 Let us begin good morning to all of you again. So, I have two sessions where we will take up elements of optics. So, I am not going to actually do things like calculation of the fringe width of Young's double slit experiment or diffraction pattern. Those things have been done from for years and I am sure you know better about the derivations than I do. So, what I will try to do is somewhat general. In some sense, it is good that we had a session on electromagnetism because the optics is basically built on electromagnetism and light as you know is an electromagnetic wave. So, therefore, propagation of light and all phenomena connected with light are also related to whatever we have learnt about the electromagnetic field. I will be talking about the light propagation. In particular the vector nature of light, we have seen that because of the fact that the electromagnetic field E and H or B are vectors or vector fields are the principle of superposition holds. It is the principle of superposition which is at the heart of our discussion connected with interference and diffraction. I will really not distinguish between them because the difference if at all they exist is essentially in the phenomena that they describe, but both of them are different manifestations of principle of superposition of the vector nature of light. Later on, I will be not in this session, I will be talking about light amplification and some basic characteristics of optical fibers and let us see how much time we have. So, in this session I plan to essentially talk about some things which could have been done in the electromagnetic theory itself provided we had a little more time, but it is a good idea to have done those things earlier so that we can review it again. So, firstly there is some you notice that yesterday I told you the physicists do not like the auxiliary fields like H and D, but it turns out that the people who do optics they are quite happy with using E and H. And since most of the time they deal with linear media, linear non-magnetic media they do not really have that much of difference to worry about the B and H in which case why keep an extra mu 0. So, therefore, they just always work with E S, but with H. So, with those modifications the Maxwell's equation that we have this is the Faraday's law equation del cross of E was minus d B by d t since B is mu 0 H. So, I get del cross of E I am talking about light propagation vacuum. So, del cross of E becomes minus mu 0 d H by d t. The corresponding Ampere Maxwell's relation which was del cross B equal to epsilon 0 mu 0 d E by d t plus a source term, but because I am working in vacuum I do not have any sources. So, therefore, this equation becomes del cross of H equal to epsilon 0 d E by d t. Remember that it was del cross of H equal was equal to d d by d t. Now, if I have a source free region the by source free region I mean region in which I do not have currents or charges in which I am looking at the electromagnetic field the sources could be somewhere else, but the region in which I am looking at if they are source free region then my del dot of E which should have been rho by epsilon 0 is equal to 0 and del dot of B which is equal to mu 0 times del dot of H which is equal to 0 which means the divergence of the magnetic field as always is equal to 0. So, I could take these two equations and do a del cross of that. So, my del cross of E was minus mu 0 d H by d t. So, if I do del cross del cross of E I get this equation and so that quantity is equal to minus mu 0 epsilon 0 d square E by d t square because of del cross of H becoming equal to E. There is a standard vector expression for the curl of curl this you must have worked out several times. It says that the del cross del cross of E is del of del dot of E minus del square E. So, this is derived in an elementary vector as you are a book. Now if you substitute this relationship there along with the fact that your del dot of E is equal to 0 substitute this year you get an equation which you called yesterday as the wave equation. Namely del square of E is mu 0 epsilon 0 which is nothing but 1 over c square times d square H by d t square where c is 1 over square root of mu 0 epsilon 0 if you put in the numbers here it works out to 3 into 10 to the power 8, but this is rigorously correct irrespective of what the values of mu 0 and epsilon 0 are. So, let us look at the a linear isotropic media. So, what we did was to say that the my wave equation is del square of E is 1 over c square d square E by d t square c is the velocity of light and now since both these equations are identical or similar their solutions will be very similar. So, let us do this let us define a small u to represent 1 over square root of mu epsilon. Now I said that I am dealing with linear isotropic media which means that the medium is may have a dielectric constant, but then dielectric constant is simply a permittivity of that medium epsilon is simply a constant times the permittivity of free space and where this k is called the relative permittivity also known as dielectric constant. To realize that we are dealing with dielectric constant and not a general dielectric function because in solids the relationship between E and d could be lot more complicated and so therefore, and they could be dependent on many other things direct isotropy if they could be anisotropic and things like that. And so therefore, what we say is that I am dealing with linear isotropic linear of course, means the relationship is just a proportionality constant isotropy means the dielectric constant does not vary from direction to direction and likewise I define something called a relative permeability which is the ratio of the permeability of the medium to mu 0 and that is incidentally though I have formally defined it the medium that we always will deal with will be non magnetic. So, as a result this relative permeability factor is equal to 1. So, my velocity u can be expressed as this was 1 over square root of mu epsilon and if you substitute these it becomes c times square root of under root of 1 over square root of k times k m and these out of that I said that I will take k m to be equal to 1. So, as a result what I get is that my refractive index of the medium now remember everything is constant here now I am not saying things change I am not dealing with dispersion which arise because of the fact that the refractive index could depend upon the frequency all those I am ignoring right now. And I am saying that the ratio of the velocity of light in vacuum to the velocity in the medium which I have represented by u that is given by square root of k into k m and most optical media are non magnetic. So, therefore, this is simply equal to square root of k and one always takes for non magnetic medium square root of the relative permeability to be the same as the index of refractive. So, I am not dealing with any functional dependence no dependence on frequencies and things like that. So, therefore, we had seen that each component of e or h which I have now represented by a capital U will satisfy this equation I have written it in Cartesian coordinate this satisfy this equation and there are harmonic solutions to these equations you can simply substitute them and find out that the equations are satisfied. The form is u 0 cosine or sin that is unimportant because all that it means is a phase difference between them times this is a three dimensional wave equation cosine of k dot r minus omega t ok where this vector k is i times k x plus j times k y plus k times k z is what is known as a propagation vector the propagation vector whose magnitude is known as the wave number. So, this is our starting point of discussion. What we are now saying is that I have an electric field or a magnetic field. So, I will be using the word electromagnetic field which has a solution in vacuum or in isotropic linear medium as a term which is amplitude which is constant times a cosine function and the argument of the cosine function is k dot r minus omega t. So, therefore, if I am looking at how this field is behaves with space and time. So, obviously, there is a space dependence and there is also a time difference. However, if you look at the surface which is determined by at a particular time for instance if you look at that k dot r minus omega t is constant. Now, which means if you fix t at a given time then you get an equation which says k dot r equal to constant if time is fixed omega of course, is the radium frequency which is also fixed. So, k dot r equal to constant which is nothing, but equation to a plane. So, therefore, what we are saying is this that if you take constant values of the argument of cosine. Now, they define at a particular time a plane, but if you now look at that plane at subsequent times which means that as time keeps on changing this surface also keeps on moving because the value of that constant that I am talking about k dot r minus omega t equal to constant. So, if I take one particular time I get k dot r equal to one constant if I take another time k dot r equal to another constant. So, with time what is happening is this surfaces they are moving and what is the property of that surface it means that the phase of this cosine function given by k dot r minus omega t it is constant at a particular time and if I am now incidentally surfaces of constant phases are known as wave fronts. So, these wave fronts they propagate with time. So, therefore, I look at I have a direction of propagation and this k by definition remember your school coordinate geometries k n dot r equal to constant is equation to a plane and the vector n is actually perpendicular to r in that case. So, therefore, what happens in such a case is that we are saying there is a direction k and the surface of constant phase is essentially a plane perpendicular to that at any given time. This plane which is perpendicular to it keeps on moving with t because that constant keeps on shifting. So, this is what we are talking about this entire argument is what I am calling as a phase and this phase of term of course changes with time. So, this is a pictorial representation of what happens that by virtue of that equation that k dot r equal to constant n dot r equal to constant represents a plane I have a sequence of planes ok. Here what you are seeing is a part of the surfaces and this is obviously the direction of the k vector and the surfaces are normal to it by definition because I have said k dot r equal to constant and that is an equation to a straight line equation to a plane with the direction of k being normal to the surfaces alright. Now, this is fine accepting that when you deal with cosine and sine functions life becomes a little more complicated it becomes complicated because when you differentiate a sine function it becomes a cosine function when you differentiate a cosine function it becomes a sine function and things like that. Mathematically it is much more prudent instead of looking at solutions in terms of cosine and sine functions to represent them by a complex function. Now, you would say what is a complex way there is nothing like a complex way. So, I can say my solution is actually a real part or imaginary part of this quantity. So, I do not really worry about which part I take I do all my mathematical calculation using the form e to the power i k dot r minus omega and since exponential functions have this attractive features that differentiation of exponential function give us exponential function back the mathematics becomes much easier. And at the end of your calculation when you are ready to calculate physical quantities then what you do is to take real part alternatively multiply your solution with its complex conjugate and take a square root. This is the mathematical technique it has nothing to do with the what actually is going to happen. Now, so therefore, the we are dealing at this moment with plane waves the phrase plane wave means you remember I talked about surfaces of constant phase or wave fronts. Now, if the surfaces of constant phase happen to be planes we call them plane waves and the these are essentially the solution plane wave solutions. Now, point is this that you would then say that what happens in three dimension. You know that I will have a similar equation in three dimension which the way I have done in either quantum mechanics or other things which we discussed I will be able to separate the solutions and then write down a wave equation in terms of the distance r. Now, it turns out you would normally say that if cosine k dot r minus omega t is a solution of the Cartesian form of the wave function then I expect cosine k r minus omega t ok not k dot r I am talking about r as a distance k r minus omega t would be the corresponding solution. Now, you see there is no problem in saying that if I look at functions like cosine k r minus omega t these things obviously, have constant values of the phase at a given time t when I take surfaces of constant radius they have r is constant which is radius then these are spheres. Now, so therefore, my wave fronts becomes spheres that is not a problem. The problem is if you write down cosine k r minus omega t this is not a solution of a wave equation. You can go back to your radial wave equation and try to see what happens if cosine k dot r minus omega t is a solution of the 3-dimensional wave equation. But if you are looking at just the radial wave equation then your expectation is cosine k r minus omega t would be a solution, but it turns out cosine k r minus omega t is not a solution of the wave equation. However, it turns out that instead of taking cosine k r minus omega t if you take 1 over r times cosine k r minus omega t that will be a solution of this spherical wave equation. Now, this is an important point because it tells you that the amplitude of this wave you see in this case which was a plane wave the amplitudes are constant. But if you are taking spherical waves the amplitudes have to depend upon r as 1 over r and that is the origin of the inverse square law of intensity of light. That is if you take the flux through a surface of radius r then as you proceed with distance because of the fact that the surface area of the sphere is proportional to 4 pi r square. And if the flux is to remain the same the same electromagnetic energy is coming out through a larger and larger sphere it says that my intensity must fall at 1 over r square that is the standard inverse square law for the intensity. So, let us look at the propagation of plane waves. The so my equation as I said this capital U stands for either electric field or magnetic field in optics the fancy of the people is to look at the electric field strength. For the simple reason it has a much larger magnitude than the corresponding magnetic field strength. And so therefore, what we do is look at this equation with 1 over small u square is the u small u is the velocity of light. And we have plane wave solutions of this form u 0 e to the power i k dot r minus omega t. But this is actually a very attractive way of writing it. If you look at this form that u is equal to u 0 into e to the power i k dot minus r minus omega t and you apply a del operator that is you are differentiating it with respect to space. All that it does is to multiply this function with an i k. Now notice that this feature will not be there if I had taken a cosine function or a sine function. Cosine functions will become sine function and vice versa. So, here what will happen is that if I take a del operator the effect of applying a del operator on u is to multiply this u with i vector k. Now likewise if you are to take a time differentiation d by dt then the effect of the time derivative is to multiply your solution by minus i omega. So, the for this situation I have a nice prescription and the prescription is that when you have to put a del operator del cross del dot whatever you feel like. So, you simply multiply your solution with i k. If you have to differentiate with respect to time multiply your solution with minus i omega and this happens because I have chosen complex functions. So, therefore, you look at this equation del cross of E equal to minus mu dh by dt this is our Faraday's one. So, since del cross is tantamount to multiplying with i k. So, this will give me i k cross i k cross E equal to minus mu and this is d by dt and d by dt is multiplication by minus i omega that takes care of this minus sign. So, I will be left with after cancelling i from both side k cross u equal to k cross E is equal to mu omega H and in the same fashion the ampere Maxwell's relation del cross H equal to epsilon d by dt gives me again del cross is i k. So, I get k cross H that is equal to minus epsilon omega E del dot of E equal to 0 i k dot E equal to 0 which means k dot E equal to 0 del dot of H equal to 0 gives me k dot of H equal to 0. So, you notice I have replaced my Maxwell's equations by replaced all these differentials by equations which look like k cross E equal to mu omega H k cross H equal to minus epsilon omega E k dot E equal to 0 k dot H equal to 0. So, this is my set of equations. Now, you will agree that they are looking much less formidable than our original equations because they are algebraic equations though it is vectors, but they are algebraic equations. Now, firstly you look at what statement that I can make k cross E equal to H I mean forget about constant then k cross H is a minus E then you see in other words H cross k would be E. So, therefore, k E and H they form a right handed system k cross E is H E cross H in the direction of k H cross E is in the direction of k. So, k E and H in that order or you want to remember it as E H and k it does not matter. Similarly, this relationship tells me k is perpendicular to E k dot E equal to 0 this relationship tells me not only k is perpendicular to E and k is perpendicular to H the previous relation it also tells me both E and H are perpendicular to k and because of the fact that k cross E is equal to H I realize that I have a system where the direction of propagation the electric field and the magnetic field are mutually perpendicular to each other. Now, this is something which is a characteristic of electromagnetic wave. Now, incidentally I want to make it very clear my discussion is only in vacuum and so, therefore, in the vacuum why I am pointing this out is that though I would not have time to discuss this in when you teach fiber optics to your students you will find that you differentiate between modes which are known as TE mode and TM mode transverse electric mode and transverse magnetic mode. There you say that if I take electric field perpendicular to a direction of propagation not worrying about what happens to the magnetic field that is one type of a mode. Now, the question always arises that if your fields are anyway always perpendicular what is this classification that you are talking about? Point to realize is that these statements that I am making that E H and k E is perpendicular to H and E and H are mutually perpendicular to a direction of propagation. These statements are valid only when light is traveling in vacuum. The moment you try to confine light which is what we do in case of an optical fiber for instance or any other form of optical waveguide then you have several other conditions to be met. We have discussed some, but this is not the place to talk about for example, you know that the tangential component of the electric field must be discontinuous and the normal component of the electric field must have a discontinuity if there are charges thereby. Now, these are the relations which govern things when you talk about a dielectric medium or a conducting medium or whatever you have they are called waveguides. So, this statement that I am making very frequently made electromagnetic waves are transverse very flip statement. The correct statement is electromagnetic waves are transverse when they are traveling in vacuum ok. If you have put a constraint on the electromagnetic waves which is what you do by making it forcing it to travel in a waveguide such as an optical fiber this statement is no longer true. This is a point which is very frequently missed, but this is a point to be realized that electromagnetic waves are transverse when they are traveling in vacuum ok or in of course linear isotropic medium which are basically in finite extent. What you do in an optical fiber is to confine them and when you confine them you have additional constraints coming in at the interface between two medium that you must have the tangential component of electric field continuous ok. So, therefore, now the fields are now now if you will once you have made the statement k is perpendicular to is k is perpendicular to h you can go back here. You can go back here and say that well in that case h and e are simply related that h is equal to epsilon omega by k from this equation and which is your omega by k is nothing, but the phase velocity of the medium the velocity of light in that medium. So, therefore, it is epsilon times u times c. Now, we had defined the refractive index as the ratio of the velocity of light in vacuum to the velocity of light in the medium and that gives me along with this equation this is the refractive index n equal to c by u you would now translate this into a relationship between h and e. You get h is equal to n e divided by a quantity which is always represented by electrical engineers by z naught and z naught is nothing, but square root of mu by epsilon or mu 0 by epsilon 0. This number is a very well known number to all electrical engineer this is called impedance of the free space and has a value very close to 377 ohms. So, what we have said is that my electromagnetic field while travelling in vacuum is transverse the k is the direction of propagation. Now, if k is perpendicular to e and h the direction of k is that of e cross h we did not have time to go through this, but the rate at which electromagnetic energy flows out of a surface of unit area that is given by a quantity which is simply defined as e cross h. So, e cross h is what is known as a pointing vector. Now, if we had time in the electromagnetic field course then we would have shown that if you look at a electromagnetic field you can talk about an energy density or an energy associated with a given volume. Now, this amount of energy is lost from a closed volume only in two ways. One is of course, by mechanical work that it might have to do on charges or things like the etcetera that might be there in the medium. And the second is just the radiation the radiative flow of the electromagnetic energy it simply goes out. Now, it is that quantity the radiative flow per unit area per unit time is what is represented by the e cross h that is called the pointing vector ok. So, therefore, in fact, I have already shown it here what I am doing. See if you took my cosine expression alright e is equal to e 0 cos k dot r minus omega t and h is equal to h 0 cos k dot r minus omega t. Then my S is given by e 0 cross h 0 times a cosine square term ok. Now, if I did the corresponding exponential form then I have to take a real path ok. Now, if you look at now. So, you notice that this has a dependence on time as well. So, if I am looking at a time average of the rate at which the energy flows out. Then I need to take a time expectation value of this quantity and because cosine square averaged over time has a value half the expectation value of the pointing vector happens to be i k by k ok. The average value of pointing vector is also known as e radians and that is simply half e 0 h 0 which is given by this expression ok. Let me come to another important aspect of electromagnetic waves and that is what is known as the polarization. We had some discussion on polarization, but I will try to assure you that what we talked about last time was right. So, we said that consider plane electromagnetic waves by definition when we say plane electromagnetic waves. So, what plane means that surfaces of constant phase are planes yes. Sir in pointing vector e cross h is there. Sir can we still call it pointing vector if e and h are coming from static field suppose e is perpendicular and h is. No, no, no we are the point actually is this that if everything is static that is an incomplete situation right. We have got these solutions only after we have got the full Maxwell's equation written down a wave equation ok. Solutions of the wave equation we received and then we said how does the energy vary with time. See the question is how can you talk about the pointing vector in a static case because the definition that I have given you is the rate at which the energy is flowing out of a unit area per unit time ok. So, if there is no time variation none of this statement is correct ok. So, let me take a plane electromagnetic wave. You could do the same calculation for for example, spherical waves, but there the amplitudes will not be constant, but it will go as 1 over r. Now, suppose this is a complex expression. Now, the constant that is front of it supposing this is a real quantity and let us suppose that these are also constant. So, if electric field is e 0 exponential i k dot r minus i omega t no dependence on r or t if they are real constant vector then the wave is plane polarized. Suppose you look at a particular time and look at the direction of the electric field at various points in space we have said e 0 and h 0 are constant vector. So, if you are looking at a freezing a particular time you are looking at the different positions then because of the fact e 0 and h 0 are constant if you take a snapshot of various position take a picture of various position at a given time the directions of the electric field or magnetic field will be always contained in a plane and because of that this is called a plane polarized light. Alternatively supposing you decide to look at the same point, but at different times there are two ways of looking at the same situation either I can take a snapshot of what is happening to the electric field at a given time in all this position. Then since the electric vectors are parallel to each other they all lie in a plane and hence the name plane polarized. Alternatively if I fix my attention at a particular position in space and look at what is happening to the electric field as time changes. Now if time changes because of the fact that there is this cosine factor there, but r is constant. The amount of the electric field that I have the length supposing I am representing them by a graph the lengths of the electric vector will keep on changing, but their consecutive positions will always be along the same line and because of this the state of polarization is also called linear polarization. They are just two different terminologies, but coming from different statements they are either called linear polarization or plane polarization. Plane polarization because at a given time in different points the electric vectors are contained in a plane. Linear polarization because if I look at a particular point at different times the all the subsequent electric vectors always happen to be along the same line. This is a general picture of what happens for a plane polarized wave. Let us now change our situation little bit and say that supposing I have two linearly polarized waves. Now remember the principle of superposition is valid. So, if I take two different solutions of a wave equation add them up whatever is the resulting wave that is also a solution of the wave equation. Now because of that if I take two waves supposing one of them I take along let us say the x direction. The electric field is oriented along the x direction ok. The another wave which is oriented along the y direction and not only that the phase part of it one of the waves has a cosine and the other one is sin meaning thereby there is a phase difference of pi by 2 between them. So, if I add them up because of the fact I have taken the amplitudes to be the same. One of them is along the x direction another of them along the y direction. The addition of these will give me a vector whose tip is going to describe a circle because of the simple reason that cosine square plus sin square equal to 1. And this direction of the circle with different time would depend upon which one is chosen as cosine which one is chosen as sin. So, I can either have a circle where the electric vector is moving in a anticlockwise fashion or in a clockwise fashion. So, corresponding to that I have what is called as the right circular repolarized wave or a left circular repolarized wave. So, this is circular polarised. The electric polarisation is very much similar. I still have the same type of representation accepting that what I do is instead of the constant here being the same in both the waves I take them to be different in which case the tip of the electric vector will describe an ellipse ok. This is the picture of what happens to a circular polarisation for instance. So, what is happening is there is this direction k in which it is progressing, but if you are looking at the tip of the electric vector at a particular point in space now what it is doing is to describe a circle. Let us come back to the question of what happens when an electromagnetic wave is incident at the plane boundary between two different dialectics. For example, air and glass which is the most common example. Now, you all of course know what are the law of reflection and refraction. So, the point is that how does one understand it from electromagnetic theory. So, I have an interface here which is perpendicular to the plane of the screen and I have an incident ray, I have a reflected ray and a transmitted ray. So, let me make the following statement supposing the electric field the space time dependence of the vectors are given by this. Only I have not written the amplitude I have simply said that this is given by I k dot r minus omega t which is the incident wave, I k prime dot r minus omega t which is the reflected wave and I k double prime dot r minus omega t which is a transient. Now, what I want is that when they come to the phase sorry to the boundary interface at all points on the boundary and at all times I must have a constant relationship which must exist between them. The irrespective of what happens whatever is the magnitude in order that that constant relationship does not change the argument of the exponential function must be the same at all times on the boundary. So, I need k dot r equal to k prime dot r equal to k double prime. Now, we using this I have assumed that k is the incident wave vector, k prime is the reflected wave vector k double prime is the transmitted wave the refracted wave vector same thing as transmitted wave vector. Now, I can now decide on my coordinate system. So, the coordinate system remember I have this relationship k dot r equal to k prime dot r equal to k double prime dot r at every point. Basically the confusion in the transmitted ray and the refracted ray is this. In the instruments which you which measure the transmittance and the refractance they measure the intensity which is transmitted out of the media itself. That is what I said that you are talking about an emergent ray. So, emergent ray that the transmittance is measured in the terms of emergent ray intensity itself. But that is why there is a confusion. So, there are two interfaces here. The first interface is on the first slab of the glass and second interface is inside the second medium when it again meets a boundary between air and the glass. Yeah, definitely. So, intensity is going to reduce. The refracted ray is going to be more intense. But you see the what we are talking about is at one interface. Okay. Okay. I am not talking about what happens to the future of that ray. No, no, but I am telling that transmittance definition itself is having some error in that sense. In optics there is no confusion. If you are talking about the interface between two media. The first medium is called incident medium. The second medium is called transmitting medium not refracted. The phenomena is called refracted. Transmission is what is transmitted to another medium. Okay. Yes. Now what the confusion is which he has partly corrected is most of the time when you are doing an experiment in a lab you are not going to have an infinite slab. You have a slab of finite dimension. So actually in that case you have two refractions. One at the interface of air and the first glass then that ray goes and hits the bottom of that glass and goes out again. Okay. In most optics book they will call the first one as the incident ray. Second one as the refracted ray. The third one as the emergent or it transmitted ray. Sir in the measurements transmitted ray and emergent ray are treated to be same when the angle of incidence is it means it is normal incidence. No, no, but we are not talking about necessarily about normal incidence. Okay. See. It means in the measurements we introduce the sample normally. We are wasting time on what I call as the English language problem rather than physics problem. I am sure all of you understand it. Okay. What I am trying to tell you is the standard language followed in any optics books. I also understand the language that you are talking about because I have also gone through the same school books. Okay. What I am trying to tell you is that if I take this as a medium when a ray came here that is incident ray. When it is inside water it is you would call it refracted ray. I would call it transmitted ray because my water amount is infinity. It is not coming out. On the other hand you would call that a refracted ray and since it is going to come out you will call that a transmitted ray. That is all. But I have said that time and again that I am using what is normally used in any book of optics. Any book you can see the you have a reflection. Refraction. Refraction what is the for the phenomena? Not ever for the wave. Not for the ray. Okay. Move into the medium that is. The one which goes into the second medium is a transmitted. Clear. Okay. Let us not waste more time on the language. Okay. So after we have agreed that we will take my understanding of the nomenclature as the standard for this class. Let us do the following. So what we said is this that I got a relationship of this type that in order that there is a constant phase relationship k dot r equal to k prime dot i equal to k double prime dot r. Now you notice that this tells me that the vectors k, k, remember that r is any point on the interface. Okay. Now it tells me the vectors k, k prime and k double prime are coplanar. Okay. And their projection on the boundary have to be equal because k dot r is the projection on the boundary. This is the basic physics way. Now suppose I take the k to lie in x y plane. Right. Supposing I take k to lie in the x y plane then k prime and k double prime will also lie in the same plane because k dot r equal to k prime dot r equal to k double prime dot r. Now so as a result now with this choice and because of the fact that k dot r is the same I can choose my coordinate axis properly and see that k sin theta is k prime sin theta prime. Remember theta prime is my angle of refraction and the angle of transmittance is phi and equal to k double prime sin phi. This simply comes from k dot r equal to k prime dot r equal to k double prime dot r. So that immediately tells me that if I am looking in the first medium, remember the reflected ray is always in the same medium. So as a result the magnitude of k and k prime are the same. So if magnitude of k and k prime are the same, the wave number is the same then this will tell me sin theta equal to sin theta prime means theta equal to theta prime. That is nothing but the law of reflection. Then again the if I look at what is my ked ratio of k double prime to k, I know that is the ratio is omega by the velocity in that medium to omega by velocity in this medium and that is equal to the velocity in this medium to the velocity in that medium which is nothing but the refractive index. And that is where you got that sin phi by sin theta is the refractive index of the second medium with respect to the first medium ok that is your Snell's law. So the thing is this that I have got the following situation. I have said that for the incident ray I have h equal to k cross e, for the reflected ray I have got h prime equal to k prime cross e prime, I have got h double prime k double prime cross e double prime that is for the transmitted. Now what is done is that we are interested in finding out what fraction of electromagnetic wave or electromagnetic energy if you like is transmitted into the second medium how much is reflected back and things like that. Now this is very important from technical point of view I will tell you a little later but in order to do that what is done conventionally is to divide the solution into two parts. Say that you talk about what is known as a TE polarization. The TE modes the electric vectors of the incident wave is taken to be perpendicular to the plane of polarization. And in the TM mode the magnetic vector is taken to be perpendicular to the plane of polarization. But let me just talk about just the TE and what I want to get is the ratio of the amplitudes of the electric field or magnetic field it does not matter of the reflected part and the incident part. So let us look at that I will not do the derivation but it is fairly straight forward see the point is this now I have said that my light is going into a second medium in other words it has met with a boundary. Now when electric field is incident on a boundary there are two conditions to be met. I need to say remember I talked about electrostatic boundary condition and we said the tangential component of the electric field must always be continuous. It is that continuity which is what helps us in determining these ratios. So look at what it means. So I say that this is the way I would write down I have said that tangential component of the electric field is continuous which means E plus E prime equal to E double prime. Now I know that the magnetic fields H H prime H double prime are proportional to this. So I can write down E plus E prime equal to E double prime in terms of the corresponding fields and the angles. Now I will not do this arithmetic but if you do that you find a ratio of E prime by E corresponding to the T E polarization here and this is the expression for the T M polarization. So this is simply telling you what is the ratio of the amplitudes of the these are known as Fresnel equations. I am not going to derive it here there in any textbook but you can see how the derivation is entirely based on saying that the tangential component of electric field is continuous. Having done that here the S is for T E polarization P is for the T M polarization. You can show these are known as the Fresnel equation E prime by E is given by such an expression and the reflectance which is defined as the square of this is E prime by E whole square and for example if you want to take normal incidence then you will have theta equal to 0 which will leave us with that the reflectance for both the types is n minus 1 by n plus 1 whole square. What does it actually tell me? It says that if you are in normal incidence I still have certain amount of field which is reflected. Now notice that I have not talked about any losses or any such thing. It is just because of the fact that something is incident on a surface. You cannot avoid reflection all to it. Now this problem this tells me that if you take for example glass with a refractive index of 1.5 it tells you that there is a minimum reflection of 4 percent because if you calculate n minus 1 whole square by n plus 1 whole square that is what you get. Now this is rather important particularly in optical instruments like for example camera which may have a large number of lenses and at every lens you will have a reflection and even though they are just 4 percent reflections these reflections can add up to a significant one ok. So, there will be a light loss at every time. So, what they do to avoid such things in cameras is to coat the lenses with anti reflection coatings ok. Now there are many other interesting things about the Fresnel equations. So, let us go back take any of these expressions. See it has this let us take just the T polarizer cos theta minus square root of n square minus sin square theta. Remember what is n? n is the refractive index of the second medium with respect to the first. I just have 2 medium I do not have a third medium ok. Now notice if it happens that sin theta is greater than n. Now obviously, I am talking about a situation where the refractive index of the second medium is less than the refractive index of the first medium ok. So, if it happens that n is less than sin theta then at that angle this square root the argument of the square root would become negative. Now if the square root argument become negative I will be able to write the reflection coefficient as cos theta by interchanging the sin theta and n in this expression as cos theta minus i times square root of that cos theta plus i times square root of that. Now notice that the numerator and the denominator of these expressions are complex conjugate of each other. As a result the absolute value of that is equal to 1. What it tells me is that in such a case the entire field is reflected back into the medium and because of that this phenomena is known as the total internal reflection it goes back into the medium ok. This is something which is known to you, but though the incident energy is totally reflected back into the first medium the field in the second medium is not 0. There is a field close to the interface into the second medium that is the field leaks into the second medium I will show you how and this field is known as an evanescent field ok. Its behavior is a little interesting look at what we are saying regarding the electric vector of the transmitted wave. So, we had this expression there ok transmitted wave is e double prime e to the power ik double prime dot r etcetera and I have written what is k double prime dot r in terms of the refract the angle of refraction. You do this you find that the cosine of the refraction angle is imaginary. So, therefore, the electric vector actually becomes like this. So, there is a part which is moving along the interface with the cosine or sine that you have taken, but as you go away from interface the field exponentially decreases ok. Now, let me let me summarize what I said. I said that when though I am showing it in a medium where I am going from air to a denser medium the actual question that I am discussing is the light is coming from a denser medium to a rarer medium ok and it is falling on an interface at an angle greater than the critical angle for that medium. You know that when light is incident at an angle greater than critical angle the light is totally reflected which comes out from the final equation. Because we have shown just now that the magnitude is equal to 1. Now, though it is totally reflected the total energy is totally reflected the field just outside the interface is not equal to 0. This is called an evanescent wave. It is not equal to 0, but it propagates on the surface, but if you go away from the surface its amplitude decreases exponentially. So, therefore, there is really not a great deal of leakage into the second medium. Though it is rather an exponentially decaying thing, but one can actually detect it by putting a detection mechanism very close to the interface. So, these are evanescent wave ok. So, let me since we have just talked about total internal reflection. Let me use this opportunity of talking to you about what is known as the acceptance angle of a wave guide. Now, since all of you have been worried about that I need more mediums I mean. So, I have here a medium which such as air. Now, this is a refractive index N 1 which is the core of the fiber and on these are jacketed by a cladding. And what we have is this N 1 is slightly greater than N 2. Typical numbers are 1.48, 1.46. It is it is not a very much different ok and of course, you have air. Now, what happens is this that ok go back to this picture. So, what happens is that if light is incident at you know supposing I have a point source of light symmetrically placed. Now, the what are the two extreme rays ok, which defines my region over which the light can be incident on the fiber optic wave guide such that the it is guided inside. Now, what is meant by guided inside? We know that in case of fiber optic wave guide. Now, general guidance is along the length of the fiber which let us call it as the z direction, but it does not mean it is travelling along the axis. What it could be doing is to go to the make an angle, go to the interface between the core and the cladding and since the index of refraction of the core is slightly greater than that of the cladding depending upon the critical angle of the medium. Yes, it will get totally reflected. Come back again go to the other phase and it keeps on doing this, but on an average it starts moving along the direction. So, what we say is this supposing I is the angle of incidence. Remember the angle of incidence is not this angle. The angle of incidence is the angle which the normal to the interface. So, supposing this is the direction. So, I have a ray which is falling like this. The angle of incidence will be I have to take a perpendicular on that interface and that is my angle of incidence I ok. The angle of refraction which you normally call is then sin i by sin theta I is in the air medium. Theta is in the denser medium, but there are two denser medium there, but theta is in core. It has to fall on the core. So, sin i by sin theta is n 1 by n 2. Let us say n 1 is the refractive index of air which will later on take to be equal to 1 and n 2 is the refractive index of the core ok. Now I know that for total reflection to take place, my incident angle at the top see what is happening is you have to imagine unfortunately I do not have that picture. See the thing is this that you have a refraction correct. Now this ray is now going and falling on the core cladding boundary. So, draw a normal there. Now that angle that angle at which it has fallen there is what I am calling as phi, but that phi obviously is 90 minus theta clear. The geometry is clear that is the only thing that I wanted. So, therefore, my sin phi is same as cos theta and this sin phi or the cos theta if it is greater than n 2 by n 1 right. Now my I have n 0, n 1 and n 2. If it is greater than n 2 by n 1, then I have a critical reflection is it clear? Because n 2 by n 1 is the relative refractive index. Now the point is this that if you look at this relation then it tells me my sin theta should be less than root of 1 minus n 2 square y. This thing and then you do a trivial arithmetic it works out to n 1 square minus n 2 square by n 0 square. I will take n 0 to be equal to 1 traditionally. So therefore, it tells me that assuming n 0 is equal to 1, the maximum value of the incident angle ok. Once incident angle is decided you can also decide what is the aperture here. Angle at which a total reflection will take place at the core cladding boundary is sin of I maximum is equal to square root of n 1 square minus n 2 square ok. So, this maximum angle say anything falling with an angle less than that the light will be guided here. It will not be guided it will be not be totally reflected if this condition is not met. So, in some sense I m the maximum value of the incident angle is a measure of the light gathering power that this is the light which is gathered together. So, that the it can be sent through a 5. Can I say like this sir that in the angle of incidence is equal to acceptance angle at that time 5 becomes 5 c and the total internal reflection taken place. See the what I have defined is what is known as a numerical aperture at this moment. But basically I have talked about in terms of the incident angle. Now, if you are looking at a light cone you see I have a point source from point source various lines go out right. There would be a solid angle at that point ok which you can easily calculate from there if the rays come out and hit the air fibre boundary at an angle less than I m then the guidance is possible is this clear. So, this is what it is there is actually if another relation is there it is equal to 1. So, this numerical aperture is defined to be a measure of the light gathering power of the at the fibre end ok which is given by square root of n 1 square minus n 2 square. Now, if you take the 1.48 and 1.46 the typical values of the fibre then you find that it is about 14 degrees. In other words the maximum angle of incidence it is a very narrow beam it should be and that is why you want a laser source or things like that ok then only it will go, but we will come back to discussion of fibre little later in the second session. So, I will now spend the rest of the 10, 15 minutes that I have in discussing the linear superposition and interference. I am not discussing the standard Young's double slit experiment or diffraction or things like that I am just going through the principle of it. Very interesting things come out of here firstly let us consider two plane harmonic waves which are linearly polarized ok. The let the first one now you know I have used a very funny nomenclature E with a subscript 1 within its bracket is E 1 times the exponential i k 1 dot r minus omega d plus phi 1. The second electric field I have taken is E 2. So, there is a difference between this E 1 and this E 1 within bracket that is one of them is multiplied with its phase factor the other one is not ok. Now, we define iridescent at a point as the square of the amplitude of the light field and I know that the superposition principle is right. So, if I have two waves E 1 and E 2 ok then my iridescence would be given by E 1 plus E 2 absolute square. Now, I have taken exponential form. So, I have to be a little more careful. So, my i is E dot E star because I need real part only and that is E 1 plus E 2 dot E 1 plus E 2 and you just multiply them you get E 1 square plus E 2 square plus 2 E 1 dot E 2 times cos theta where theta is defined by this. Now, notice E 1 and E 2 cos E 1 dot E 2 cos theta. It is not E 1 into E 2 into cos theta because the cos theta has come from the cosine part. I have electric factor which I am multiplying. This if I look at it has a E 1 square which is my I 1 E 2 square which is my I 2 and this is the interference term that I have. It is an interesting point. There are very many things that come out of that interference term. Only a part of it is explained in the school. First thing that you realize is this that my irradiums is given by 2 E 1 dot E 2 times cos theta where theta is k 1 dot r minus k 2 dot r plus phi 1 minus phi 2. Look at this depending upon whether this term is greater than or less than 0. The net intensity that I get at a point could be greater than or less than the sum of the intensities due to a single that of course, you know ok. So, this is the reason why you get familiar fringes ok. The second point is suppose I take the sources to be mutually ok incoherent meaning thereby the 2 waves that you have taken they have a phase factor phi 1 and phi 2 which are random. Now if that happens then the cos theta which has phi 1 minus phi 2 it is randomly varying and other results its average is 0. So, therefore, I will not have any interference. So, incoherent waves the waves which have no definite phase relationship between the 2 waves they do not contribute to the interference term. The third point is this supposing the 2 waves are differently polarized then E 1 dot E 2 factor will also be different. Normally when we do this calculation in school or college we assume that the direction of polarization of E 1 is the same as the direction of polarization of E 2. So, that this expression is always written as E 1 times E 2, but what comes out of the vector addition is not E 1 times E 2, but E 1 vector dotted with E 2 vector where E 1 is the amplitude part of it along with the direction. Now in particular if E 1 and E 2 are orthogonally polarized perpendicularly polarized then E 1 dot E 2 is by definition 0. Now since that term becomes 0 it tells you that you cannot have interference between 2 waves whose polarizations are perpendicular. This is not only true for linear polarized wave you cannot have interference between 2 circularly polarized light if one of them has a left handed polarization and the other one has a right handed polarization. The experiments are very easy to perform actually. So, you do interference experiment in your this thing labs and frequently near one of the slits you put a small mica. They say let us determine the thickness right a standard experiment done. Why? Because light in moving through the mica travels an extra distance. So, there is an additional phase difference which depends upon the thickness through which it is gone. Now supposing you are to put in any other phase giving thing there. You first polarize them then as I was talking to you yesterday supposing you decide to put a quarter wave plate there. So, that one of them has an additional phase there. Then when you add you have to put this additional phase also and if you manage to give a full half wave between the two ok. Then you are also having something else, but suppose I now put in polarizers there. So, that the two waves that you have got are differently polarized. Then you will find there will be uniform illumination on the. So, what you have said is incoherent waves do not give us interference pattern that is number one. The polarize now mostly what we have is you get an interference pattern with the light that comes in. Because the even when they have the polarization in similar direction there would always be a component in the same direction. Because you are never able to perfectly make them in orthogonal way ok. In the second session I will be discussing some properties of the optical fiber this will be again a hurried session yes. You talked about this evanescent wave. Yeah, evanescent wave. Yeah, but this is related to the case when you have this total internal reflection. Yes. And you said that the energies actually completely get reflected. So, but there is a still a bit of kind of leakage outside. The signal is there. Signal is there. So, is this really like this quantum mechanical tunneling kind of. Well, this is basically classical tunneling. Classical it looks like actually. Where the fluid field is there, but you remember if such a evanescent wave was not there you will never be able to couple a fiber optic to an outside cable ok. Well, ok let us come back to it again. Ok. So, second session remind me that question we will talk. Ok. Yesterday we talked about we had a discussion about equivalence of the electric field and the magnetic field that you talked about. But today obviously you showed these two equations which are completely identical. But if electric field and magnetic field were really truly identical or equivalent to each other then instead of having h I should have an equation in terms of b. But instead of h instead of b why you have an equation with h. So, I cannot really say that probably that they are not truly equivalent that is my. I said that the equations that they satisfy is identical ok. So, therefore. No, the identical equation are being satisfied by h and not b. b also satisfy similar equation with the constants being different that is all. Yeah. Ok. You can write down that also. Yeah. The same constants will come there. Not because I have talked about here of a linear isotropic medium. So, h wherever h is there write b by mu 0 and cancel out the problem. Yeah. Equation is the same. Ok. Thank you. How is the equation will not change. Yeah. Ok. Thank you. Ok.