 Welcome to the 18th lecture in the course Engineering Electromagnetics. Today we continue what we started towards the end of the last lecture, the topic listed in the course content reflection and refraction of plane waves. The topics that we have on the agenda today are reflection at the surface of a conducting medium and the reflection at a perfect conductor when the wave is incident obliquely. But before we take up any of these two topics, let us continue with a small part of the previous lecture and we consider the normal incidents of a uniform plane wave at a perfect conductor and placing the perfect conductor at x equal to 0 and considering that the incident electric field has an expression E i equal to E naught e to the power minus j beta x oriented in the y direction. The total electric field taking the reflected field also into account was found out as minus 2j E naught sin beta x so that it is 0 at x equal to 0 and at periodic distances from the interface. Similarly we found out the total magnetic field as 2 h naught cosine beta x oriented in the z direction and we said that these constitute a standing wave in front of the perfect conductor interface. Now what will be the power carried in this standing wave? That is the point of small interest and if we consider the pointing vector corresponding to these field components and we write the average pointing vector as half real of E cross h star we immediately find that since E t and h t are in phase quadrature they have no component which is in phase the result here is 0 although the medium that we have has a real intrinsic impedance or characteristic impedance. So this is another situation where the pointing vector comes out 0 when we consider the real part of E cross h star and it is a result of the fact that the total electric field vector and the total magnetic field vector have no component which is in phase they are so to say in phase quadrature. This will be the situation inside resonators and when we have this kind of perfect standing waves. Now we can go on to considering the topic that we listed for discussion today the first topic that is the reflection at the surface of a conducting medium. The difference that we have in this case from the previous case of perfect conductor is that we do not have a perfect conductor now we have in fact a general medium it is an interface between two general media which could be represented in the following manner. Let us say this is the interface between two media say medium one and medium two with constitutive parameters as mu one epsilon one and sigma one the permeability, permittivity and conductivity and similarly mu two epsilon two and sigma two and we consider that uniform plane wave is incident normally on the interface. So this is still a case of normal incidence. For this direction of propagation the electric and the magnetic field vectors must be in a plane which is normal to this direction we may represent these as follows. Let us say this is the electric field vector associated with the incident wave and let us say this is the magnetic field vector associated with the incident wave so that E cross H is in the direction of propagation. Now unlike the perfect conductor where there could be no power carried in a propagating wave or there could be no power lost because of the absorption in the medium in a general medium those restrictions are not there and therefore in general we will have part of the power of the incident wave transmitted through this interface and the part which is not transmitted should be reflected back. On this basis we can put down a transmitted wave with the corresponding electric field vector as E t and the magnetic field vector as H t and similarly for the reflected wave which is going to propagate in the opposite direction we say that let this be the electric field vector associated with the reflected wave E r and let this be the magnetic field vector associated with this way. Then corresponding to the wave propagation we will put down appropriate relations between these. These are nominal directions for the reflected field vectors and now we apply the boundary conditions at the interface boundary conditions on the tangential components of the feeds. For a general interface like this the tangential field components both electric and magnetic field components should be continuous and therefore we can write E i plus E r equal to E t and similarly H i plus H r equal to H t. However the field components within a particular wave say incident wave or reflected wave are not independent they are related through the intrinsic impedance of the medium and therefore we can write the relations E i by H i equal to E t 1 where E t 1 is the intrinsic impedance of medium 1. Similarly E t by H t should be E t 2 and as far as E r and H r are concerned they exist in the same medium that is medium 1 but this reflected wave is propagating in a direction which is opposite to that of the incident wave and that should be taken into account by writing this as minus E t 1. Now we can process these equations to find out what is the fraction of the incident electric field which is reflected back which will become the reflection coefficient. From 1 we are going to get let us say we divide by E i throughout so that we have 1 plus E r by E i equal to E t by E i. Let us say that is our equation 3 then from 2 we substitute for H in terms of E from these relations. We are going to have therefore eta 1 into E i minus E r which should be equal to it should be 1 upon eta 1 that is right and similarly here we are going to have E t by eta 2 from which by dividing through by E i and multiplying by eta 1 we can get 1 minus E r by E i which should be E t by E i into eta 1 by eta 2 which becomes our equation 4. Now both equations 3 and 4 contain E r by E i and E t by E i we can eliminate these 1 by 1 and obtain the values of these unknowns respectively. For example from 3 and 4 we can get E r by E i equal to eta 2 minus eta 1 upon eta 2 plus eta 1 considering that we are dividing 3 by 4 and then doing the standard manipulation component or dividend. The similarity with the transmission line is unmistakable. If we consider the reflection coefficient based on the magnetic field components one can see easily that this is going to be minus H r by H i and similarly E t by E i is going to be twice eta 2 by eta 2 plus eta 1 whereas H t by H i is going to be twice eta 1 upon eta 2 plus eta 1 which respectively become the reflection coefficients or the transmission coefficients depending on which field we are interested in and these are coming out in terms of the intrinsic impedances of the 2 media involved. For the case of normal incidence the relations are really quite rate forward. To understand these better let us take a particular example yes or no. We have nominally put these directions their values will come out appropriately from the reflection coefficients etc and that they constitute a wave propagating in the opposite direction that we have taken care of here. In fact this is something quite similar to what we did for the transmission line right. We take an example and we can do that on the projector. We consider to a particular media. Let us say medium 1 is air alright and let us say that medium 2 is copper. So that here we have mu 1 equal to mu naught epsilon 1 equal to epsilon naught and sigma 1 equal to 0. If we have written free space that would have been more accurate. Here we have once again mu 1 and epsilon 1 that of free space as for good conductors mu 2 is mu naught and epsilon 2 is epsilon naught. However sigma 2 is very large it is 5.8 into 10 to the power 7 mose per meter. The corresponding impedances can be calculated we will have eta 1 equal to mu naught by epsilon naught which value is 120 pi or 377 ohms. For eta 2 some calculations need to be done. What will be the expression for the intrinsic impedance? The general expression is j omega mu upon sigma plus j omega epsilon for conducting media in general which can be simplified as j omega mu upon sigma for good conductors. And therefore the intrinsic impedance for this medium for copper requires specification of a particular frequency before we can numerically calculate the intrinsic impedance. So let us say that omega corresponds to 1 megahertz okay. So 2 pi into 10 to the power 6 radians per seconds and for this value of omega eta 2 comes out to be 3.69 into 10 to the power minus 4 with the associated angle of 45 degrees. A very small value okay it is only very slightly different from a short circuit alright. So let us see what kind of results we get for the various parameters that is reflection coefficient, transmission coefficient etc. So for the reflection coefficient involving the electric fields E r by E i which we have written as eta 2 minus eta 1 upon eta 2 plus eta 1 substituting the values of eta 2 and eta 1. We get a value which is minus 0 point and then 5 nights 1, 2, 3, 4, 5, 8, 6 at a very small angle minus 0.4079 or simply 8 so many degrees. The reflection coefficient is practically minus 1 which is what we expect from a perfect short circuit or something which is very close to a short circuit. The reflection coefficient based on the magnetic field components is going to be similar simply the negative of this. We can also calculate what is transmitted through such a virtual short circuit and we get E t by E i which is twice eta 2 upon eta 2 plus eta 1 the calculations can be done and we get result which is 0.5012345196 at an angle of 45 degrees okay. So there is very little transmission quite as expected. h t by h i can also be calculated and that comes out to be 1.5986 at an angle of minus 0.40 and then very close to 4 degrees okay and therefore one can say that the copper shield is virtually a short circuit. Of course it is conceivable that if we increase the frequency sufficiently this behavior may change which is what happens at very high frequencies alright. With this example we have considered the reflection and transmission at the surface of a conducting medium and now we go on to consider reflection at a perfect conductor okay. We have already considered this reflection at a perfect conductor for normal incidents and now we consider the more general case of oblique incidents. We consider let us say this is the interface between one medium and the other medium and let us say the lower medium is the perfect conductor okay. This is the normal drawn at a point where a ray representing an incident uniform plane wave strikes this interface let us say at an angle theta. Now in this case we need to distinguish between two possibilities. What are these two possibilities? This is the direction in which the uniform plane wave is incident. The electric and the magnetic field vectors have to be in a plane which is normal to this direction of propagation utilizing the properties of uniform plane waves. One possibility is one of the many possibilities is that the incident electric field is in this direction normal to this direction okay. Let us say this is the it is normal to the plane of the board. Let us say if this is y and this is z then the incident electric field is oriented along the x direction alright. And the corresponding incident magnetic field intensity vector is in this direction consistent with the fact that the vector e cross h should have a direction which is the direction of propagation. This is one of the many possibilities this possibility can be identified or specified as the case of perpendicular polarization because the polarization of the incident wave is normal to the plane of incidence. What is the plane of incidence? Plane of incidence contains the incident ray and the normal drawn from the point at which the incident ray strikes the interface okay. So in this case we have perpendicular polarization. Other possibilities are also easy to see. For example the magnetic field vector could be oriented like this that is in the direction x and the electric field vector may be like this. This would be the case of parallel polarization because the polarization of the incident wave would be in the plane of incidence or would be parallel to the plane of incidence as we shall see there are differences in these two cases that arise okay. Now a question can come up that well we have a uniform plane wave incident in this direction why do we have to consider only these two specific cases there could be a whole lot of cases in between right. Now the simple answer to that is that any other general case can be considered as a superposition or as a combination of these two orthogonal polarizations. So one could consider the components of the incident wave with these two linear polarizations okay. And once we have understood the behavior for these two cases we will be able to analyze or understand any other general case alright. In fact that becomes the basis of some very interesting phenomena which we shall talk about later. So to start with we consider the case of perpendicular polarization okay. Now since we have once again come back to the perfect conductor the arguments that we invoke for normal incidence would again apply and therefore there is not going to be any power which is transmitted through the interface either as a propagating wave or as something which supplies power to power losses. And therefore whatever power is incident or is associated with this incident wave must remain above the interface. In fact one could go a little more further one could consider this incident wave also as composed of two parts okay. Some part of the power is falling normally on this interface and some part of the power is flowing parallel to the interface and physically what we expect is that the power flow parallel to the y direction parallel to the interface should not be affected. There is really no constraint on that and the power flow which is normal to the interface should be completely reflected back and as our experience is it should form a standing wave. Physically that is what we expect let us see how we can obtain those results proceeding more systematically. The first question that needs to be answered is that what is the direction of the reflected wave okay. Intuitively we feel that the angle of reflection should be the angle of incidence but there is a way out we can handle this a little more rigorous. For that purpose we make the following construction maybe we can make some space here. We consider the wave front associated with the incident wave and similarly the wave front associated with the reflected wave. And for that purpose let us say this is the interface and let this be one of the rays representing the incident wave and let this be the associated second rate both incident in this manner and let us say this angle is theta 2. The angle of incidence is theta 2 and as a result there is a reflected wave which is generated. Let us say which makes an angle theta 1 with the normal at that point. There will be a reflected ray corresponding to the second ray also which we may draw like this okay. Now let us draw the perpendicular I think we will have to redraw this there is a slight problem in this we have to consider more inclined waves so that we can show this better. This is angle theta 2 these two rays are incident in this manner and let us say this is angle theta 1 okay. We begin with the assumption that the angle of incidence and the angle of reflection are different. And now we consider the wave fronts associated with the incident wave and the reflected wave and let us say that these are drawn like this okay. So let us say this is point A and this is point D. A D represents the wave front associated with the incident wave on this wave front there is a uniform phase. Similarly there will be a wave front associated with the reflected wave let us call that BC okay where this angle is going to be theta 2 and this angle is going to be theta 1 alright. Now let us consider the distance AC and the distance DB or BD. This is the reflected wave wave front this is the incident wave wave front during the time the ray striking at point A travels to point C okay. In that time the other ray should travel from point D to point B that is how the wave fronts would be constituted like this okay. And therefore AC must be equal to BD because these are travels in the same media okay. So the travel is with the same velocity and therefore in the same time the distance travel must be the same and one can see that this is going to be AC is AB sin theta 1 and BD is AB sin theta 1. And therefore we get the simple relation that the angle of reflection is equal to the angle of incidence okay. So using this kind of arguments one can make out what will be the angle of reflection given the angle of incidence in such a situation or later on what would be the angle of refraction when we consider a general medium other than perfect conductor okay. And therefore we can now safely say that the angle of reflection here is the same as the angle of incidence. And now let us put down an expression for the incident electric field which is going to be EI equal to let us say some magnitude EI and then we will have e to the power minus j beta y sin theta okay. And minus z cos theta considering the direction in which the wave is propagating the direction cosines of this direction are sin theta and minus cos theta. And therefore multiplying the phase shift constant with that we get this expression for the incident electric field which is going to be in the x direction. Now we also want to put down the expression for the reflected wave electric field. And we are aware that at the interface say at y and z equal to 0 the total tangential electric field must be 0 this being a perfect conductor. And therefore EI and ER at the interface must completely cancel each other. And therefore the direction for ER that one may put down is this. Now this is slightly different from what we did last time. Now we are taking the direction into account in putting the electric field vector for the reflected wave. And once we have identified the direction of the reflected wave electric field to be consistent with the direction of propagation we must have the magnetic field vector in the reflected wave which is perpendicular to both of these. And E cross H should be in the direction of propagation and therefore we have the reflected wave magnetic field coming out like this. The magnitude of ER is to be the same as that of EI as we have just found out. And therefore HR magnitude must also be the same as that of HI using the power conservation arguments that we mentioned a little while ago. And therefore now we have ER equal to minus EI e to the power minus j beta the y direction remains unchanged consistent with our intuitive arguments it is y sin theta. But the z direction is now reversed so we put plus z cos theta here and this remains x cap. And now we can consider the behavior of the total electric field which will be a combination of the fields electric fields in the incident wave and the reflected wave. And we are going to have EI into e to the power minus j beta y sin theta. And then we have e to the power j beta z cos theta minus e to the power minus j beta z cos theta x cap which gives us 2 j EI e to the power minus j beta z cos theta y sin theta. And then sin of beta z cos theta overall direction remaining x cap from which one can make out the behavior of the overall electric field. It will be somewhat easier if we consider the corresponding time varying total electric field as against the phasor notation in which we have put down the total electric field. We have the time varying total electric field it will be twice EI sin of beta z cos theta. And what else do we have sin of omega t minus beta y sin theta. Which indicates that in the y direction we have a propagating wave with an argument which is t minus y by v that velocity we will identify shortly. And in the z direction it is a standing wave with the field value total field value which is 0 at z equal to 0. And as we have progressed up it is going to show periodic nulls and maxima nodes and anti nodes which are the characteristic of standing waves. We can introduce new symbols for briefer notation it is going to be twice EI sin of beta z z and sin of beta z cos theta. Omega t minus beta y y x cap where beta z is beta cos theta whereas beta y is beta sin theta. One can put down the corresponding wave lengths and the phase velocities lambda z would be 2 pi by beta z. So, that it is 2 pi by beta cos theta or in other words it is going to be lambda by cos theta where lambda is the wave length associated with free space. The corresponding phase velocity let us call it v z is going to be omega by beta z and one can see that that is going to be the velocity of the wave in free space divided by cos theta. It is going to be greater than the velocity of light in free space, but this is just the phase velocity. Energy or information is not going to be transmitted at this velocity. Correspondingly we have lambda y as lambda by sin theta. So, this is going to be the and v y as v by sin theta. The reason for these z directed and y directed velocities being greater than v can be understood by considering an example like this. And the distance swept along the y direction or along the z direction is greater than the distance that the wave travels in its direction of propagation and the time involved is the same. Therefore, these phase velocities come out greater than the velocity of light in free space. This kind of expression in the z direction indicates a standing wave with the periodic nodes and anti nodes and the distance between adjacent nodes is going to be lambda by 2 cos theta or lambda z by 2. Quite consistent with what we have done earlier in similar situations. And as we had argued out intuitively, the incident wave can be considered to be composed of two parts. One part incident normally which should be completely reflected back this being a perfect conductor forming a standing wave and other part travelling parallel to the interface and going unaffected by the interface. So, our final result is quite consistent with those intuitive arguments. Now, we are in a position to consider the second case corresponding to this, the case of parallel polarization. We make the following construction. These are the y and the z axis. The interface is coincident with the z equal to 0 plane like last time and this is the wave incident at an angle theta and this time the incident wave electric field is parallel to the plane of incidence and the corresponding magnetic field is like this. As far as the direction of the reflected wave is concerned, the angle of reflection that should from similar arguments come out the same and therefore, we can mark the reflected wave in this manner and similar intuitive arguments are going to apply to this case also. Now, of course, we expect the reflected wave fields to be in a plane normal to the direction of the propagation of the reflected wave. And from this point of view, one can consider some simple cases. To be able to put down more clearly the field orientations, let us consider that the incident electric field is composed of two parts. One is the y component, the other is the z component which we label as E z i E y i. At the corresponding point here, now E y i is a field component which is tangential to the interface. And when we apply the boundary condition that the total tangential electric field at the perfectly conducting interface must be 0. E y i must come out E y r must come out as negative of E y i. So, incorporating this direction in the reflected wave y component of the electric field, the field component is going to be like this. This becomes our E y r. Then as we have been arguing out the part of the wave, the component of the wave, notional component of the wave which is travelling in the y direction should remain unaffected by this interface, because this is not imposing any constraints on that. And therefore, the z component of the electric field in the reflected wave must remain unchanged. And this is how E z r must look like, so that we get the reflected wave total electric field which is oriented in this manner, normal to the direction of propagation with these components. Once the electric field has been obtained, then since h must be normal to the electric field and to the direction of propagation and E cross h must be in the direction of propagation, we can say that the reflected wave magnetic field should be like this. And now, one can combine the various field components and consider their behavior along similar lines which part we shall postpone till the next lecture. In today's lecture, we have considered the case of normal incidence at an interface between two general media. And we have considered the reflection coefficient and the transmission coefficient. And next we considered the case of oblique incidence at a perfect conductor. We saw that we need to consider two different cases. We have considered the case of perpendicular polarization today. And we shall consider the case of parallel polarization in the next lecture. Thank you.