 Welcome to the 21st lecture in the course Engineering Electromagnetics. In this lecture we continue with our discussion on reflection and refraction of waves and the topic for discussion today is what is called the plane slab. You would recall that so far what we have discussed is the case of a single interface on either side of which we had infinite media. But if let us say the medium on which the plane wave is incident is of a finite extent then one can easily visualize that the wave is going to encounter two interfaces instead of one. And therefore, the interaction of the wave with the interfaces is going to be more complex and such a case is known of the case of a plane slab. The slab could be made of any material in general, but as we will see to derive some special results we will consider that the slab is made of perfect dielectric material. The slab is represented in the following manner. We consider that this is an interface between say medium 1 and medium 2 and medium 2 is in the form of a slab. So, that this is an interface between medium 2 and medium 3 and medium 2 extends over a width which may be D. And therefore, this is the thickness D is the thickness of the plane slab in medium 2. We will consider the case of normal incidents. The case of oblique incidents can also be considered, but is going to be more complicated. So, we consider the case of normal incidents of a uniform plane wave on such a slab from medium 1. Let us say that the incident wave represented by say E i is incident like this and we are showing it incident in this manner for the sake of clarity. Otherwise, the wave is incident at a normal angle. Now, part of the wave is going to be reflected back at this interface which we may call are 1 2. Please take note of the convention or the notation. 1 2 represents the reflection at the interface between media 1 and 2 for a wave travelling from medium 1 to medium 2. While part of the wave is going to be reflected back part of the wave is going to be transmitted into medium 2. Let us say it is transmitted in this manner and what is transmitted may be called as E t 1 2 following our convention. So, that when this transmitted wave travels to the second interface, it will have got changed in its phase by an amount equivalent to the phase shift constant of medium 2 and the thickness or the distance that is travelled in medium 2. Let the phase shift constant medium 2 be k 2. So, that by the time it reaches the other interface it becomes E t 1 2 E to the power minus j k 2 times d. Now, at this interface again part of the wave is going to be transmitted which may be called E t and following our convention E t 2 3 and part of the wave is going to be reflected back. Let us say like this where this may be called E r 2 3. So, what reaches here is therefore, E r 2 3 E to the power minus j k 2 d where once again part of it is going to be transmitted through becoming E t 2 1 following our convention and part of the wave is going to be reflected back in this manner and what is reflected back may be called E r 2 1 which when which upon reaching here becomes E r 2 1 E to the power minus j k 2 d and part of it is transmitted through becoming E t 2 3 prime and part of it is reflected back becoming E r 2 3 prime and becoming E r 2 3 prime E to the power minus j k 2 d and this is E r 2 1 prime etcetera. This becomes E t 2 1 prime and this series of reflections and transmissions is going to continue and as I mentioned it is only for the sake of clarity that the wave parts are being shown inclined to the interface plates otherwise all these reflections transmissions and the wave incidence are normal to these interfaces. See the convention that we have been following for the phase shift is always that in the direction of travel the phase is going to decrease. So, we are being consistent in following that convention otherwise you have rightly notice that the direction of propagation of this wave is opposite to that of the other wave but the wave you put down the phase shift it is alright. Now since the various reflections and transmissions are going to be governed by the reflection coefficient and the transmission coefficient at the various interfaces let us put down these coefficients. We will have first let us say r 1 2 using the notation that r 1 2 is the reflection coefficient for a wave travelling from medium 1 into medium 2 which will be equal to eta 2 minus eta 1 upon eta 2 plus eta 1 assuming that the eta's are the intrinsic impedances of the two media. Similarly, t 1 2 as we know is going to be 1 plus r 1 2 on the other hand what will be r 2 1 reflection coefficient for a wave incident on the interface between media 2 and 1 travelling from medium 2 that will be eta 1 minus eta 2 upon eta 1 plus eta 2. So, that it is minus r 1 2 and accordingly t 2 1 the transmission in a similar situation is going to be 1 plus r 2 1. So, that it is 1 minus r 1 2 we are also going to require r 2 3 which is similarly eta 3 minus eta 2 upon eta 3 plus eta 2 and t 2 3 which is 1 plus r 2 3. So, armed with this picture and these coefficients reflection and transmission coefficients we can now consider what is the overall reflection back into medium 1 and what is the overall transmission through the plate slab. We will have to consider these various contributions for reflection let us say and the various contributions for transmission, but if you proceed systematically it will be possible to come to a simple result. Let us consider what is reflected back in medium 1. The first term in the series of various reflections that are going to be added to find out the overall reflection is going to be E r 1 2 which is going to be r 1 2 times E i. The second term is going to be E t 2 1 which is going to be t 2 1 into E r 2 3 e to the power minus j k 2 d and then E r 2 3 can be traced back to E i through a reflection here and transmission here and two phase shifts with magnitude k 2 d. So, on this basis E t 2 1 can be written as t 2 1 times r 2 3 times t 1 2 and e to the power minus 2 j k 2 d times E i this becomes our let us say equation 1 this is equation 2. Next let us put down E r 2 1 in terms of this we will be able to get E t 2 1 prime and therefore, let us put down E r 2 1 which is going to be r 2 1 and then E r 2 3 e to the power minus j k 2 d and therefore, in the above expression if we interchange t 2 1 and r 2 1 we should get E r 2 1 and doing exactly that we get this as r 2 1 2 3 t 1 2 e to the power minus 2 j k 2 d into E i which may be equation 3. In terms of E r 2 1 now we can get E t 2 1 prime E t 2 1 prime is going to be t 2 1 E r 2 3 prime e to the power minus j k 2 d and tracing the series of reflections and transmissions one can make out that E t 2 1 prime can be written in two parts as t 2 1 r 2 3 times t 1 2 e to the power minus 2 j k 2 d this becomes the first part in the expression for E t 2 1 prime and the second part becomes r 2 1 r 2 3 e to the power minus 2 j k 2 d times of course, E i which may be equation 4. Similarly, one can put down E r 2 1 prime which is going to be r 2 1 r 2 3 t 1 2 e to the power minus 2 j k 2 d times r 2 1 r 2 3 e to the power minus 2 j k 2 d into E i which is equation 5. If you have any questions we can tackle those now. It is the product of this term and this term. Let us say we are talking about E t 2 1 prime E t 2 1 prime is this and we are writing it in terms of E i. So, you can start at this point t 1 2 then r 2 3 then r 2 1 and then r 2 3 and then t 2 1. So, this will be the coefficients and then the travels that we made 1 2 3 4 travels that we made phase shifts corresponding to those. So, this way it is possible to check the final result that we have reached. It will come you just check t 2 1 is here in any case. This is the final term that will come final coefficient. Sir, the phase shift because of reflection there will be some phase shift. No, whatever phase shifts are there because of reflection and transmission those are taken into account in the coefficients. So, we do not need to talk about those phase shifts. We are multiplying by the coefficients. If they are real there is no phase shift. If they are complex they will contribute their phase shift appropriately. So, now this way if we consider E t 2 1 and E t 2 1 prime it is possible to see a pattern. There will be a common term and there will be a common ratio of these various terms which constitute an infinite series. And therefore, we can say that E r which is going to be E r 1 2 plus E t 2 1 plus E t 2 1 prime etcetera. So, that E r by E i will be equal to E t 2 1 prime e i which is the reflection coefficient that we are interested in. We may call it rho or small r. Rho is the symbol we have been using earlier and small r is what we have used for the other component coefficients in this case. That is only a matter of notation. So, E r by E i is going to be equal to as one can see from there r 1 2 plus A plus A s squared etcetera. Where A is the first term of this series with the value t 2 1 r 2 3 t 1 2 e to the power minus 2 j k 2 d and s is the common ratio with the value r 2 1 r 2 3 into e to the power minus 2 j k 2 d. So, that the summation is quite straight forward and it is going to be r 1 2 plus A upon 1 minus s. So, that it is t 2 1 r 2 3 t 1 2 e to the power minus 2 j k 2 d in the numerator and in the denominator we have 1 minus r 2 1 r 2 3 e to the power minus 2 j k 2 d. Next, we are going to be looking at the summation we write the transmission coefficients in terms of the reflection coefficients just to reduce the number of variables or the quantities which need to be calculated in this expression. So, that this becomes r 1 2 and then t 2 1 is 1 minus r 1 2 t 1 2 is 1 plus r 1 2. So, that we will have 1 minus r 1 2 square into r 2 3 into exponential of minus 2 j k 2 d in the numerator and in the denominator we will have also recognizing that r 2 1 is minus r 1 2 it will be 1 plus r 1 2 r 2 3 exponential minus 2 j k 2 d. Using the space on the left hand side this can be simplified and we get the expression for rho as r 2 3 times the exponent and then r 1 2 in the numerator r 1 2 plus r 2 3 e to the power minus 2 j k 2 d divided by the same numerator denominator that is 1 plus r 1 2 r 2 3 e to the power minus 2 j k 2 d which expression is quite compact even though we have added an infinite series. To interpret this expression and to see how it could be utilized in practice let us simplify it further and let us now make the assumption that the media that we are dealing with are perfect dielectrics. What will be the simplification caused by this? If the media are perfect dielectrics then these various coefficients are completely real and therefore, we can now consider the magnitude of rho or magnitude of rho square. Since, these coefficients are all real now we need to worry about only the exponential term and we get r 1 2 whole squared plus r 2 3 whole squared plus 2 r 1 2 r 2 3 cosine of 2 k to d in the numerator considering that magnitude squared is the square of the real part plus the square of the imaginary part. And similarly in the denominator we get 1 plus r 1 2 squared r 2 3 squared plus 2 r 1 2 r 2 3 cosine of 2 k to d which is the expression for the magnitude of the square of the reflection coefficient. And now one can derive or obtain a very useful application of this expression. We pose the question that for such a plane slab for such a situation involving three media is it possible to eliminate reflections completely that is is it possible to have magnitude rho squared or rho equal to 0. And for that purpose we will just put the we will try to see when the numerator becomes equal to 0. And it is possible to make the numerator 0 if we have r 1 2 equal to r 2 3 and also if we have cosine of 2 k to d equal to minus 1. Both conditions are required to make the reflection coefficient numerator equal to 0. Let us consider the implication of these conditions. The expressions for r 1 2 and r 2 3 are here and therefore, we just write down eta 2 minus eta 1 upon eta 2 plus eta 1 should be equal to eta 3 minus eta 2 upon eta 3 plus eta 2. And since we have already made the assumption that we are dealing with perfect electrics eta can be replaced by the corresponding epsilon or the permittivities square root of the permittivities. And therefore, this becomes root epsilon 1 minus root epsilon 2 upon eta 2 plus eta root epsilon 1 plus root epsilon 2. And here we have root epsilon 2 minus root epsilon 3 upon the summation of the same terms which is the simple result that epsilon 1 upon epsilon 2 is equal to epsilon 2 upon epsilon 3 or epsilon 2 is equal to the geometric mean of epsilon 1 and epsilon 3. We have obtained this kind of result in a different context earlier. In the context of a quieter wave matching section on a transmission line and here also you see that somewhat similar result is coming. So, this correspondence between the plane wave phenomena and the transmission line phenomena we shall draw upon we shall make more formal a little later. But, right now let us just notice the similarity. So, if the permittivity of medium 2 is related to the permittivities of the medium on the either side then this is one of the conditions for eliminating reflections back into medium 1. What is the implication of the second condition? This amounts to saying that twice k 2 d is equal to an odd multiple of pi radians. So, that it is 2 m plus 1 into pi yielding a condition on d that d is equal to 2 m plus 1 into pi by 2 k 2, where what is k 2 d 2 k 2 is the phase shift constant in medium 2. So, that it is 2 pi by lambda 2 and using this result it becomes 2 m plus 1 by 4 into lambda 2, where lambda 2 is the wave length in medium 2 and can be further related to the wave length in free space as lambda naught by square root of the dielectric constant of medium 2 for the simple case that we are considering that the media are perfect dielectrics. So, that this is a standard relation for the wave length in a certain medium for a uniform plane wave to the wave length in free space in terms of the square root of the dielectric constant. Not only for perfect dielectric constant. Well, wherever the velocity can be written as the 1 by square root of mu epsilon there this result will hold. Therefore, just to be on the safe side we are saying that for perfect dielectrics this is the wave length. k 2 is serving the same role as beta for this medium. So, that it is omega square root of mu 2 epsilon 2 for the perfect dielectric that we are considering it to be. That is lambda naught by epsilon 1 2 epsilon r 2 relative permittivity of the second medium lambda naught is the wave length in free space for the frequency of the incident phase. The result that we have derived here says that in such a situation the reflections can be completely avoided provided the permittivities are related like this. And the second medium has a thickness which is an odd multiple of quarter wave length where the wave length is the wave length in the medium 2 which again is a result very similar to what we obtained for the quarter wave matching section. And as we mentioned there this concept is utilized very frequently in eliminating reflections from lenses and other optical items. And this is the basis of what is called the anti reflection coating. Although as we consider this more deeply it becomes clear that the anti reflection coating will have 0 reflection only at the design wave length. Only where the thickness of the coating satisfies this relation that it is it is an odd multiple of quarter wave length where the wave length is the wave length in that medium. Apart from acting as an anti reflection coating in the context of optical elements this concept can be utilized for reducing reflections from aircraft and missiles also. So, that they become less easily detectable by the enemy radars used in this context that is for reducing the reflections from the point of view of radar takes us into what is called the radar absorbing material field. A ram is a radar absorbing material and this concept applied to that will imply that if we have a suitable coating on the surface of the aircraft or on the surface of the missile satisfying these conditions it will be possible to also become obvious as we pondered about it. First of all the reflection will be 0 only at particular wave lengths or only at particular frequencies. And secondly the results have been derived for the case of normal incidence and when these conditions are not maintained obviously the coating is going to be some reflection. And therefore, I am guessing a little bit here what may be more useful from the point of view of radar absorbing material which acts over a broader range of frequencies is perhaps an arrangement which would be like this. So, we have a missile and we provide a coating, a coating is hand insaturated for the sake of the similarity which has a mu and epsilon such that their ratio is equal to the intrinsic impedance of the three states. So, that eta naught is equal to square root of the mu by epsilon or if the medium is conducting in the more general case it is j omega mu upon sigma plus j omega epsilon. In which case reflections on this interface can be avoided over a very broad range of frequencies further if the medium is made lossy. So, these mu and epsilon down the r of the coating. If you like you could put subscript c to show very clearly that these are the properties of the medium which is being used for coating purposes. Now, this only ensures that there is no reflection at this interface, but part of the way but the way that enters here will be reflected back and will be available for the reflection purposes. So, this does not serve the purpose. So, if the medium is lossy and if the medium is such that it provides sufficient loss. So, that the way which is travelled this part in this medium is sufficiently reduced in amplitude. Then the reflections can be considerably avoided. So, development of this kind of data and all the materials which are effective for a broad range of frequencies is the of a great strategic importance from the point of view of difference. And this active area research. Now, let us proceed further and consider once again the expression for the reflection coefficient that we have derived earlier magnitude of those squared. And this time we write the reflection coefficient magnitude squared for another special case where let us say medium 1 is the same as medium 3. What is the effect on these coefficients for this? We will have R 2 3 equal to 1 2 for this case. You just replace eta 3 by eta 1 and the result is obvious. So, for this conditions the magnitude rho squared turns out to be a frequently figured expression twice R 1 2 squared and then we have 1 minus cosine of 2 k to d divided by 1 plus 1 plus 1 minus R 1 2 raise to the power 4 minus 2 R 1 2 squared cosine of 2 k to d which can be simplified by expressing the cosine in terms of half the angle. And it simplifies to be 4 R 1 2 4 squared and then the sine of k to d divided by the same. Thank you, sine squared 1 plus R 1 2 to the power 4 minus 2 R 1 2 squared cosine of 2 k to d. And in this case that is when media on either side of the plane slump are identical. The reflection coefficient is going to be 0 when k to d is going to be a multiple of pi ray beams that is k to d equal to n pi f equal to n integer. So, that d is the point is equal to n times pi by d 2 and as before since k 2 is 2 pi by lambda 2. What we get here is n times lambda 2 by 2 with lambda 2 once again be equal to lambda naught square root of r. And we find that if the thickness of the second medium for this condition is a multiple of half paper half lambda 2 then also there are no reflections back into medium 1. This kind of a slab plane slab satisfying this condition is called a half wave plane or a half wave plane plane. And what we saw earlier which could serve as an anti reflection protein is called a quarter wave plane. What could be the applications of this observation? What is the observation? The observation is that if we have to if the middle medium on the either side of a plane slab and the plane slab is multiple of half wave plane thickness then there are no reflections either for a wave incident from medium 3 or for a wave incident from medium 1. So, this observation is utilized in the making what are called ray domes. A ray dome can be considered to be a derived from a combination of two words. Let us say dome for a radar and it provides environmental protection to the antenna, antenna of a radar or of a communication system. You know that the antenna is exposed to the severities of the weather and therefore its performance can degrade with course of time or let us say when rain or snow fall on the antenna surface. And these problems can be avoided by putting what is called a ray dome. And let us consider a usual dish antenna which would have seen in many places let us say it looks like this. It is actually a parabolic reflector antenna then it has a feature which is called a baffle which we can discuss later. And now if you put a window here which at the operating frequency or which at the operating frequency is a half wave plane or multiple of a half wave plane then it will be completely transparent through the signals that are transmitted or through the signals that are received providing at the same time environmental protection to this antenna. The rados can take many different shapes depending on the application or depending on the requirements. For example, if the antenna is required to be rotated then one could put a spherical ray dome covering the entire antenna or else to reduce cost it may be there is a window in this manner. And sure you can recognize this shape and you have seen this on top of many buildings in detail. So, the purpose of this dielectric window is to provide environmental protection and to remain transparent at the same time. So, the artwork that we had done in the dielectric reflection coefficient had these two obviously very interesting and important applications. One is in providing anti reflection coating and the other is in providing protection to antennas in communication systems and radars. Now, we can proceed further and consider the total field that is transmitted through the laser. In this case we proceed in a similar manner and consider the addition of ray types which are t equal to t to 3 plus e t to 3 prime plus e t to 3 double prime plus e t to 3 double prime etcetera which again is an infinite series. And the expressions for these various components in the series various terms in the series can be worked out in a very similar manner as we did for finding out the reflection coefficient. And on that basis when we try out the transmission coefficient which is going to be e t y in this case also it turns out to be a dielectric progression with terms reading as a 1 plus a 1 s plus a 1 s square etcetera. Where a 1 is t 1 to t 2 3 e to the prime power minus j k 2 t and s remains the same value as earlier that is r 2 1 r 2 3 e to the power minus 2 j k 2 t. So, that the transmission coefficient is a 1 upon 1 minus s that is t 1 to t 2 3 e to the power minus j k 2 t t 2 minus j k 2 t a 2 single without the factor of 3. That can be made out from the first term here there is a single phase shift in the first term. So, that is what will come out as a common factor and then 1 minus s so that is 1 minus r 2 1 r 2 3 e to the power minus 2 j k 2 t. Making substitutions as before expressing the transmission coefficients in terms of the reflected coefficients one obtains that some of the simpler expression for the transmission coefficient means 1 plus r 1 2 into 1 plus r 2 3 e to the power minus j k 2 t upon 1 plus r 1 2 times r 2 3 e to the power minus 2 j more or less complete our discussion on the feature called slag. Now, here we have derived these results for the reflection coefficient and the transmission coefficient considering a series of reflections and transmissions which provides a somewhat clear inside into what is actually happening inside the plane slab where we have interactions of the wave with two interfaces. However, there are alternative methods available which are less tedious, but they do not incorporate this kind of clear understanding. Before we stop let me pose a question. Supposing we are required to find out the total power transmitted through such a slab and what are we going to be required to be 2. One straight forward method as the answer has come is to calculate the transmission coefficient and hence calculate the pointing vector, but if we already have calculated the reflection coefficient then it is not really necessary. We just calculate what is the power reflected back and if you assume that the plane slab is non-absorbing, it is not lossy then whatever is not reflected back is going to be transmitted through the plane slab. So, this is where we like to stop unless we have some questions. Sir, this is 1 minus r mu 2. No. Sir, yes. Just write out. Sir, after a different number of reflections, the entire layer has been detected on pressure proof. So, e t plus e r should be e. Okay, it is fine if we can discuss that now and then we can go ahead. Let us stop here.