and then I get the cosine of half the sum so I get the cosine of Omega t minus Delta R divided by two times the cosine of half the difference so that becomes the cosine of Delta divided by two and now you see why the light intensity is proportional to cosine square Delta over two because when you calculate the mean value of the Poynting vector you have to take across B and so B also has this term and also has this term and so you get the square of this term and the square of this term but the average of the square of this term is 1/2 to forget that for now and so you see it is going to be proportional to the square of Delta over 2 and so this function that you see Delta is linearly proportional to sine theta this curve that you see is a cosine square curve the experiment that I'm going to show you has a separation of the slits which is 1/4 millimeter my wavelength is 633 nanometers which is very much the same and the distance to the screen is about 5 meters and so the way that we have this arranged is that the two slits they are vertical they are like this so this separation is 1.4 millimeter and then we put over there a laser beam and then the result is what you're going to see there and so I now have to think about this is where it is you're going to see it there and in order for you to see it I have to make it completely dark when you look here on the screen you will see Ares which I distinctly dark and you see areas whereby you see the light in other words you see the pattern that we just calculated you see constructive interference and you see destructive interference and you're looking here at an experiment that is historically of great importance it is one of the most mysterious and puzzling issues in all of physics this interference pattern which was first shown by young in 1801 is clearly convincing evidence that light is a wave phenomena however 20 century physics has shown that light comes in the form of individual photons with well-defined energies and photons can behave like bullets they have momentum they produce radiation pressure and they can be localized in space we can detect individual photons in a way that we detect individual Tomatoes photons behave like particles however the interference pattern that you were staring at on that screen can only be explained if we assume that each photon went through both slits but how on earth can one particle go through both slits it will have to be one slit or the other and here lies the great mystery light is both it is a wave if you want it to be a wave and it is a particle if you want it to be a particle if you manage to determine for every photon through which of the two slits it went and that is possible if you do it at very low light intensity you can really determine through which of the two slits each photon went then you will not see any interference anymore the the screen will be like that of the tomatoes you get two piles so the moment that you establish the particle character of the light by determining through which slit it when you have destroyed its wave character and it will not behave like a wave it will behave like a particle however as long as you do not determine through which slit the photon went those as long as it remains your guess through which one it went light will reveal to you its wave character and you will see interference the choice is yours but you cannot have both ways so you're looking now at something that had a huge impact on physics clearly at the time that this experiment was done it was accepted that light wave that we now look at it in a very different way I want to show you a slide of this what we call double slit interference in white light we did it in red light that's the easiest thing you get very sharply defined Maxima minima and the reason why you get them so sharply defined is because you don't have the interference of the other colour but of course if you do it with white light then you will see the blue light has its own spacing and the red light has its own spacing and the net result of course is that you see a somewhat different pattern and that I will show you here I will make it a little darker for that so the upper panel is then the double slit interference in blue light and you see the clearly dark lines which I D the destructive interference locations and you see as I already indicated on the blackboard that the separation of the red is larger because its wavelength dependent and if you then do it with white light then you see something like this so it's more difficult to actually see the maxima and the minima now what you can do with light if you can turn light plus light into darkness then obviously you should also be able to turn some plus sound into silence it's just a matter of scaling up the whole experiment and that is what we have set up for you here we have two loudspeakers and so all the equations that you have here apply verbatim except that we scale it a little bit so our D is no longer a quarter millimeter but our D is now one and a half meters so here is a source of sound and here is a source of sound and they are fed by the same electronics and so the D the separation is 1.5 meters and the frequency is 3000 Hertz so we do this experiment now for sound and so Landa is about eleven point three centimeters if you take 340 meters per second for the speed of sound so we have an away monochromatic light monochromatic sounds more or less one wavelength and so you can calculate what lambda divided by D's it's about 0.075 and so you can calculate where now those surfaces are they are really hyperboloid all surfaces they're coming out from the center here and they bent in your direction where all those hyperbola little surfaces are where there is no sound and then you can calculate where all those hyperbole orbital surfaces are where there is sound let's first calculate how many hyperbole orbital surfaces there are so in order to do that we simply have to stick in our equation D sine theta equals and lambda we just put in theta is the maximum value possible and that is 90 degrees so you put in theta max and that is 90 degrees PI over 2 radians and so when you do that you'll find that n max that is the highest number of n in that series 0 1 2 3 4 5 that n max is then D divided by letter which in our case is about 30 it is 1 over this number so that means there are 13 of those surfaces going in this direction but they're also 13 going in this direction member you have plus and minus so you really have to multiply this by 2 and so there are 26 surfaces no more and probably no less in this audience of constructive interference and then they are probably also around 26 destructive interference so now I want to be as qualitative as I can be and I would like to know the students who are sitting about 5 meters away from me what the distance is between one maximum the zero order maximum right at the center and the next maximum we get an idea where to search for those maxima and where for those minima so I will go now for the Maxima so I know that at state a 0 right here all of you here sitting here must hear a maximum there in this plane now the question is where is the next maximum well the next Maxima depends of course on the distance how far you were sitting away from me so I take the students which I about 5 meters away from me the sine of theta 1 this one now makes reference to N equals 1 that is the first surface away from the center so the sine of theta 1 is then 0.075 and that translates into a value of theta 1 I think it's about 4.3 degrees that's right and so now you can calculate what the separation is between one maximum and the next maximum that separation x1 I call it x1 is roughly L times that 0.075 and that is about 38 centimeters so right there are those guys that I'm looking at now you will have a maximum sound here and when you move 38 centimeters you will have another maximum and that is a minimum in between so the minima and the Maxima right there where eligio is sitting is about 19 centimeters now for you there in the back it is larger and for you here in front it is smaller because remember those surfaces go like this and so here the separation is smaller I just give you this as an example that you know roughly an order of magnitude of what you are looking for it so I'm going to turn on this sound now and then I want you to wiggle back and forth as you did when you were in nursery school and I want you to find the locations of maxima and minima and I can assure you that the ones with minima the ones with sound silence are extremely well defined because of course we have done this several times ourselves don't move too fast move very slowly and there will be points where you hear no sound and there will be points where you hear a lot of sound so are you ready for this keep moving most of you move too fast who was able to clearly distinguish the Maxima from the minima not too many and the reason is that all of you allow the scientists if the difference in space between maxima and minima is 19 centimeters that is this much what is the separation between your two ears it's about 20 centimeters it's not an accident that I chose that so if you have a maximum ear you would have a minimum there so what have you have to do you have to close one here and then you will be real scientists so I'm going to turn it back on again I want you to close one ear and then you will be able to find those minima and those Maxima move again back and forth now there may be some reflections from the wall which may interfere with this experiment no pun implied go slowly go very slowly mortal unbelievable I hear nothing here here here a lot of noise so who heard it now there you go alright this is the great moment for the the mini break with a mini quiz so if someone is going to help me I owe you something regarding exam 2 you see a histogram of exam 2 and what is interesting and in a way helpful to me it's almost bimodal now we have 47% to go in the course if it's 40% is for the final alone and so clearly I can't say much about dividing lines between ABC D's and F's I want to make only a global statement and only warn those of you who I put in the danger zone the danger zone for me are students whose cumulative assuming that they've taken all the exams then whose cumulative is less than 30 or near 30 and those are really the students almost exclusively who have less than 45 for exam 2 or approximately 45 so this is the danger zone now that doesn't mean that you will fail the course it's just a matter of probability I would say is high and then those of you who are dying to know whether they go to get an A all I can say is that if you cumulative now is more than 45 assuming that you've taken all exams then you will probably end up with an A will you end up with an A I do not know it's all a matter of probability but the chances are you may but that's as far as I can go regarding the dividing lines of course grades so now I will continue the idea of interference which has very far-reaching consequences and I will turn to what we call thin-film interference and I will cover normal incidence it really doesn't change very much if you change to a different angle the concept is the same thin-film I have here a thin film of oil and we will shortly see what mean by thin so this is air so we have plane waves of light coming in this way so it's normal incidence and this is n1 which is air so that's about 1 and this is n 2 which for oil is about 1.45 let's make it one point five so we round it off and this is n 3 which is again air in this case so let's make it 1 it doesn't have to be air but I will make it air and let the separation or the thickness I should say of the oral-b d and i call this surface surface a and this surface between oil and air I'll call that B so that light come in normal incidence comes in from above over a large surface I just put in one arrow here and that that intensity be i0 and now I'm going to calculate how much is reflected back into the air and I put the arrow here so I offset it just for the purpose of clarity so some of it is reflected and some of it goes straight through and all of you are more than capable of calculating how much is reflected at that point a the are the reflectivity of the e vector is n1 minus n2 divided by n1 plus n2 and that is minus 0.2 the minus sign tells you that there is a flip here in the e vector by 180 degrees the intensity of the light that is reflected is then 4% the square of this number and if 4% is reflected then clearly 96 percent goes through so if this is 100 then this is 4.0 and then this is 96 so this light here hits the surface at B the boundary between oil and air and some of that light goes through which I'm not interested in and some of it comes back and so I can calculate again the R value for that surface near B which is now n 2 minus n 3 divided by n 2 plus + 3 + 2 this is n 3 and so there is plus 0.2 so there is no flip in the e vector as it returned and the fraction of the light that is reflected is again 4% however it is 4% of the 96% because only 96 percent goes through and 4% of 96% is 3.8 for when it reaches this surface here again 4% will be reflected and 96% will go through so what comes out here then ultimately is 96 percent of this number which is 3.7 I now want to ask the question what happens when you are looking here from above and you see this light coming back straight from a and this light coming back at you which made this journey through the oil whether or not the combination of those two can constructively interfere and whether they can destructively interfere now as far as destructive interference is concerned let me make a qualitative statement the intensity of this is 4 and the a pencil T of this is 3.7 so it can never be completely darkness but you can never kill 3.7 intensity with 4 but if they 180 degrees out of phase which we will be able to do there is very little light left there will only be the difference which is 0.3 left so it will be quite dark but I will still call that constructive and destructive interference so the question now is what is the phase difference between this light here and this light there phase difference that's what matters well let's first look at the difference in path the difference in path is clearly 2d because this light has to go in this direction that is DNA has to come back so that's 2d so now is the question how many wavelengths fit on that separation on that distance 2d wavelengths in oil of course that's what matters so there is this many so what now is the phase difference now we have to multiply this by 2 pi because if this was 3 then the phase difference is 3 times 2 pi for each wavelength you get a phase difference 2 pi so now you have to multiply this by 2 pi so now you may think and that would not be unreasonable that this is the phase difference between this radiation and this radiation but what have you overlooked when you do that that's right you have overlooked the fact that there is here the flip of 180 degrees which does not occur here so there is an additional PI in the in the phase difference so I'm going to write down here now the total Delta so Delta is now 4 pi times D now I can write this as lambda oil it's perfectly fine I can do that I prefer to write it as lambda air and then all I have to do of course is define the lambda air by the index of refraction of oil and so yes I could have written lambda oil here but I prefer to write here lab there and then all I have to do is multiply this by n - yeah that is at 1.5 it's the same thing and so this is them the Delta that we calculated from here and then we have to add the PI and the PI then is the consequence of this minus sign here which you don't see here keep in mind if someone wants to be nasty to you during an exam of course not I but if someone wanted to be nasty to you on an exam they could make n3 larger than m2 what would change in this equation the PI will disappear that's all the rest will remain the same so don't think of this as a universal equation there's no such thing so this pi is a result of the fact that here and 2 is larger than n1 that that here and 3 is smaller than m2 that's why you get only pi so now we are in business we now have here the criterion for constructive interference you want me to have it higher what is your oh yeah thank you very much how could I it is d right yeah oh my Greek heritage somehow all right so we agree right that this is the in this case this is the the face angle so now we're going to ask ourselves the question where are what thickness are required to get minima and to get Maxima the first very interesting cases that if you make D equals zero that you kill this whole term and you only have pi so that's destructive interference question destructive now you may say yeah zero zero film you know what kind of nonsense is that is that meaningful yes it is very meaningful I will demonstrate it to you I cannot make an oral film zero thickness but that's not necessary all that is necessary that this D overlapped our oil becomes exceedingly small for instance 100 so if I make D 100 of the wavelengths of light in oral then I'm close enough and you will indeed see them I will do this with soap that in reflection you will see no light you get completely destructive interference apart from the fact which I mentioned that there is an imbalance between the four point zero and the three point seven but that's a different issue so we will even be able to enjoy the D equals zero but now I want to be more general and I want to give you feeling qualitatively for what you're going to look at and also a feeling for what thicknesses are really necessary to see these colors and so I am going to address now only the case of destructive interference and I'm going to choose a lab that in air of 400 nanometers that's my choice so that means that my lab that in oil is 400 divided by 1.5 which is about 2 than 67 nanometers and now I want to calculate for what thickness is d I'm going to get destructive interference so I'm going to set this equal to 2n minus 1 times pi that's all I do and I choose N equals 1 and what do I find I find D equals 0 of course we knew that already we just discussed that because if N equals 1 then that requires that Delta is PI and if Delta is PI then this term is 0 so you see the D is 0 they can immediately see that so now we take N equals 2 and if you substitute for N equals 2 in this equation you get 3 PI and this is already PI so this has to be 2 pi and so if now you calculate what these you'll find that D is 133 nanometers well that's obvious it obviously must be half of the wavelength of the light in oil because if the wavelength of light in oil is 267 you have to travel this distance twice and if you travel this distance twice that is exactly one wavelength and one wavelength corresponds to a phase angle of 2 pi and so here you get 2 pi and it is this pi that kills you that gives you the darkness and then N equals 3 you'll find that D is approximately 267 nanometers so now 2 D translates into two wavelengths now I'm going to choose this thickness and with that thickness now I'm going to work and I'm going to radiate white light onto that thickness so I select that one for which I already know that in blue light I kill it destructive interference so we now have white light and we have these hundred thirty three nanometers so that is non-negotiable D is 133 nanometers I'm going to write here lambda air in nanometers I'm going to write here Delta and I'm going to write here the cosine squared of Delta over two that is a measure for the light intensity it's proportional to that we discussed that and I show that I derived it for you okay 400 nanometers blue light Delta is 3 PI this is 0 we already calculated that that's the destructive interference if you make it that layer if you make it hundred thirty three nanometers now I need green we have green so now I'm going to also shine on it green light 500 nanometers so I know what B is so that's non-negotiable and now my lab that air is now 500 nanometers and so I can calculate now what Delta is untouched by human hands very simple and I find now two point six pi and if now I take the square root not the square root if I take cosine square of Delta over two I find 0.35 but now I go to read and I do 650 nanometers and now I calculate what Delta is and I find 2.2 pi now I get a point nine zero now I am very close to a maximum and so this is very convincing that if you make your film this thin that it will look distinctly red maybe not perfect red but it will look distinctly red the red color dominates and the blue practically absent at all apart then from the imbalance that you may have between the 4.0 and the 3.7 a question that I have asked over the years at the final exam for 803 is where a thin film interference is the result of the difference between the index of refraction between the different colors in the ends with absolutely not the index of refraction I have taken the same for all colors for all practical purposes that difference does not explain the colors the color is explained by the different path length divided by lambda that is why you see course it has nothing to do with the index of refraction you can see thin film interference in soap and I will try to demonstrate that you can see it in oil spills on the road I will show you an example only thin films give you colors thick films do not give you colors if you go through the exercise which I want you to do to make D 0.1 millimeter which is by practical standards still very thin you will not see any colors and the reason is that if you go to this equation and you calculate for which colors you will see constructive and destructive interference there are so many colors in the visible spectrum for which you get constructive interference so many you will see that your brains will tell you see white lights there is not one caller that dominates the end values that you will need by the way are going to be very high there in the range 500 to 700 huge values of M are necessary in order to get the destructive and the constructive interference go through this exercise on their own and you will see very quickly that so many colours constructively interfere that the film will look white so the first thing that I want to do want to try it's not so easy to make you see these colors with soap which of course all of you must have seen in fact if you just take a shower and you soap yourself than the bubbles themselves in reflection already have these colors I'll try to make one that is slightly larger in size we do not always succeed but let's try that so the idea is I want you to see colors oh well maybe you did but there was a little quick now you see colors thank you that's all I wanted you to see you did you see them if you said if you'd said no I would have done it again but now I want you want me to do it once more it's not easy though now you saw it right yeah thank you okay so so bubbles give these colors because they are exceeding this it also gives you a feeling of how thin they have to be really they can be much thicker than a few times the wave length of light if they become much thicker you lose the colors for the reason that I just mentioned to you that there are too many colors that constructively interfere and then you don't see it anymore now I have a demonstration which is very tricky it works most of the time but not all the time we're going to make a soap film on a metal frame going to be done in this box so we have a metal frame and we dip it in soap and so you get a soap film there and then gravity will make the film thinner at the top and thicker at the bottom so as you make a cross-section through here then the film for this is then the soap I can make this so thin that it's completely dark in reflection that is what I promised you that D is not zero but D divided by lambda is so small maybe 120 s of 130 s of the wavelength of light that it will turn completely black and as this proceeds in the beginning the top will show you colors would go through these phases of colors and then it gets thinner and thinner and thinner and then it turns completely black I will show it to you upside down but don't get confused and that has to do with the way that we project it so the thinnest part of the film will be low on the screen and then the part of the screen that is the thickest of the film will be the upper part as look at it appreciate the fact that when we go into the dark face if we succeed that you were talking about the thickness of the film which is many times smaller than the wavelength of light which is typically half a micron okay let me first get the film up actually let's not get the film up let's just get the light up let's turn it off here has to be completely dark didn't we agree this to be completely dark here so are we going to dip it in here okay so the bottom is the thinnest and you see the colors you see them and you see that the thickness changes over the film and that's why they're areas where the red dominates and then other areas where the red dominates again and so the bottom gets thinner and thinner and thinner the white light that you see now here is the result of Marco's trick in order to avoid that the film burst he has to put glycerin in the soap and what you see here is a reflection of the layer of glycerin which forms on top of the soap so we'll have to wait a little bit for that glycerin to also go down and then you will see the pure soap and that is then the area and boy look at it black that area that you're looking here now is way thinner than the wavelength of light and so the reflection is well when I say zero of course that depends again on the asymmetry that you have that we discussed here on the blackboard between the 4.0 and the 3.7 it's a little different for water than it is for well for oil what you see black you can just look at it for a while it may burst you also see some very interesting thin film interference in the glycerin itself which is the oil layer which is on top of the soap this is not uncommon you may remember when you see oil spills on the road that you sometimes see these very nice structures brownish reddish blueish but here look at this look at this isn't that incredible see that fabulous boy the film is holding up Marcos we can't complain ah you see these nice oil structures here in the glycerin the nightmare that we have always is that it pops too fast and then you have to do it again again again and our experience is if somehow the soap is not right then in general no matter how often you try it it keeps popping I look how beautiful look how incredibly nice the syn film interference in the glycerin and then here the total darkness because the thickness of the film of the soap is many times smaller than the wavelength of the light in in the soap I never get bored when I look at it isn't it gorgeous think about the physics that is going on in there all right let's look at some other phenomenon of interference try to remember this this is a gorgeous picture so there's a whole family of ways that you can see film since film interference and in your problem sets we give you two problems to work on that quantitatively and there is also this this take-home experiment which supports the problems and I want to concentrate on those two because they can they are very nice to demonstrate because I can demonstrate them to you in monochromatic light whenever you have monochromatic lights that means you don't have white light of course the fringes we call these fringes the dark and the bright areas we call them fringes show up so much better than when you have white light because then you get the superposition of all the colors and the maxima and the minima depend on the color remember from well in the same thickness you have you may have destructive interference for one color but you have constructive interference for the other so if you do it in monochromatic light you can see this very dramatically and then one of the classic examples is you can do that at home and you should do that at home is the two microscope slides so you have a microscope slides which itself is way too thick to give you thin film interference don't even think about the fact that of this glass there will do nothing way too thick probably in tens of a millimeter it's ridiculous and then there is another microscope glass here and I exaggerate highly this area in between because they are lying flat on top of each other but there is an air gap between them and this is that air gap and it is this and this air gap now that acts like a thin film so you're going to get interference now of light that bounces of this layer and bounces of this layer and so it is this thickness D now of the air layer that determines the darkness and the brightness destructive and the constructive interference are you being asked in your problem-set to calculate where those maxima and minima are and you can also do that then with the take-home experiment and I will demonstrate it very shortly there is another one which is a classic which I will also demonstrate and which is also part of your problem set this is a lens the glass lens the curvature is highly exaggerated and then here is a glass plate and so here is now an air gap and now if you shine monochromatic light on this then of course seen from above the symmetry about this line you will see rings of darkness and rings of brightness and these rings of darkness and brightness will be very far apart here and they will be very closely together there this is part of your homework assignment to show that and it has to do with the fact that the change here in thickness is very little because of the curvature and the change here is much faster this is called Theia the Newton these are called the Newton rings and then when you just walk out on the street preferably on a day when it is not sunny when it is cloudy you look on the ground and you see oil spills and they have phenomenal course I'd like to show you one of my own pictures that I took some time ago of an oil spill first and then we'll make it a little darker well I'll make it a little darker so that you can see it better I photograph countless oil spills and actually the reason why I photographed so many is that always I tried to reconstruct thickness of the oil on the road and I was never very successful for one thing you can only get a spot like this you would think that a car loses some oil so does one drop of oil that goes and then it starts to run out so you would always think that it is thickest here and then gradually tapers off near the edges and I've always been wondered why then here do you see so much white light you would expect that perhaps it would be much darker and so many attempts that I have made to convince myself that it's really thicker here than there many have failed but there is no question what you are looking at of course is a striking example of thin film interference and not only that but you know that this film has to be extremely thin clearly very close to that in that range of 100 to 200 nanometers otherwise you wouldn't see over the entire area such beautifully equal blue and here you see this this reddish which is not pure it never is very well defined so clearly it has a spherical symmetry so it is probably a drop that then spread out and so now I want to demonstrate and your microscope so the microscopes we have here microscope glass turn only see the TV there if everything works and we do it with near monochromatic light this light from a mercury lamp which is largely in the green so you see very well-established maxima and minima that's the beauty if you do it with monochromatic light and so these two microscope slides are just on top of each other but yeah there is always a little bit air in between and so you can sort of make up your own mind that they may be touching each other really touching where they have practically no air for reasons that are not so clear which must be somewhere here and that's where the separation between the dark lines is the largest and then somehow here you see these fringes the areas of darkness and maximum constructive in the fields and when I push on one side of these slides then of course I change this very subtle structure of light in between these microscope and then you will see an immediate change in the separation of the lines so I'm just pushing on it now changing the configuration you know this is the comp the point of contact more or less and you can really see that I what I'm looking at right now I see it just as well as you can see it there and so you should really do that experiment at home because it's part of your take home experiment so I mentioned in the write-up that it's not necessary that you turn in a written summary of your experiment but so it won't count for your problem set as such but I do want I will hold you responsible on the final so I advise you to do the experiments because I may ask a few questions which you can only answer if you actually did this experiment here I have two flats like the microscopes but thicker sorry a little little sturdier and this is the way that opticians actually test their optics so there is here a an air gap that grows in size and when you push I push now on it you see now I make them flat now I see almost no fringes now I push so hard that I open them again like this so I make one go up like this and now you see many fringes look you got a huge angle and then I want to show you this which is called the Newton rings we have an arranged in such a way that with some set screws we can actually push down on this so we can also make the the air gap larger and smaller and that we're going to show you there I can actually make it even darker that may be nicer for you you already see there at the center portion of the contract where the contact is here the air gap doesn't change very much and therefore you will see that the spacing of the rings is large but when you go further out then this patient gets smaller and it becomes difficult to see perhaps but if you close here I can really see the the Rings here much closer than the Rings here and I can put some pressure on there so in order to change the geometry a little bit so now I squeeze a little harder I hope I don't break it and when you squeeze harder they touch each other over a larger surface in the center you see therefore it opens up so over this very large area now is the gap the same and when I loosen it up then you expect those rings to become smaller and this is something I believe that is also part of your take-home experiment all right I can't wait to see the result of the mini quiz I'll be proud of you if you did well let me tell you and I see you then I hope on Thursday

Channel: Lectures by Walter Lewin. They will make you ♥ Physics.

Views: 56,875

Rating: 4.9613733 out of 5

Keywords: Walter Lewin, Physics, Interference, Huygen's Principle, Young, Thin Films, Soap Films, Oil, High Resolution

Id: LnJIZ_lwSBM

Length: 76min 42sec (4602 seconds)

Published: Wed Feb 11 2015