 What I am going to do is the following, I will start with a quick review of propagation of light, this and vector nature of light, both of them are based on the properties of electromagnetic field that we have talked about and then I touch upon coherence and interference emphasizing the first part I do not talk about things like Young's double slit or single slit diffraction which of course all of you have learnt properly from school onwards. I will also talk a bit about amplification of light in particular laser sources and also I will try to give some focus some or give some attention on propagation of light in optical fibers. It will all depend upon how much I cover will depend upon the time that I have in these two lectures. So let us proceed with that. So if you recall, I am going to be primarily interested in propagation of light in a non-conducting and dielectric medium and a medium which does not have any sources of charge and current. So my equations, Maxwell's equations which will be necessary for me will be del dot of B equal to 0 which of course is still true but in addition today I will have del dot of E equal to 0 because there are no charge sources Faraday's law of course remains the same that is del cross of E is equal to minus d B by dt and finally I have del cross of H which is d D by dt I do not have any current term because there are no sources there and since I have assumed linear dielectric and non-magnetic medium what I have is del cross of H is equal to epsilon 0 d E by it should have been d D by dt but I have it as epsilon 0 d E by dt. So basically what I have done is to write my Maxwell's equation in terms of only these variables E and B instead of bringing in H and this thing. I mean I will remove that H also in no time. So if you look at that what I am going to do is to see how does one what is the solutions of these equations. So I have supposing I take a curl of this equation I get del cross del cross E and this I have told you is a very standard thing which is del cross del cross of E is del of del dot B minus del square E and if you look at this thing here that is I have got del cross of E is minus d B by dt. So this is equal to minus d B by dt let us supposing I decided B is equal to mu 0 H. So I could write this as d by dt of mu 0 and del cross H because I have taken a del cross of both sides. So that gives me minus mu 0 d by dt of now I put this in place of del cross H I put epsilon 0 d E by dt. Now I use the fact that my del square of E sorry this term is equal to 0. So I am left with since minus sign cancels from both sides I am left with del square of E is equal to minus mu 0 epsilon 0 d square E over dt square alternatively I mean this is nothing but a wave equation because I have got del square of E plus mu 0 epsilon 0 happens to be as we have talked about several times 1 over c square. So this is this plus 1 over c square d square E over dt square that is equal to 0. You can check this by putting in numbers also it works out that way. So the thing now is that I have a wave equation and you could get a very similar equation in so therefore if you like this quantity here is of course 1 over c square I have written down but supposing I had a linear dielectric with epsilon coming in here normally mu 0 remains mu 0 then I would have brought in a dielectric constant to the medium. So which tells me that my speed of light or speed of the electromagnetic wave in whatever medium I am talking about is given by 1 over square root of mu epsilon well normally it is mu 0 epsilon. So that was my equation that and this quantity because see the reason why I have not changed I am not particularly keen on changing that mu 0 to mu is because we normally deal with non-magnetic medium and so therefore mu remains mu 0 but in propagation of light we are frequently interested in how light travels through for instance glass when we talk about reflection, refraction and things like that. So therefore we bring in this 1 over c the epsilon factor. Now this ratio of epsilon to epsilon 0 that is permittivity of whatever medium you are talking about to the permittivity of the free space that is the relative permittivity or the dielectric constant. So therefore what I have written here is this that we write down what is the velocity of light and it scales according to the as we know that in a medium the velocity of light reduces by a factor of the dielectric constant. So the by the refractive index and most transparent optical media as I said are non-magnetic and so therefore the index of refraction is square root of k. Now so when we write down this equation del square u equal to 1 over u square d square u over dt square with the minus sign of course which has been forgotten here. And so if you look at each component that is whether it is x component y component or z component whether it is e or h I have represented by the same letter u. Now this satisfies this equation this is what we have seen and the harmonic solution which we are normally interested in goes as some constant u 0 cosine k dot r minus omega t. This k as we know I am going through this section a little fast this section this k as you know is the propagation vector whose magnitude is called a wave number. So suppose I have constant values of this argument that is suppose k dot r minus omega t is constant I look at these surfaces this is surface this will define a set of planes k dot r minus omega t this would represent a set of plane and so this is the way it would be. So this propagation vector that we are talking about then is normal to the surface of the constant phase. And these surfaces of constant phase are normally known as the wave fronts and we have said that these wave surfaces they would move in the direction of the k vector with a particular velocity. Now what one does because dealing with sine and cosine functions they become somewhat difficult because of the fact that sine cosine on differentiation mix up. But mathematically what one finds very convenient to do is to take this expression that is instead of taking cosine or sine of k dot r minus omega t the one takes the corresponding complex quantity. The corresponding complex quantity and you would say what is a complex wave well there is nothing, but what we do is for mathematical simplicity we take this form and later on at the end of the calculation because exponential functions are very you know nice to deal with we simply take the real part of that that is what I am trying to say is my wave is supposing you are taking a cosine function my wave is nothing but the real part of this function because e to the power i theta is cos theta plus i sin theta or you could take a sine function in which case it is an imaginary part of this function. So what we do is we carry on all our calculations using the complex factors and complex numbers are much easier to deal with complex arithmetic is much easier than sine cosine arithmetic. For instance when you differentiate an exponential function you get an exponential function back but when you differentiate a sine function you get a cosine and vice versa. So that makes the algebra a little more difficult, but this is a very neat way of doing things. Now let us look at a slightly different function. So what about cosine k r minus omega t instead of k dot r minus omega t? Now you can see immediately that just as the k dot r minus omega t had constant value over the wave front that is over a surface supposing I wrote k r minus omega t. Now then k r minus omega t at a given time t would have constant value on surface of the radius r. Now so there is no dot here k r minus omega t, but problem with this function is this is not a solution of the wave equation. If you substitute it into the wave equation k dot r minus omega t is a solution of the wave equation but k times r minus omega t is not a solution of the wave equation. So however it turns out that if you write down the equation in spherical polar coordinates and look at the radial portion then you would find this to be a solution of the wave equation provided you divide it by a factor of r. So 1 over r cosine k r minus omega t is a solution of the radial wave equation. So in principle this wave is the wave fronts are spherical and it is a spherical wave. Now so look at this structure del square of u is 1 over u square d square u by dt square. Now suppose I am dealing with plane waves as I said we will take u 0 e to the power i k dot r minus omega t. Now when I do apply a del operator on it. Now if I apply a del operator on it this is same as equivalently as if you are multiplying this exponential with a factor i k vector because what you are doing is to differentiating in three dimension. So what you will get is that effectively taking the gradient operator is same as multiplying this with the factor i times k vector and if you are taking the time derivative this is equivalent to simply multiplying with a minus i omega factor because d by dt of that there are some minus sign changes there. So if you come back to now this equation del cross of e is minus mu d h by dt. Now that tells me that since del cross is equivalent to multiplying with an i k. So I get i k cross e and since d by dt is equivalent to multiplying with a minus i times that. So I would get that k cross of e equal to mu omega h and similarly I get del cross of h equal to minus epsilon omega e and the two divergence equation del dot of e equal to 0 gives me i k dot e equal to 0 and i k dot h equal to 0 i cancelling out i I get this. So you notice i because of the fact that I because of two things one is I decided to use the complex phase part they so what I did and the secondly because we have decided to look at the plane wave solutions only. The radial equations as I told you are little different but then that 1 over r factor is important because that is the one which gives you the inverse square way in which the intensity changes in a spherical way. So the because of these two reasons my equations which were in terms of derivatives I got simple algebraic equation k dot e equal to 0 k dot h equal to 0 k cross e equal to mu omega h k cross h equal to epsilon minus epsilon omega So these are my replacement of the entire set of Maxwell's equation for the case of the electromagnetic waves which are plane wave solutions to the Maxwell's equation and look at these equations. So obviously I am dealing with much simpler situation and because of this two things that I did one is the exponential representation of the plane wave solutions instead of cosine or sine I could not have done that if I had done kept the cosine or the sine because the derivatives would give me complicated results. So my results are my k cross e is mu omega h k cross h equal to minus epsilon omega e and k dot e equal to 0 k dot h equal to 0. This tells you two or three things firstly the vector k where vector k is perpendicular to the electric vector it is also perpendicular to the magnetic field vector. See in while dealing with the electromagnetic waves people prefer to write take the electric vector e and the magnetic field vector h rather than taking e and d but it is immaterial there is a factor of mu because we have taken just the linear magnetic field. So k vector is perpendicular to the electric vector as well as to the magnetic field vector. Now you look at these equations any of them it tells you k cross e is proportional to h or k cross h is proportional to e that is minus e. Now that tells me that the vector k vector e and vector h they form a right handed triad system that is k cross h e will be in the direction of h h cross k will be in the direction of e which is shown with a negative sign because I have written there as k cross h. So I have a system in which the electric vector is electric vector and the magnetic vector are both perpendicular to the propagation vector k and the propagation vector k well these form a right handed triad. Now look at the ratio of the field the magnetic field h to the electric field e. Now you can see that we had written down earlier here this is what we have got here k cross h any one of them you take. So if you do that and take their magnitude because now that we have shown that they are perpendicular to each other so my cross products will not give me any other factor. So I will be able to write that magnitude of h is epsilon omega divided by k but omega by k since I am writing down a wave equation is nothing but the velocity of light. Now mind you I have taken epsilon so therefore the I am talking about a velocity in whatever medium we have. So my definition of refractive index is always the ratio of the velocity of light in vacuum which is of course the largest value divided by the velocity of light in the medium in which you are talking about. And that quantity if I substitute it here it tells me that h is equal to remember c is equal to 1 over square root of mu epsilon. So if you substitute it here you find that the magnetic field strength is proportional to the electric field strength but there is a factor there which is square root of mu 0 by z 0 it is reduced and that amount if you can calculate it is normally called the impedance of the medium or vacuum in this case and its value is very close to 377 ohms. So having done that we recall that we had shown that the time rate of flow of the electromagnetic energy per unit area is given by the pointing vector. So s is equal to e cross h and s specifies both the direction and the magnitude of the energy flux. So let us look at this pointing vector which is giving me information about the energy transport and let me talk about it for the case of plane electromagnetic wave. For the case of plane electromagnetic wave I have taken e is equal to its amplitude e 0 and we have said cosine k dot r minus omega t and h was correspondingly this I could have written down in exponential notation but in this case it is not necessary. So if I take e cross h so I get e 0 cross s 0 because these are the directions of the vector the times cosine square k dot r minus omega t. So if I am looking at that time average of the pointing vector if I am looking at the time average of the pointing vector I am going to get s is because we know cosine square or sine square functions of time on an average has a value half. So therefore the average pointing vector which is what we normally are interested in is given by half e 0 cross h 0 and this quantity is what is normally called the irradiance. So the what we are talking about is you can write it like this. So we define irradiance to be the magnitude of the pointing vector the and so irradiance is nothing but half e 0 h 0 and this quantity is given by this. Having talked about the propagation of plane waves in a linear dielectric medium because or air vacuum they are the same really because all the difference that it makes is that there is a change in the speed of light when it enters a dielectric medium and that is done by a refractive. There are other changes which we will be talking about later. Let us look at what is this electromagnetic wave we have taken. We have written this as e equal to e 0 vector exponential i k dot r minus omega t and correspondingly this. Now we did not quite talk about that whether these things they remain constant all that we know is we have said that they are not dependent on time. But suppose we say that these are constant real vector these are constant real vector then the waves would be called plane polarized wave and see why. Because you see if this amplitude factor if you like has a constant magnitude and direction. So, this look at this yellow curve pictures. So, I have said that that there is a constant vector e and supposing I am taking a cosine or a sin does not matter. Then that is why I keep on switching between the two. Then you see the my actual electric field is the direction does not change. But this thing the strength of the electric field gets modified by either a cosine factor or a sin factor and that is the reason which you are seeing it here. And the corresponding red one which represents the magnetic field vector is perpendicular to both the electric vector and the magnetic vector. So, the thing is this that we are talking about that such a wave in which now let me let me tell you how does one picturize this what is this picture. There are two ways of doing a picture let me let me though I have written down the exponential consider a cosine k dot r minus omega there the real one. Now there are two ways you can look at how the wave looks like or how do you picturize that way. There is a one picture which I can call as a local picture. What local picture means is that you look at a fixed point in space, but let time vary. Now supposing you do that that you say that this is instead of exponential I said you take cosine k dot r minus omega. Now I am saying supposing you make the position to be fixed. So, in other words the first factor there does not change. Now what you then do is this that since that factor is not changing that. So, if you are looking at a particular position the electric field vector will then have a variation of the phase factor because of the fact that with time the thing is changing. So, in other words if you concentrate at a particular point in space and look at how does the electric field vary with time. Then you will find that the amplitude of the electric field keeps on fluctuating according to a cosine or a sine function that you have taken, but the argument of the cosine or sine function is a phase factor which is a constant phase factor because we have said k dot r is constant minus omega t and that is giving you the time variation. So, you will find that direction is not changing, but with time the amplitude is going up and down and oscillating in some sense. The other picture is what we will call as a global picture. Now this is the picture that is normally written down here. So, basically in the global picture what you do is you freeze in time. So, you say suppose I look at what does the electric field look like at a particular time, but everywhere in space. So, here is a picture which I am plotting for example, in the direction of propagation. So, we have said here that supposing I say cosine k dot r minus omega t and we say that we choose a particular direction of propagation. Now in this particular case I have chosen the direction of propagation to be the y direction does not matter. Now when I do that then the picture is because the time factor is now frozen. So, that gives me a fixed phase there and I have cosine in this case k y y minus omega t. So, what I see in space as I go along the propagation vector and supposing I am looking in the plane which contains the propagation vector and the electric field. This is the picture the yellow picture that you will get. On the other hand if I look into the plane which contains the propagation factor and the magnetic field direction and that the red one is the picture that I will get. So, and it is because of this that there are two nomenclatures which are common. Now one calls them either by the name linearly polarized or by the name plane polarized. Now if you look at the nomenclature plane polarized what does it actually mean? It means why is it called a plane what has it got to do with the plane? So one is that if you look at the direction of either electric vector or the magnetic vectors at various points at a given time. Now since their directions are remaining the same they are always contained in a plane and hence the name plane polarized. However if you are looking at a local picture I do not have a plane. There what I am doing is to say that I have fixed point and with time my electric vector or the magnetic vector is oscillating along the direction of the electric field or the magnetic field. And so therefore the corresponding picture is a linear picture and that is the reason why both the names are interchangeably used. Now there is another form of polarization I will be talking about the most general form of polarization is the elliptic polarization. But let us talk about the ones which are much easier to draw and picturize. Now supposing I take two linearly polarized way and both have the same amplitude but E 1 E 2 both have the same amplitude. But supposing they have a phase difference of pi by 2. So what I have done is this that I have a situation where I have to remember that any linear combination of solution is also a solution. Now I choose by coordinate axis such that the two waves the electric vectors of the two waves because I said pi by 2 difference. So they are in the x direction and in the y direction. So one component in the x direction another component in the y direction but I choose the E 0 to be the same. So if I do that the since these two are solutions of the wave equation the total electric field is also a solution of this wave equation. Now look at what is this? Now supposing I look at this plus this. This is also a wave the say notice supposing I had a cosine here instead of a sin. Then I would simply say this is I E 0 plus J E 0 times the cosine but all that it means is your direction of polarization has now shifted from one of the axis to 45 degrees.