 Welcome to the 19th lecture in the course Engineering Electromagnetics. We initiated our discussion on the topic of reflection and refraction of plane waves. We continue with this discussion today. The first topic that we take up today will be that of perfect conductor with oblique incidence of a plane wave on it followed by the case of perfect dielectric with the plane wave impinging on it at an oblique incidence. We start with the first topic that is the perfect conductor oblique incidence. You would recall that in the last lecture we considered the same case for one of the two possibilities and the possibility that we considered in the last lecture was the case of perpendicular polarization that is the case in which the electric field in the incident plane wave is polarized such that it is normal to the plane of incidence and we considered the overall behavior of the total electric field and the magnetic field. We considered the parallel case for that situation today and we considered the case of parallel polarization and we still consider the case of oblique incidence on a perfect conductor. We had reached up to a certain point in the last lecture and we will utilize that information today. We consider that the incident plane wave is incident at an angle theta with respect to the normal and as we saw last time the angle of reflection is equal to the angle of incidence and therefore we have the reflected wave also travelling at an angle theta with respect to the z axis. For the case of parallel polarization the electric field in the incident wave is in the plane of incidence that is the plane constituted by the incident ray and the normal at the point at which the incident ray hits the surface and since the fields electric field and the magnetic field should be perpendicular to the direction of propagation we put down these orientations for the incident electric field and the magnetic field. Considering the incident electric field to be consisting of two components a y directed component and a z directed component we were able to make out that the reflected electric field should have this kind of orientation and by considering the power flow we were able to make out that the magnitudes of the field vectors in the reflected wave should be the same as that in the incident wave. This is the point from which we can start today and we put down the expressions for the incident magnetic field and the reflected magnetic field in the following manner. We write h i equal to let us say h i e to the power minus j beta y sin theta minus z cos theta and what is the orientation of the incident magnetic field it is oriented in the x direction. Similarly for the reflected magnetic field using the information that the magnitudes of the field vectors are the same in the incident wave and the reflected wave we can continue to write the magnitude as h i and then e to the power minus j beta y sin theta plus z cos theta x cap taking into account the change in the direction of propagation particularly with respect to the z direction. And therefore the total magnetic field which is going to be a superposition of the incident and the reflected wave magnetic fields comes out to be twice h i e to the power minus j beta y sin theta and cosine of beta z cos theta oriented in the x direction which could be written introducing new symbols as twice h i e to the power minus j beta y times y and cosine of beta z z x cap and now we are in a position to interpret the behavior of the total magnetic field in the region z greater than or equal to 0 and we see that in the y direction the behavior is that of a propagating wave with the phase shift constant beta y in the y direction being equal to beta times sin theta whereas in the z direction the behavior is that of a standing wave it does not have the characteristics of a travelling wave it has the characteristics of a standing wave with the corresponding phase shift constant beta z being equal to beta cos theta and the total magnetic field being maximum at z equal to 0 and at periodic distances which are separated by lambda z by 2 where lambda z is equal to 2 pi by beta z this is the wave length in the z direction and the separation between the corresponding points in the z direction is lambda z by 2 corresponding point means the successive nodes and anti nodes or the adjacent minima and maxima they are separated by the distance lambda z by 2 along the z direction the magnetic field is tangential to this perfect conductor and it has a maximum value at the perfect conductor this result is going to be useful to us somewhere later on alright so that is as far as the behavior of the magnetic field is concerned what about the behavior of the total electric field and it turns out that we have to consider the electric field in terms of its two components separately because their behavior is slightly different in fact if you recall our discussion while E z components are involved in the propagation in the propagation of power in the y direction which remains more or less unaffected by this interface E y components are involved in propagation in the positive z or negative z directions therefore there is a difference in the behavior of these two components also E y components are tangential to the interface and they are restricted to have a certain value that is 0 value at the perfect conductor so these two components behave slightly differently the difference should come out as we consider their total behavior we write E i by h i equal to eta the intrinsic impedance of the media now since E z i E z i is this and this angle is theta therefore E z i is E i sin theta and therefore we can write E z i by h i equal to eta sin theta. I hope you can see the strategy since we have got the h i and h r written down we will try to obtain the expressions for E z and E y components in terms of the magnetic field components alright similarly what is E z r E z r should be related to h r similarly as eta sin theta as we have just argued out the z component of the electric field is involved in propagation along the y direction wave propagation and the corresponding power flow in the y direction which direction remains unaltered. And if you consider the rule that we had stated that the sin here depends upon whether the indices the subscripts of the field components appear in the cyclic order with respect to the direction of propagation or not so that rule is satisfied z x y so the cyclic order is maintained and we do not need to change the sin therefore the total z directed electric field E z t which will be E z i plus E z r it simply becomes eta times sin theta into h i plus h r which expressions we have already put down and therefore we can readily use those and it becomes eta sin theta times the total magnetic field so that it is twice h i E to the power minus j beta y y cosine of beta z z of course the direction is the z direction and we find that the behavior of the z directed electric field component in the overall wave constituted by the superposition of the incident wave and the reflected wave is similar to the to that of the total magnetic field it constitutes a wave propagating in the y direction with the phase shift constant beta y beta y equal to beta sin theta and in the z direction it is a standing wave with phase shift constant beta z beta z equal to beta cos theta. Just like the total magnetic field this field component is also maximum at z equal to 0 and then the field values repeat at distances of lambda z by 2. This field component is normal to the perfect conductor perfectly conducting surface and now we can see that the tangential magnetic field and the normal magnetic field behave in a similar manner which relationship comes about through the Maxwell's equations alright. Let me consider the y directed field electric field component and for that we need to write the following relations first of all E y i by h i is equal to eta cos theta the y directed electric field and the x directed magnetic field these are propagating in the negative z direction. So, overall you see that the sin convention is followed however for the reflected wave y component E y r upon h r we need to write minus eta cos theta since it constitutes a change in the direction of propagation and therefore the total electric field E y t can be written as eta cos theta and then we have h i minus h r the expressions are available to us the overall result can be written down fairly easily it will it should be eta cos theta and then twice j h i e to the power minus j beta y times y and then we should have sin of beta z z which can be verified without much difficulty and we see that the behavior of the y component of the total electric field is different from the behavior of the z component of the electric field along the z direction along the y direction this also constitutes a wave propagating with the phase shift constant beta y in the z direction remains a standing wave, but now at z equal to 0 it has a 0 value and at a distance lambda z by 4 from the perfectly conducting surface it will have a maximum value and the field values will repeat themselves at similar distances that is lambda z by 2. And therefore we see that there is a difference in the behavior of the normal field component and the tangential field component and this is consistent E y i and E y r as we are obtaining here with the considerations that we made earlier in making out the direction of the electric field in the reflected wave that also was obtained on the basis of the consideration that the tangential component of the electric field should be 0. So this is the kind of considerations that we make when we deal with a perfect conductor and in the general case of oblique incidents. In general we will see that whenever we are dealing with oblique incidents we will need to consider these two cases separately that is the case of perpendicular polarization and parallel polarization. We can take up the next topic listed for today now which is the case of oblique incidents on a perfect on a perfect dielectric. Now as far as the restriction that is implied by the perfect dielectric is concerned it is not very severe as we shall see it only amounts to a certain simplification in the expressions that we will obtain for the reflection coefficient and the transmission coefficient. And we are anticipating that as a uniform plane wave hits a boundary between two dielectric media part of it is going to be reflected and part of it is going to be transmitted. So apart from the simplification that we will have in the expressions the procedure that we are going to utilize is going to be applicable to general conducting media for the case of oblique incidents. And where in practice are we going to have this kind of situations that plane waves are impinging on different surfaces or different media many situations for example, we could consider the radio waves hitting the surface of the earth or the radio waves impinging on the surface of sea water or on lake water or at different layers in the atmosphere. For example, what happens when the radio waves interact with the ionosphere or else going to the optical frequencies what happens when the plane waves pass through different media. So these are situations which can be encountered in many cases and these results are going to be of general significance. First we try to make out some general characteristics about the case of oblique incidents on a perfect dielectric. Let us say this is the interface between two media which we are assuming as perfect so that we can write their constitutive parameters as mu naught and epsilon 1 and mu naught and epsilon 2. First of all we will need to make out the relations between the angles of reflection and angle of incidence and the angle of transmission and the angle of incidence for which we use a procedure that we utilized earlier for making out the angle of reflection equal to angle of incidence. And we consider that there is a ray representing the incident wave impinging on this interface in this manner. Let us call this 1i, ray 1 of the incident wave. Similarly let there be a second ray representing a part of the incident wave. Let the angle of incidence be theta 1. Let these rays be reflected in this manner. Let this be the ray corresponding to 2i so that this is the ray 2r and similarly let this be the ray 1r where this angle let us say is theta 3. We extend these normals that have been drawn into the second medium and consider that the transmitted rays are of this nature. So that this is 1t and this is 2t where the angle of transmission is theta 2. Now we consider the wave front corresponding to the incident wave, the reflected wave and the transmitted wave by dropping perpendicular in the following manner. Let this point be A, let this point be B, let this be C, let this be D and let this be E. So what is the situation like? The situation is as follows, a uniform plane wave with wave front AC impinges on the interface and due to the interaction with the interface a reflected wave with the corresponding wave front BE is generated and a transmitted wave with the corresponding wave front BD is generated. Whatever is the time that is taken for say the ray 2i in travelling from C to B should be the time taken by ray 1i in travelling from A to E and in travelling from A to D in the second medium that is how these wave fronts would be constituted and therefore we consider the time taken for let us say the incident wave in travelling from C to B. Let us call this time taken as delta t. So that C B by the velocity of wave in medium 1, let us call it V 1, this is equal to delta t which should be equal to A E by V 1 since these travels are made in the same medium. Now however C B and A E are related to A B in the following manner, this angle one can make out is theta 1 and this angle between the interface and the normal drawn for the wave front BE is going to be angle theta 3 whereas this angle is going to be theta 2 and therefore this implies that since C B is equal to A B sin theta 1 and A E is similarly equal to A B sin theta 3 we can say that the angle of incidence is equal to the angle of reflection. Let us consider the relationship between the angles of incidence and transmission and we make a similar consideration that is C B by V 1 should be equal to A D by V 2. A D is a distance travel in a different medium so in general the velocity is going to be different and let us call it V 2. So that A B sin theta 1 by V 1 is equal to A B sin theta 2 by V 2 recognizing that the velocities are given by 1 by square root of mu epsilon for perfect dielectrics. We can see that sin theta 1 upon sin theta 2 is going to be V 1 by V 2 in general and for the case of perfect dielectrics it will be square root of epsilon 2 by epsilon 1 the so called law of science or the Snell's law. So this is how the angle of transmission and the angle of incidence that the transmitted ray makes with the normal and the incident ray makes with the normal at the interface. This is going to be a basic information that we will utilize in considering the case of perfect dielectric oblique incidence and let us retain this law here in the form sin theta 1 upon sin theta 2 is equal to square root of epsilon 2 by epsilon 1 and as I mentioned the procedure is applicable to general conducting media but these relations we are writing for the special case of the perfect dielectrics where the expressions become somewhat simple. Next we consider the power flow normal to the interface earlier we have been using this kind of arguments but we never use these arguments for quantitative results and this time to show the possibility of such a thing we use the power flow arguments to incorporate these results into the expressions. We consider the power flow associated with the incident wave with the reflected wave and with the transmitted wave. The power flow that will involve the characteristics of the interface will be the power flow normal to the interface. The power flow tangential to the interface is not restricted by the interface characteristics therefore, we consider the power flow normal to the interface. Alternatively one could tie up this argument with the fact that it is the tangential field components which satisfy certain boundary conditions at such an interface and it is the tangential field components which will be involved in the power flow normal to the interface. So, when we are considering power flow normal to the interface actually in an alternative way we are using the fact that the tangential field components are continuous at the interface. So, the procedures are going to be equivalent it is just that they are done in different manners. So, let us consider the power flow normal to the interface and what is the pointing vector or the power density that is impinging normally on the interface corresponding to the incident wave. It will be related to E cross H of the incident wave and assuming that this is a linearly polarized wave. We can say that the power density that is impinging normally corresponding to the incident wave is E I squared by eta 1 into cos of theta 1. We do not worry about the average or the instantaneous because same consideration is going to be made for the other two waves also. This is the power density impinging normally which must be equal to the power flow away from the interface in the reflected wave and in the transmitted wave which is a basic relation which we shall use for treating the case of oblique incidents for perfect dielectrics. We are basically putting the condition that the power flow normal to the surface should be conserved. Now, we can divide through by let us say this quantity because as your experience would show we are basically interested in finding out the reflection coefficient E R by E I. So, we divide the entire equation by the quantity on the left hand side to get us 1 minus E R squared by E I squared which should be equal to eta 1 by eta 2 E t squared by E I squared and then cos theta 2 by cos theta 1. We call this equation 1. By the way as I have been trying to explain similar result would have been obtained if we had considered the continuity of tangential field components. To proceed further now we need to make a distinction between the two possibilities the case of perpendicular polarization and the case of parallel polarization as we needed for the case of oblique incidents on a perfect conductor. Let us make some space here and we draw the simplified diagram draw the normal consider that this is the incident wave this is the reflected wave this is the transmitted wave this is theta 2 these are both theta 1 and now we consider the case of perpendicular polarization. So, that the incident wave electric field may be put down like this and the corresponding incident wave magnetic field can be put down in this manner. So, that the cross product of E I and H I is in the direction of propagation of the incident wave similarly for the transmitted wave we can consider that this is the transmitted electric field vector with the corresponding transmitted magnetic field vector oriented in this manner normal to the transmitted wave direction and similarly here normally we put down this as the direction of the reflected wave electric field and of the reflected wave magnetic field a priori we are not in a position to say what will be the orientation of E R I mean perpendicular to the direct perpendicular to the plane of the board all right because that is what the direction will come out when we consider the boundary conditions, but otherwise whether it is coming out or going in that we cannot say that will depend upon the actual value of the reflection coefficient right. Now for this case considering the continuity of the tangential field components at the interface and considering that these are the field electric field vector magnitudes at the interface we can say that E I plus E R is equal to E T. So, that 1 plus E R by E I is equal to E T by E I which is the second relation which is of significance in this context basically from equation 2 we can substitute for E T by E I in equation 1 and eliminate E T by E I to obtain the value of E R by E I right. So, substituting from 2 in 1 we are going to have 1 minus E R squared by E I squared equal to now here also we can incorporate a small simplification considering that these are perfect dielectrics this would be the situation for general conducting media, but for perfect dielectrics it will simplify to epsilon 2 by epsilon 1 E T squared by E I squared cos of theta 2 by cos of theta 1. So, that here on the right hand side we get epsilon 2 by epsilon 1 and then for E T by E I we substitute from 2. So, that we have 1 plus E R by E I whole squared multiplied by cosine of theta 2 by cosine of theta 1 which equation involves only E R by E I. So, rest is a matter of manipulation we can factorize this and then 1 factor will cancel with this and therefore, we are going to get 1 plus E R by E I upon 1 minus E R by E I as square root of epsilon 1 cos theta 1 upon square root of epsilon 2 cos theta 2. We have divided both sides by 1 minus E R by E I which implies that E R by E I is going to be equal to square root of epsilon 1 cos theta 1 minus square root of epsilon 2 cos theta 2 upon the summation of both the terms that is square root of epsilon 1 cos theta 1 plus root epsilon 2 cos theta 2 which becomes the expression for the reflection coefficient for the case of perpendicular polarization. We could listed here for comparison E R by E I is equal to square root of epsilon 1 cos theta 1 minus root of epsilon 2 cos theta 2 in this form it is quite easy to remember also and summation of both the terms in the denominator. Now we can manipulate this final expression in different ways depending on our convenience. It is in terms of the angle of incidence, angle of transmission and the permittivities of the two perfect electric media that are involved. We can manipulate this expression in different ways. For example, we can eliminate epsilon 2 epsilon 1 by considering the fact that root epsilon 1 by epsilon 2 is equal to sin theta 2 by sin theta 1 and when that is done and substituted in this expression you find that it comes out in a particularly simple form as sin of theta 2 minus theta 1 upon sin of theta 2 plus theta 1. Now does this mean that the reflection coefficient has become independent of the permittivities of the two media? Not at all because theta 2 and theta 1 are related through the permittivities. In fact that is the relation we utilize to eliminate those two quantities. Similarly one could eliminate for example theta 2 because normally what would be given to us is two media an interface between two such media and the angle at which the incident wave is incident that is theta 1 would be known and epsilon 1 epsilon 2 would be known. So that can also be done easily. We have cos theta 2 equal to 1 minus sin square theta 2 whole square root which is 1 minus epsilon 1 by epsilon 2 sin theta 1 whole square. Which substitution can sin square theta 1? Thank you. Which substitution can also be made here? And we are going to get the reflection coefficient equal to root epsilon 1 cos theta 1 minus root epsilon 2 into cos theta 2 which is 1 minus epsilon 1 by epsilon 2 sin square theta 1 divided by a summation of two similar terms. Now one can manipulate this to read as this is equal to cos theta 1 by epsilon 2 by epsilon 2 minus epsilon 2 by epsilon 1 minus sin square theta 1 divided by same terms with a plus sin between them. So that it is root of epsilon 2 by epsilon 1 minus sin square theta 1. The verification is straight forward. So the same expression can be put in alternative forms depending on which one is more convenient for usage. And one can make out that this reflection coefficient is going to be real may be positive or negative depending on the relative values of epsilon 1 and epsilon 2. Once we know the reflection coefficient the transmission coefficient can be found out through the intermediate relations that we had written earlier. That should not be difficult. For example for this case for the case of perpendicular polarization we have E t by E i equal to E r by E i plus 1 a relation that we had written just a short while ago. That more or less completes the consideration for the case of perpendicular polarization. Next we can consider the corresponding case of parallel polarization and we need to make some space here. This is the interface between the two media. This is the incident wave, the reflected wave, the transmitted wave, the angles are as shown and the waves propagating in the directions of the arrow. And now we consider the case of parallel polarization that is the electric field is going to be in the plane of incidence or parallel to the plane of incidence. And therefore let us say that the incident wave electric field is normal to the direction of propagation and is oriented in this manner with the corresponding incident wave magnetic field like this. And normally the other field components can also be put down. Let us say this is h t and this is E t. The angle that E i makes with the horizontal is going to be theta 1 and similarly the angle that E t makes with the horizontal is going to be theta 2. This we are going to require shortly and similarly we say that let this be the orientation of the magnetic field in the reflected wave and what about the reflected wave electric field? We can draw it like this. So that this is E r and this is h r. And now we once again consider the continuity of the tendential electric field components at the interface. These angles we have already shown. This angle is also theta 2 and therefore we should have E i minus E r cos theta 1 equal to E t times cos theta 2 which becomes the corresponding starting relation for this case. And we can try to make out the value of E t by E i for subsequent substitution in equation 1. That gives us 1 minus E r by E i into cos theta 1 by cos theta 2 equal to E i minus E r by E i into cos theta 1 by cos theta equal to E t by E i which becomes the second equation for this case and can be used for substitution in equation 1. And therefore 2 and 1 combine to give us 1 minus E r squared by E i squared which should be equal to square root of epsilon 2 by epsilon 1 and E t by E i whole squared is in this case 1 minus E r by E i whole squared and we are left with cos theta 1 and cos theta 2. So that 1 plus E r by E i upon 1 minus E r by E i is now equal to root of epsilon 2 cos theta 1 divided by root of epsilon 1 cos theta 2. You can compare this with the previous situation and see that epsilon 1 and epsilon 2 have interchanged places. With that difference the entire procedure is going to be similar. Now we do not need equation 1 anymore and therefore we have E r by E i E i by E i whole squared. Equal to root of epsilon 2 cos theta 1 upon root of epsilon 1 cos theta minus root of epsilon 1 cos theta 2 plus root of epsilon 2 cos theta 1. The terms in the denominator have got interchanged but that is of little consequence. Like for the case of perpendicular polarization we can write alternative forms. We can write this expression in alternative forms. For example this one will come out to be tan of theta 1 minus theta 2 upon tan of theta 1 plus theta 2 and similarly the expression where theta 2 has been replaced in terms of theta 1 comes out as follows. It will be equal to epsilon 2 by epsilon 1 cos of theta 1 and then the other things remain as in the previous expression. That is this is epsilon 2 by epsilon 1 minus sin square theta 1 and in the denominator both terms are repeated with a change of sin epsilon 2 by epsilon 1 cos theta 1 plus epsilon 2 by epsilon 1 minus sin square theta 1 which becomes the corresponding expression for reflection coefficient for the case of parallel polarization. One small step is left. What about the transmission coefficient in this case? Can we still use this expression that we wrote for perpendicular polarization? We cannot. In fact E t by E i is given by the second equation. So in this case if we need to find out E t by E i the transmission coefficient it must be obtained from such an expression. So we see that depending on the case that we have whether it is the case of perpendicular polarization or parallel polarization the expressions can be slightly different. We stop this lecture here. In this lecture we considered the case of oblique incidence on a perfect conductor, the case of parallel polarization and then we went on to consider what happens when a uniform plane wave impinges on an interface between two perfect dielectric media and we saw what would be the expressions for the reflection coefficient and the transmission coefficient in such a case. Thank you.