noc19-ee21 Lecture 19-Oblique Incidence of Plane Waves-I

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noc19-ee21 Lecture 19-Oblique Incidence of Plane Waves-I

Hello and welcome to NPTEL MOOC on Electromagnetic Waves in Guided and Wireless Media This is Module 19 and in this module we are going to consider a different type of incidence, which is called as oblique incidence. Okay This case is important because in many scenarios, especially, when we talk of waves in guided media, we can think of the waves or the modes propagating in that guided media as being composed of successively reflected obliquely incident waves. Okay. So whatever we are going to learn here, we can apply it to the study of waves in the guided media, which we are going to do in the next, I mean, not next, but some other modules after we finish this properties of plane waves, right? So with that, let’s actually look at what the physical situation is. The physical situation for the problem is kind of the same. So you have this plane of interface wherein you have a medium one and a medium two. Previously, we considered angle of incidence in such a way that the propagation vector was coinciding with the normal to the interface plane. So the normal to the interface plane was the Z axis, and then the angle of incidence was coinciding, sorry, the propagation vector of incident reflected and transmitted media were coinciding with exactly the same Z direction normal to the interface. Okay Now instead of this propagation vector coinciding, what happens when the propagation vector is at an angle theta, we will call as theta1 because we want to distinguish two angles, what happens when the incident wave arrives at this plane of interface with an angle theta1 as measured from the normal? So this is a normal. Move theta1 here and this is the angle of incidence now. So what happens? We know from Snell’s law, two things are going to happen. One is that there will be a transmission into the second media whose angle of refraction theta2 can be related to theta1 by the following Snell’s law. So you have n1 Sin theta1 equals n2 Sin theta2 where n1 and n2 are the refractive index of the media, indices of medium one and medium two. Okay. So this is a law which allows you to determine what is theta2 given n2, theta1 and n1 There is another law which states that angle of reflection, which we will call as theta reflection would be exactly equal to the angle of incidence, right? So what it means is that if I consider again this obliquely incident wave, this wave as it approaches, okay, part of that one will be reflected onto this side, right? So part of the wave is actually reflected onto this side and some portion of the wave is transmitted into the second medium with an angle that is different from the angle of incidence Now nowhere with the Snell’s law you actually are specifying or you are actually able to determine how much of the power that has been incident is actually being reflected and how much of the power is being transmitted, right? So to obtain that important information, we have to go back to the electromagnetic perspective of this problem. Moreover, it is not just, you know, the amplitudes that are or rather the power relationship that are important, but also something interesting happens with the amplitudes as well. Okay. Moreover, this Snell’s law, the so-called Snell’s law actually fails for certain scenarios, which will be very important when you consider what is called as optical waveguides. Okay So for all these reasons, we need to go back to the electromagnetic perspective, okay, starting with electromagnetic waves and then apply boundary conditions to really understand how much power is being reflected and what exactly happens if the medium of first, if the first medium has a refractive index or equivalently the permittivity greater than the medium, second mediumís refractive index So all those things can be answered by looking at Maxwell’s equations or wave behaviour at the boundaries Now before we go further, we actually have to have two kinds of waves that we can think of. Okay. So letís return back to our picture This is my interface plane and I have this angle of incidence. I mean, I have this incident, this one. This is the propagation vector The black one is the propagation vector. Please imagine that this is at an angle. Okay Now I can have two cases. I can have electric field in this plane, okay, which would be if you look at it in this manner, so it should be perpendicular. So it should be like this let us say. So this electric field lies in the same plane of interface. Okay. Or I can

have the other way round. I can actually have the magnetic field in that plane, okay, may be in that slightly different, this one We define the incident plane as the one that would be concerned with this particular plane, right? The one that would involve that normal and one of the tangential components. So if we take the tangential component to be along the X, then any, the electric field can lie along in the XZ plane, okay, or the magnetic field line can lie in the XZ plane. Okay Depending on these two choices, you have what is called as transverse electric polarised waves. Okay. In this case, you have the magnetic field in the same plane as the interface plane, which we have, I mean, as the plane of incidence, which is X and Z. Okay. Or you can have transverse magnetic wave which is written as TM when it is the electric field, which lies in the plane of incidence, which we take as X and Z So as you have seen here, in this diagram, I am assuming that the electric field lies in this plane of incidence and therefore I am describing what is called as transverse magnetic because the magnetic field will be perpendicular to these two lines, right? So the magnetic field will be perpendicular and that is why it is the magnetic, sorry, transverse magnetic waves that we are considering. Okay So the rest of the ideas are quite simple All you have to do is to find out appropriately the boundary conditions. There are four boundary conditions. Et1 will be equal to Et2, no doubt Ht1 will be equal to Ht2, no currents. This medium has refractive index or equivalently the permittivity epsilon r1. This medium has a permittivity epsilon r2. And this time I have switched x and z axes. I have taken the z-axis downwards and x-axis along the horizontal thing, and the waves are given by these green lines and the electric field components are shown By the way, I have taken the electric field component for E2 to be completely arbitrary Equations will tell us whether the direction is this one or the direction of electric field E2 should be reversed. Okay. So donít worry about that. This is the incident wave. This is the reflected wave and this is the transmitted wave. Okay. So you have these three waves and you have these two boundary conditions You can also have, of course, the other body condition, which is Bn1 equals Bn2 and finally, epsilon 0 En1 or rather epsilon r1 En1 to be equal to epsilon r2 En2 coming from the D field normal relationship So these are the four boundary conditions that you have and you have to use these boundary conditions to tell us or to find out what would happen to the reflection, reflected power and transmitted power. Okay So let us go with this. So I have this k vector here, which is k1. This k vector is also k1 whereas this k vector is k3. k2 is equal to k1 because or rather Iíll write it as k2 = k1 because these two actually in magnitude they are in the same medium, right? Okay And this is k3. Okay We will now look at the tangential component for the electric fields in medium one and tangential component of electric field in medium two and equate the two, right? So I have blown up this portion of the picture here, okay, because I want to talk about the electric field angles. So you can see that if this is the x-axis, this is the z-axis, this is the plane of interface, okay, I have this electric field E1 itself having two components, which is Et1 and En1. Okay The tangential component, of course, is given by E1 Cos theta1 that is the amplitude, but there is a also a phase. Now what is the phase here? In so far what we have considered, our direction of k was exactly equal to, you know, it was actually equal to one of the normal or one of the unit vectors. It could be z, x or y. We have taken it to be z. So our k vector could be written as whatever the magnitude of the k vector, so medium one times z, right? So I could have written this k1 in a vector form as k1, which is the magnitude times the angle which is z In this case, that is not true. In this case, I have the k vector itself at an angle. Okay So the k vector should actually be written kt1 along say x plus kn1 along z, right, because you can take this line and then, you know, decompose this into two lines of this particular nature. One will be along x. One will be along z. And what is the value of kt1? kt1 will be, so this k vector can be written in terms of the, so along z it would be k1 Cos theta1 z plus k1 Sin theta1 x. So this would be the k vector, k1 vector. Okay

What about this e-jk1z that we were writing earlier? We were writing earlier, you know, as, you know, very simple as k1z because you could take this k1 vector and the position vector r. In the previous case, the position vector r was simply z z-hat because z was the only direction in which the wave was propagating So when you take the dot product of k1 and r, this phase factor was simply equal to e-jk1z, but in this case the position vector can be in the X and Z planes. So at any point that you can consider, okay, which would be described by X and Z, so that point can be described by both X and Z values, the actual, the phase factor that should be written will be e-jk1.r where r is given by x x-hat + z z-hat, okay, meaning that if you now combine everything, so what we have here is the position vector r given by x x-hat + z z-hat Now I will erase here itself so that these can be written correctly. The phase factor corresponding to the incident wave, okay, would actually be given by e-jk1 Sin theta1x + k1 Cos theta1z. Okay. So this would be the phase that should be appended to the amplitude So this e-jk1 Sin theta1x + k1 Cos theta1z should be accompanying the tangential component that we have for electric field E1. Okay See the electric field E1 would have the tangential component, which is given by this Et1. So this is Et1, which is of course making an angle of theta1 with respect to the electric field in the medium one, sorry, with respect to this axis. So you write down this in terms of tangential as well as a normal component Normally, you are not worried at this point Tangential component has an angle of theta1 with respect to this axis. So you simply have E1 Cos theta1, E1 being the magnitude of the incident electric field; Cos theta1 giving you this one. Okay. Now k3 which is in the same direction as k1 will also be given by the same expression except replacing theta1 by theta3. Okay So the tangential component in second medium, okay, the tangential component of the electric field in the second medium, of course, has the same angle also. So if you look at this electric field, you see that this electric field will be in the same angle. Everything is same except theta1 will become theta3 So I can write down Et3 as E3 Cos theta1 e-j(k3Sin theta3x + k3Cos theta3z). Okay. So that is for the transmitted electric field How about the electric field E2? Now E2 is slightly different because k2 vector itself will be given by k2 magnitude, which, of course, actually is equal to kl because magnitude wise they should be the same. They are in the same medium, right? So you will have kl itself Now look at this. The direction along z will be opposite to the incident wave. That is how, of course, the wave is propagating along -z direction. Therefore, you can write this as kl Sin theta1x. Why theta1? Because theta or rather we will write it as k2x at this point. Okay. So kl Sin theta2x, sorry, this would be x-hat, that is the vector along the x direction, plus or rather minus because now k2 vector is -k 1 Cos, so I have this k2 Cos here, sorry, along x Hold on. I’ve made a small mistake here. It should have been k1 Cos everywhere. We put in Sin instead of Cos. So this actually should be Cos, right? Let me write down this correctly. So this is your kl at an angle theta1. So along the x-axis will be kl Cos theta1, kl Sin theta1 Okay. This is fine. I mean, whatever we wrote earlier was actually fine. So we’ll go back and write it. kl Cos theta1 along z, k3 Sin theta3 along x and k3 Cos theta3 z So this is all right. I mean, I thought Cos and Sin should be inverted, but no. It is actually correct. So kl Cos theta1 will be along z, kl Sin theta1 will be along x, but for the reflected wave which is making an angle theta2 here, you have k1 or -kl Cos

theta2 because this is along the -z direction, right? So this would be along the -z direction and along x direction it would be positive I mean, it would be along the same +x direction as kl. Okay. So the reflected wave vector k2 can be written as k1 Sin theta2 x – k1 Cos theta2 z. Okay So I can write down Et2, which is the tangential electric field, what is the amplitude of a tangential electric field? Well, you have to again rewrite this picture and then determine what is the responding amplitude there? So let’s put down the amplitude here. The picture that I am now looking at is when electric field is making this angle E2. This angle is theta2. So, clearly, onto this one, this line makes an angle of theta2. Between this line that is electric field E2 and this normal will be 90 – theta2. Correct? Because this E2 is perpendicular to k2, so this angle will be 90 – theta2 However, these two lines, which is the perpendicular line and the horizontal line, they themselves are 90∞ apart. Therefore, this angle is basically theta2. Okay. So you can write this as E2 Cos theta2 along x direction that would be along the -x direction and E2 Sin theta2, that could be along the z direction, and it would be along -z direction However, we are interested in the tangential component. Therefore, I can write this as E2 Cos theta2, but this is along the -x direction because of the way that we have taken the electric field to be. So this would be – E2 Cos theta2 multiplied by this k2 phase factor, right, multiplied by this phase factor, which is given by -jk2 and k2 Cos will be along z, k2 Sin will be along x. So we have already written that one. So -k2 Sin theta2 x – k2 Sin or rather Cos theta2 times z So we have the three individual components It took us a little bit of effort to find out the correct amplitudes and other things That is because we are dealing now with oblique incidence. Normal incidence would be very simple You also have seen that there is a small change in our electric field coordinates because for the normal incidence theta1 will be equal to zero, okay. We assume that theta3 is also zero and theta2 is also zero. Why? We will see it shortly, but if you assume all these thetas to be zero, then you will clearly see that all the x dependent phase factors will go away as it should because there was no e-jk something times x in the normal incidence case. All the waves were propagating either along plus z direction or along -z direction All Cos theta factors will become one. Et1 will be along the +x direction. Et3 will be also along the +x direction. Okay However, Et2 has now become along the -x direction Okay. If you did not want this -x direction, all you could do is to simply switch this direction of the electric field, so instead of considering the electric field in this manner, you can select the electric field to be in the same direction as E1 and then adjust this minus sign in the H case. Okay So when you do that you can remove the minus sign and make this one plus without changing any of the other arguments. Okay. This is just kind of consistency between this module and the previous module. If you wish that, you could do this. Okay. If not, you can continue with the original assigned directions and work throughout. The questions will anyway tell you that you will have a minus sign or a plus sign. Okay Now all I am doing here is to try and make everything to be consistent with the previous module. Therefore, even though I start off with an electric field in the direction that I showed in the green line, I have now switched it over to the orange line simply because I want to be consistent with the previous one However, our original expressions that we started out in this module are also valid I would encourage you to take this as an exercise and continue that next part of the development with the original diagram as well. Okay. For now I’m simply assuming that electric field is going to be in this direction just to be consistent with the fact that when theta1, theta3, and theta2 are all zero, we land up back into the normal incidence case. Okay Anyway, so this is k2 we have written. Everything we have now written. So we have a set of equations which are valid for electric field. Okay Now what should we do? Well, we know that Et1 + Et2 should be equal to Et3. That is total tangential electric field medium one should be equal to medium three, right? And where should this equality be present? This equality should be present at z = 0 plane, but unlike the previous case because the phase

factors will be dependent on x now, this should be valid for all x. Okay This expression that the tangential electric field should be valid for all x, what it means is that if I consider the electric field to be landing at this point and then calculate the tangential electric field, it doesnít matter where the tangential electric field lies on the x-axis, right? At every point on the x or rather every point on the x should satisfy this equation. Okay. So that is the critical part of it And now when you impose the condition that at z equal to 0 and for all x, the sum Et1 + Et2 should be equal to Et3. You can write this as, okay, with z equal to 0, all these terms will go to 0. You don’t have to worry about it. So the expression will actually become even E1 Cos theta1 e-jk1Sin theta1x + E2 Cos theta2. Okay. Please note that I have switched the convention here. It doesn’t matter, and then I have e-jk2 Sin theta2 x That should be equal to E3 Cos theta3 e-jk3Sin theta3 x. Okay. And because these equations have to be valid for all x, the only way this can happen is that these individual phase factors are all equal to each other. Okay So these individual phase factors being equal to each other means that I have k1 Sin theta1 x should be equal to k2 Sin theta2 x, which should be equal to k3 Sin theta3 times x Now, obviously, in this expression, k2 is given by omega square root º 0 e p s i l o n r 1 , w h i c h i s a c t u a l l y e q u a l t o k 1 m a g n i t u d e . W h y ? B e c a u s e b o t h i n c i d e n t a n d r e f l e c t e d w a v e s a r e i n m e d i u m o n e a n d m e d i u m o n e i s c h a r a c t e r i s e d b y p e r m i t t i v i t y e p s i l o n r 1 . O k a y , a n d t h a t i m m e d i a t e l y i m p l i e s t h a t S i n t h e t a 1 m u s t b e e q u a l t o S i n t h e ta2 And what is the limits on theta1 that we can have? Well, this theta1 can be zero which corresponds to normal incidence and all the way to theta1 equals pi by 2 in which case the wave will be riding along the plane of interface. It would be riding along the plane of interface with the electric feel appropriately directed. Okay So this kind of a wave, right, where the electric field is along the ground kind of a thing and this is riding along the parallel, you know, interface plane is called as grazing angle incidence, okay, whereas this wave was called as normal incidence; this is oblique incidence; this is called as grazing incidence The wave is kind of gliding or grazing the surface and electric field line actually lies in that particular plane. Okay. Electric field is parallel to this ground. Okay The other way would, of course, be that electric field is perpendicular. The magnetic field would be gliding along this surface, right? So that would correspond to transverse electric or vertical polarisation. This would correspond to parallel polarisation. Okay So with that in mind, if you look at the equations, sorry, theta1 can go from 0 to 90∞. So if you sketch Sin theta1, it would look something like this up to 90∞. So this is 90∞ done So if you have two Sin functions equal to each other over 0 to 90∞ interval, the only way that can happen is when theta2 is equal to theta1. Okay. This is, in fact, so called Snellís Law of reflection In fact, what we have seen is there is nothing like a law. It is a simple, you known, consequence of boundary conditions, right? So all these laws that we have learnt so far are nothing but consequences of boundary conditions. Okay So that is the boundary condition The second equation that you have k1 Sin theta1 = k3 Sin theta3, right, is also Snell’s law, but this equation tells you that for the same frequency omega, this would be square root epsilon r1 Sin theta1 = square root epsilon r2 Sin theta3. Okay. But because epsilon r1, epsilon r2 square roots are nothing but refractive index, this is n1 Sin theta1 = n2 Sin theta3 Okay. So this is another of Snell’s Law. This is called as Snell’s law of refraction Okay. So we have recovered the Snell’s law as a consequence of boundary conditions, but we are not done yet, right? What we have done so far is to simply equate the tangential electric field and get an equation in this particular manner. So we will recollect, write the equation for our use now. So we have E1 Cos theta1 + E2 Cos theta2 where theta2 and

theta1 are actually equal to each other is actually given by E3 Cos theta3, right? So this is one equation that you need to keep in mind Now we need to supplement this equation with the magnetic field equation. How would the magnetic field equation be? Well, you go back to this picture here. You assume that this is H1 magnetic field, and this would be say H2 magnetic field, and then you will have H3 magnetic field. In all these cases, the magnetic fields are in the perpendicular direction except that for the reflected field, we will make the magnetic field go in the opposite direction so that E x H would be propagating along -z direction E x H, if you take H to be upwards in this manner, E x H would be propagating in this direction. E x H in the second medium would also be propagating in the given k3 direction If you reverse the direction of H2 to downwards, E x H would actually propagate in the k2 direction Okay So with that, I can write down the expression for H as since H is transverse or rather tangential everywhere, it is actually along the plane of interface, I will have H1 – H2 = H3. Okay So I will have this particular equation, but H1 is basically E1 by eta1. This would be E2 by eta2. This would be E3 by eta3 And I need to, you know, group the terms together, solve again for E3/E1, which we will call as the reflection coefficient. Okay, and then you will have E3/E1 is called as the transmission coefficient because this tells you the ratio of the electric field that has been transmitted to the electric field that has been incident E2/E1 we will call as reflection coefficient, and we will get the expressions for these two. Okay Since I’m running out of time, I will stop here, and we will, you know, start from these expressions, derive the transmission and electric, transmission coefficient