 Welcome to the 20th lecture in the course Engineering Electromagnetics. Like the last few lectures, we continue with the phenomenon of reflection and refraction of waves and the topics that we plan to discuss today are the following. We discussed the phenomenon known as Brewster angle and then consider the phenomenon of total internal reflection. Considering the first topic that is the Brewster angle you would recall that for the two cases of perpendicular and parallel polarization we have written down the expressions for the reflection coefficient and the transmission coefficient considering an interface between two perfect dielectrics. You would recall that we consider the following situation. This is the interface and this is the normal to the interface and we consider that these two media are perfect dielectrics with permittivities epsilon 1 and epsilon 2 and a uniform plane wave represented by a ray let us say like this incident at an angle theta 1 with respect to the normal at the interface is partly in general is partly reflected back at the same angle and is partly transmitted into the second medium and let us say this is the way the transmitted wave looks like and this is the angle of refraction. Now considering the expression for the reflection coefficient for the case of the parallel polarization you would recall that we wrote E r by E i equal to epsilon 2 by epsilon 1 cos theta 1 minus whole square root of epsilon 2 by epsilon 1 minus sin square theta 1 divided by a denominator consisting of the same terms, but with the change of sin. So, that it is epsilon 2 by epsilon 1 cos theta 1 plus epsilon 2 by epsilon 1 minus sin square theta 1 and we said that in general part of the incident wave is reflected back into the same medium and part is transmitted. Now one can raise a question now is it possible for some angle of incidence that there is nothing which is reflected back and the transmission is complete that would be a very interesting possibility. So, let us see if that is indicated by this expression we just apply the condition that the reflection coefficient should be 0 and try to find out the corresponding angle of incidence if it exists. And from that point of view let us say that for E r by E i to be equal to 0 we should have epsilon 2 by epsilon 1 cos theta 1 equal to square root of epsilon 2 by epsilon 1 minus sin square theta 1 and we square up both sides and write cosine theta 1 as 1 minus sin square theta 1. So, that we get epsilon 2 by epsilon 1 whole square minus epsilon 2 by epsilon 1 whole square into sin square theta 1. Let us say that this happens when the angle of incidence theta 1 is equal to some angle say theta b. So, that we can change these 2 theta b this is going to happen if it does at some particular value of the angle therefore, we specialize the angle of incidence theta 1 to theta b which should be equal to epsilon 2 by epsilon 1 minus sin square theta b. Now, one can collect the terms and one can see that sin square theta b is going to be equal to epsilon 2 by epsilon 1 whole square minus epsilon 2 by epsilon 1 on one hand and 1 minus epsilon 2 by epsilon 1 whole square minus 1 on the other side which can be simplified further and you see that we are going to have the final result as epsilon 2 by epsilon 1 plus epsilon 2 which makes sense and we see that there is an angle of incidence equal to theta b where the reflection coefficient for this polarization can be 0. One can process it further. So, that let us say cos square theta b that is 1 minus sin square theta b is going to be epsilon 1 upon epsilon 1 plus epsilon 2 and therefore, one can say that the tan of this angle theta b is going to be epsilon 2 by epsilon 1 whole square root or this theta b is going to be equal to tan inverse of epsilon 2 by epsilon 1 which angle is known as Brewster angle, angle is theta 1 equal to theta b. What is going to happen at this angle of incidence? If the incident wave is parallel polarized that is it has an electric field vector which is in the plane of incidence then the wave is going to be completely transmitted. If there is no reflection then the entire wave is going to be completely transmitted which can have very interesting applications. But before we discuss the applications perhaps it could be a question what about the other polarization the perpendicular polarization. Let us look at that for a short while for the case of perpendicular polarization continuing with our simplifying assumptions of perfect dielectric media. We have E r by E i equal to cos theta 1 minus epsilon 2 by epsilon 1 minus sin square theta 1 divided by cos theta 1 plus epsilon 2 by epsilon 1 minus sin square theta 1. And now looking for a possible Brewster angle for this case we will apply the condition that the reflection coefficient is 0 which will give us the condition that cosine of theta 1 is equal to whole square root of epsilon 2 by epsilon 1 minus sin square theta 1. Let us say this happens at theta 1 equal to theta b. Now squaring up and writing cosine in terms of sin we will have 1 minus sin square theta b equal to epsilon 2 by epsilon 1 minus sin square theta b leading to the condition that epsilon 1 is equal to epsilon 2 which is a trivial situation just indicating that there is really no interface there are no two different media it is just one single media which case is perfectly alright. Otherwise in the case of perpendicular polarization we do not get an angle corresponding to the Brewster angle as we have got for the case of parallel polarization. One can think of interesting applications of this phenomenon related to Brewster angle. For example if the incident wave is circularly polarized it can be considered to be composed of two components of perpendicular polarized component and a parallel polarized component. If the angle of incidence is equal to the Brewster angle then the parallel component will not be reflected back it will be completely transmitted. Only the perpendicular component will be reflected back and therefore a circularly polarized wave can be converted into a linearly polarized wave or in general an elliptically polarized wave can be converted into a linearly polarized wave. We consider another interesting application of this phenomenon known as Brewster angle and for that we consider we go to the overhead projector. This is the schematic of what is known as a gas laser. Now laser is short into prominence many years ago because of its some unique properties. This is a device which can be used for generating laser radiation. It consists of a gas discharge tube which is filled with a suitable gas such as a mixture of helium neon at low pressure. It is a very low pressure almost close to vacuum and then a discharge is arranged in this gas discharge tube. As a result of this there are there is a complex series of phenomena that is initiated. Electrons are knocked off their shells they are raised to higher energy levels then they come back to lower energy levels and in that process emitting light. Now this light is to be coupled out of this gas laser in a suitable manner without allowing the vacuum or low pressure to leak and therefore this gas discharge tube is closed with what are called glass windows which have to be made of a low loss glass at the wavelength which this device is generating and quartz is a popular material for that purpose. So let us say these are quartz windows which would be fixed so that they seal the gas discharge tube properly. This entire assembly is then placed between two mirrors as shown here. So these are the mirrors. One of these mirrors has a very small pinhole. Now the light which is generated or radiated by the atoms and molecules inside the gas discharge tube can pass through these quartz windows and is going to strike these mirrors and in general is going to be reflected between these two mirrors and only the light which is exactly along this axis and travelling in this direction is going to be let out of the system. And before this happens a ray which comes out like this may have got reflected back and forth back and forth several thousand times and of course that requires that these two mirrors have a very good alignment, alignment to within a few seconds of parallelism. So that is one thing and that explains how the laser light generated through this kind of devices say gas lasers is highly collimated. It will have very small angular divergence and therefore it can propagate over very long distances. Where is the Brewster angle phenomenon coming in here? We would like that the reflection and reflection that takes place at these interfaces. These are now two different media. So there is an interface between this medium and this medium. Let us say this is very close to free space and this is that of quartz and once again from quartz to another medium say free space. So these interfaces are going to be encountered by the light rays as they reflect back and forth between these two mirrors and we would like that the transmission through these quartz windows is as efficient as possible. One way of doing that is that we arrange these windows at an angle to the axis shown here which is the Brewster angle. In such a case what is going to happen? We consider let us say a ray of light which is incident in this direction along the axis on this interface. Now we draw the normal at the interface. This is the incident ray and depending on the Snell's law it will make some angle with the normal here and let us say it will come out in this direction. Since this medium has a greater dielectric constant than this the angle of refraction will be less than the angle of incidence and for this kind of a symmetric interface this medium and this medium have identical properties. This ray is again going to come out in a direction which is parallel to the direction of the incident ray or the axis of the system. And this is going to be reflected here and then once again the same path is going to be traversed back and similarly it will pass through these and then we reflected back at this mirror. If this angle is the Brewster angle then the parallel polarized component the light that is generated here has a random polarization and in general all these randomly polarized emissions or radiations can be broken up into two components perpendicular and parallel. The parallel polarized component is going to be perfectly transmitted completely transmission since this is Brewster angle but the perpendicular polarized component is not going to be completely transmitted. Partly it will be reflected back and partly it will be transmitted. This part transmission will take place at this interface for a ray travelling in this direction at this interface and as it is reflected back once again at this interface and at this interface and so on. And since there are going to be several thousand reflections before a ray really comes out through this narrow pin hole one can make out that it is the transmissions of the parallel polarized component which will be complete which will be efficient and transmissions of the perpendicular polarized components are not going to be efficient. Then if we consider a transmission coefficient for the perpendicular polarized component say to be 0.9 which is a very good value of transmission coefficient. A single pass for a ray at this point travelling to this direction and then coming back at this point will require the field amplitude to drop by a factor of 0.9 to the power 4. Since there are 4 transmissions taking place at these 4 interfaces. So if we start with an amplitude of 1 we will end up with a reflected field which is of the order of 0.65. So one pass reduces the value of the perpendicular polarized component by about 35 percent and this way one can make out that what comes out here is essentially going to be of parallel polarization and the parallel polarized component will not suffer this kind of partial transmission. So that is a very interesting application of the phenomenon that we have called related to the Brewster angle. If there are any questions here we can consider those otherwise we proceed further. The second topic that we have listed for this lecture is the total internal reflection and that is what we will like to deal with next. Let us consider the diagram that we have made representing an interface between two perfect dielectrics. Now the angle of incidence and the angle of refraction are related through the Snell's law. So that we have sin theta 1 upon sin theta 2 equal to square root of epsilon 2 by epsilon 1. Let us consider the somewhat restricted category of situations where epsilon 1 is greater than epsilon 2. Then what happens? Then theta 2 is going to be greater than theta 1 and there will come some particular value of theta 1 where theta 2 is 90 degrees. What will be that particular value? That will be sin and let us say that particular value is theta c sin theta c equal to square root of epsilon 2 by epsilon 1 beyond which value of the angle if theta 1 exceeds theta c then the incident ray is going to be completely reflected back into the same medium. This is how we have been introduced to the phenomenon of total internal reflection and therefore, we say that if sin theta 1 is greater than epsilon 2 by epsilon 1 then the incident wave is totally internally reflected back when epsilon 1 is greater than epsilon 2. Let us see how this phenomenon corresponds to or is consistent with the detailed expressions that we have derived based on field theory. For this case let us consider the second term that is appearing in both the expressions for the reflection coefficient both for perpendicular polarization and parallel polarization. And we see that if we consider epsilon 2 by epsilon 1 minus sin squared theta 1. Now, sin squared theta 1 for these situations is going to be greater than epsilon 2 by epsilon 1 and therefore, this term is going to be the argument is going to be negative and the square root is going to be imaginary. And therefore, we can write this as going to let us say plus minus j sin squared theta 1 minus epsilon 2 by epsilon 1 which is what we can write for the reflection coefficient for both polarizations. And therefore, using this we get E r by E i equal to cos theta 1 and then plus minus j sin squared theta 1 minus epsilon 2 by epsilon 1 divided by cos theta 1 and let us say minus plus j sin squared theta 1 minus epsilon 2 by epsilon 1. If you notice we have interchanged the order of sin compared to this expression it should have been minus plus according to this, but we are writing plus minus the reason for this will become evident later. However, now we have made sure that the quantity within the square root sin is positive and therefore, we can write this as let us say A 1 plus minus j B 1 and A 1 minus plus j B 1. Of course, the choice of sin in both numerator and denominator will go together it will be either upper sin or lower sin which will go together. What is the magnitude of this complex expression? Magnitude is 1, the phase could be anything depending on B 1 and A 1 and therefore, let us say that this has a value which is e to the power j phi perpendicular. Since this corresponds to the perpendicular polarization, phi perpendicular will be the phase shift between the incident wave and the reflected wave electric fields. And since the magnitude is coming out 1 it means that the entire electric field is reflected back completely and which is what is meant by the phenomenon of total internal reflection. And we see that whatever we had learnt earlier in simple terms is coming out from these expressions also. Let us see what happens for the parallel polarized component. We are going to have in a similar manner this equal to epsilon 2 by epsilon 1 cos theta 1 and then plus minus sin square theta 1 minus epsilon 2 by epsilon 1 divided by epsilon 2 by epsilon 1 cos theta 1 minus plus sin square theta 1 minus epsilon 2 by epsilon 1 which also can be written as A 2 plus minus j B 1 and A 2 minus plus j B 1 which again when these conditions are satisfied implies that the magnitude of the reflection coefficient is 1 and they could be a phase difference between the reflected wave electric field and the incident wave electric field which is represented as e to the power j phi parallel. So, in this case for the phenomenon of total internal reflection there is no so to say discrimination between the two polarizations for both types of polarizations total internal reflection can take place. And the condition for total internal reflection remains the same for both polarizations. However, the phase shift that occurs for the two polarizations is different which again can form the basis of some very interesting applications. For example, an elliptically polarized wave where the magnitudes of the two polarization components are equal, but the phase differences are not proper for circular polarization could be converted into a circularly polarized wave by an appropriate design by choosing a suitable angle of incidence and choosing proper materials. So, this can become a basis of very interesting design for this purpose changing the state of polarization. This phenomenon of total internal reflection is used very frequently in optics work. In fact, if you recall the optical fibers support propagation of light through the phenomenon of total internal reflection through very thin glass fibers consisting of a core and a cladding where the core index, core refractive index which can be related to the permittivity is greater than the cladding refractive index. So, that light remains guided between the core in this manner which I have mentioned earlier. Which optical fibers are revolutionizing the entire field of communication. In fact, optical fibers are just one kind of optical waveguide. They can be waveguides of many different types. Another type of waveguides occurs in the field of what is called integrated optics where one makes the optical waveguides on the surface of a substrate, a suitable substrate which could be say lithium niobate or even glass or silicon or gallium arsenide depending on the final application and there are various special features of each of these. So, there what one would be doing is that on a substrate let us say like this in a narrow portion one would be increasing the refractive index through some means. And then this becomes a waveguide where through total internal reflection between this core and the rest of the material which acts as cladding light rays can be guided into this waveguide. And if anyone of you takes up this kind of work he will see that these expressions utilizing these phase shifts which occur when total internal reflection takes place at different interfaces are very useful in that field. While one can use reflectors say metal coated reflectors for changing the path of light rays one can do the same thing using the phenomenon of total internal reflection also which concept I am sure you are familiar with. The change of the path of light rays using total internal reflection is supposed to be somewhat more efficient than accomplishing the same thing using reflectors. The loss that may be associated with the reflecting material is avoided in the total internal reflection based light ray path changing schemes. Now, we go a little further this seems to indicate this may be an impression that we may form that when total internal reflection is taking place the entire electromagnetic phenomenon is restricted to the region above the interface or just at the interface that impression is not correct. Why is it not correct? One simple argument that can be given is we consider the total tangential field let us say total tangential electric field at the interface in region 1 it will be E i plus E r or combination of these two depending on which polarization we are talking about. If that is the total tangential field in medium 1 at the interface there must be a corresponding tangential electric field in medium 2 as well. So, that the boundary conditions are satisfied therefore, there is field at the interface in medium 2 it is required from the continuity of tangential electric field components. If there is field in medium 2 even when total internal reflection is taking place in medium 1 what is the nature of this field? That is what we like to investigate next. Now, you would recall that we were writing the expression for the field related to the different waves the incident wave the reflected wave and the transmitted wave in the following manner. So, in medium 2 we write the field as let us say E t into E to the power minus j beta 2 beta 2 being the phase shift constant for a uniform plane wave in medium 2. And then let us say these are the y and the z directions then for the angles shown we have here y sin theta 2 minus z theta 2 minus z theta 2 minus cos theta 2. If there is a transmitted wave this is what we would write. We separate the y portion and the z portion our experience tells that these two components behave somewhat differently. So, we consider the y part here and let us say the z part here and the y part would be written as ignoring the amplitude factor E to the power minus j beta 2 y into sin theta 2. If time permits we can comment on the nature of theta 2 later on, but right now we are not sure about the nature of the role of theta 2. We are talking about the situation when total internal reflection is taking place. So, let us write theta 2 in terms of theta 1 using snails law that is we have sin theta 2 equal to epsilon 1 by epsilon 2 into sin theta 1. Theta 1 we are sure of that is the angle of incidence and the angle of reflection and therefore, this becomes E to the power minus j beta 2 y and instead of sin theta 2 we write root of epsilon 1 by epsilon 2 sin theta 1 which indicates or can be interpreted as a wave which is travelling along the y direction. This is not a subscript this is beta 2 times y with the phase shift constant beta y equal to beta 2 into square root of epsilon 1 by epsilon 2 into sin theta 1. It indicates a wave travelling in the y direction with this phase shift constant that is with the corresponding phase velocity v y equal to omega by beta y equal to omega by beta 2 square root of epsilon 1 by epsilon 2 times sin theta 1. Mathematical what is the nature of the factor multiplying beta 2 that is under the conditions of total internal reflection. Total internal reflection means that sin theta 1 is greater than epsilon 2 by epsilon 1 square root and therefore, this factor is greater than 1 under total internal reflection conditions implying that this phase velocity is less than omega by beta 2 or the velocity of light in this infinite medium 2. Otherwise it travels along in the y direction with this phase shift constant and with this phase velocity. What is more important from the practical point of view is the z variation that the wave this is the behavior of the combined wave also in medium 1 travels along the y direction is quite alright. Let us see what is the z behavior and there we get e to the power plus j beta 2 z times cos theta 2 which would simplify to e to the power j beta 2 z and then cos theta 2 we write in terms of sin theta 2 which is further going to be written in terms of sin theta 1 and therefore, we have cos theta 2 as 1 minus sin square theta 2 or e to the power j beta 2 z and then we have here sin square theta 2 is now written in terms of sin theta 1. So, that it is 1 minus epsilon 1 by epsilon 2 sin square theta 1. Under the conditions of total internal reflection the second term under the square root sign is going to be less than 1. Since, sin theta 1 is greater than root epsilon 2 by epsilon 1. Therefore, to make the argument real we take the negative sign out and that will give us e to the power j beta 2 z and then we have plus minus j and then we have sin square theta 1 times epsilon 1 epsilon 2 minus 1. Now, we have ensured that the quantity under the square root sign is positive this simplifies to e to the power minus plus beta 2 z into epsilon 1 by epsilon 2 sin square theta 1. Now, comes the question of choice of sign mathematically both signs are admissible. This wave is propagating in which direction in medium to in the negative z direction minus z cos theta 2 is what we use and since it is a passive medium with no source of energy the wave amplitude could decay in the direction of propagation it could not build up. Since, other quantities are positive here we must choose the lower sign that is the positive sign. So, that for decreasing values of z or increasingly negative values of z the wave amplitude goes down exponentially. And therefore, we choose a sign such that it is e to the power plus beta 2 z then epsilon 1 by epsilon 2 sin square theta 1 minus 1 anticipating this choice of lower sign in the square root signs we had reversed the order of signs in the earlier expressions. So, what is the behavior of the field in the z direction in medium to it is an exponentially decaying field it is not a wave it is not propagating in the z direction it does not have the characteristics of a propagating wave in the z direction for a propagating wave it is this kind of exponent which is required with a j sign. So, this is just a decaying wave exponentially decaying wave in the z direction. In fact, this is considered to be an example of a non uniform plane wave because in the plane normal to the direction of propagation the amplitude is not uniform. So, this is considered to be an example of a non uniform plane wave there is a very interesting consequence of this observation what is the observation the observation we have made is that even under total internal reflection conditions there is a field in the second medium total internal reflection is taking place in medium one, but there exists a field in the second medium which is exponentially decay. And therefore, it is necessary for efficient total internal reflection that this surface in medium two is kept absolutely clean and free of dust particles and dirt etcetera. If that is not so then the total internal reflection would not be 100 percent efficient which otherwise is indicated theoretically. With this observation which is of practical importance we can stop here unless there are some questions. Sir, can you clarify why root of epsilon by epsilon 2 sin theta 1 is greater than 1. That is quite straight forward you see for this situation and above angles greater than theta c only total internal reflection takes place otherwise it is the normal situation of partial reflection and partial transmission. So, since theta 1 is going to be greater than theta c for total internal reflection sin theta 1 is going to be greater than this quantity. So, what happens to the power flow? There is no power flow in the second medium perpendicular to the interface and whatever power was incident normally on the interface is reflected back. Since, the reflected wave electric field magnitude is the same as that of the incident wave electric field magnitude. So, it will form a standing wave in the z direction and in the y direction the entire composite wave a standing wave in this direction exponentially decaying wave in this direction will travel in the y direction. So, I suppose we can stop here.