Welcome to the lecture module on optics, we have been discussing wave optics in the last few lectures we discussed about two important phenomena in wave optics, namely interference and diffraction. Today we will discuss polarization, an important characteristic of light. This is the last topic that we will have in this module of wave optics. Polarization is an important characteristic of light, so we will discuss polarization and polarization of light in this lecture. We will see what polarization is and the state of polarization of light. How is a state of polarization of light defined? Why do we need to know and define the state of polarization of light, how to obtain polarized light polarized light by reflection? This is one of the techniques for obtaining polarized light by reflection at the Brewster angle, and then we will discuss propagation of plane, polarized light through one or more polarizers. So first, as polarization of light, polarization of light is a property of light. It is an important property of light is an electromagnetic wave comprising of rapidly varying electric and magnetic fields, and the electric and magnetic fields are perpendicular to each other and perpendicular to the direction of propagation.

We have studied this in electromagnetic theory. Em theory that the light comprises of electric and magnetic fields which are perpendicular to each other and perpendicular to the direction of propagation polarization of light refers to the direction of oscillation of the electric field of light. It refers to the direction of oscillation of the electric field. We will see what these statements mean. Light waves light is an electromagnetic wave. I have shown here an electromagnetic wave propagating in the x direction. We see the blue colored one here. Variation shows the electric field vector, the variation of the electric field vector along x at any given instant and the red color shows variation of the magnetic field vector. So, we can see that the magnetic field vector in this diagram is along the z-axis here. This is the x-axis. This is the y-axis and is the z-axis. The electric field variation is in the y direction, so it is increasing and decreasing its varying sinusoidally and, along with that, the magnetic field also varies sinusoidally, but in a perpendicular direction and both the electric field and the magnetic field are perpendicular to the direction of propagation electric Field is along y direction. Magnetic field is along z direction and the propagation is along the x direction, as we discussed, polarization refers to the direction of oscillation of the electric field.

Therefore, in this diagram here, the electromagnetic wave that we have shown now, we forget about the red, colored variations here, the magnetic field. If we look at only the electric field variation, we can see that the electric field is varying in the y direction. Only it is confined to the xy-plane, the xy-plane, and therefore this is a y polarized wave. This is a polarized wave now lets. Look at this more carefully, the state of polarization is now p. I have dropped the magnetic field variation and I have shown only the electric field variation here and the electric field is varying sinusoidally in this direction. Now, if you look from the direction x here, a wave which is coming towards you when you look in a direction perpendicular, so in an if you look at this in a plane perpendicular to the direction of propagation, so that is what is shown here. This is a plane. Perpend x is the direction of propagation and a plane perpendicular to the direction of propagation. Is this: it is the yz-plane. The yz-plane is perpendicular to the direction of propagation.

What we see is, as the electric field varies in this direction, the electric field is becoming positive. It is becoming negative, then positive, negative, and so on, because we know that the electric field can be represented as a sinusoidal wave, so we can write. For example, the electric field e is equal to some amplitude e zero into is the direction of propagation, and in this case the direction of propagation is x. So, is the time is the angular frequency. So, this is equal to Angular frequency nu is the frequency of the wave which is equal to c by lambda. Where c is the velocity of light and lambda is the wavelength of light. So, this is an electromagnetic wave propagating in the +x direction. So, that is what is shown here that the state of polarization. So, if we look at a projection here, the electric field is varying in this fashion, but in the projection, we see that the electric field is becoming positive negative, but all along this line y and therefore the projection on a plane perpendicular to the direction of propagation, is a line and therefore such an electromagnetic wave is called linearly polarized wave the projection of the electric field on a plane, perpendicular to the direction of propagation is a line.

Hence, the name linearly polarized wave, the state of polarization of any wave is given by the projection of the locus of the tip of the electric field. You can see it is the tip of the electric field. The tip of the electric field is always lying on. This line, as it becomes maximum, then reduces then becomes negative, but so it is, the tip of the electric field is the projection of the locus of the tip of the electric field, vector on a plane perpendicular to the direction of propagation. One need not remember. This definition, but this explains any state of polarization in this course we will see primarily linearly polarized light, but there are different states of polarization, namely circularly polarized, light elliptically, polarized light and so on. So, this definition given here will help in identifying the state of polarization of light, so we will primarily discuss linearly polarized light. Now, I have taken a wave, an electromagnetic wave where the electric field varies, is varying in the xz-direction. This is the z direction, so now electric field is varying in the z direction like this. Obviously, the magnetic field varies in the y direction, but I have not shown the magnetic field. So, if we look at the projection now in a plane perpendicular to the direction of propagation, then the electric field will vary in the z direction and therefore this is a linearly polarized wave. This wave is a linearly polarized wave, but it is now z polarized wave. If we look at this plane now, the electric field is confined here to the xz-plane, xz-plane. The dotted plane here represents the xz-plane and if I show the x that plane in two dimensions now xz-plane, the electric field is varying like this. I have flipped this up like this, then. What you see is this. So, y direction is now into the board into the paper here and z is here and x is in this direction and what we see is the variation of the electric field and the electric field is confined to the x z, plane and therefore linearly polarized light Is also called plane, polarized light. The e field oscillations are confined to the xz-plane in this wave here this is the two-dimensional picture, and then these are called plane, polarized light and therefore linearly, polarized or plane. Polarized means the same thing. Now, let us look at unpolarized light a little bit carefully and then we appreciate the plane pole rise light. So, what I have shown is a beam of unpolarized light from common sources such as sun, electric bulb or a fluorescent lamp etcetera are unpolarized in nature.

What is this unpolarized light? So, for example, what I have shown here is a torch, a battery torch, and a beam of light coming out from a battery torch. The beam comprises of many component waves. These are the component waves. These are waves which are emitted by various parts of the emitter. This torch bulb has a filament and various parts of the filament give out different component waves. These are all independent waves which are emitted by various parts of the filament and therefore, what I have shown here is the component waves. Now, the beam of light comprises many component waves. The component waves are emitted by different atomic oscillators.

Different atomic oscillators from various parts of the source electromagnetic radiation are emitted by atomic oscillators or dipoles oscillating dipoles. This concept is a little advanced to our level here, but they are small, tiny oscillators which emit electromagnetic radiation and various parts. So, the component waves are emitted by different atomic oscillators and therefore what would happen is that, let me show it here that a particular oscillator which is emitting would have the plane of polarization like this. Another oscillator, which is oscillating like this would oscillate. The wave emitted will have the electric field varying in this. Another oscillator, which is oscillating, could be oscillating in a different direction. Another oscillator oscillating in this direction may have a different plane of polarization and therefore, if you look at the cross section here, that is the x axis is here, so we are looking at the cross section so here. So, this is the y, and this is the z axis and x are coming out.

Then we will have linearly each one of them is linearly polarized, but we will have vibrations varying like this, some in the y direction, some in the z direction, some at various angles. In other words, the polarizations are random. Each of the component waves each of the component waves. So, this is what is explained here. The component waves are emitted by different atomic oscillators from various parts of the source. They may have different planes of oscillations and therefore the combination forms a randomly polarized beam or an unpolarized light. So, that is what I have been explaining here, that different oscillators will have plane of polarization or the line at in different angles, and therefore the net result is a randomly polarized light. So, if you see the cross section again, let me show it so here then some of them may oscillate like this. Some of them may be in a plane like this, so this is the representation of randomly polarized light, and this is also called unpolarized light.

So, the unpolarized light will have the plane of oscillations of different components in different directions and therefore it is sometimes called randomly polarized light or commonly called unpolarized light now, the electric field, if you look at the electric field vector here, so it is varying in this direction field is an electric field, is a vector and therefore we can always resolve it along two perpendicular components. So, if we have an electric field which is varying like this, then we can resolve it into two components. So, this component here so one component here and one component, so this comprises of so when it becomes reverse.

So, this component will come here, and this component will be negative here and therefore what we will have been the component is varying like this and it is varying like this, so this one, electric field variation can be represented, equivalently by components varying like this. So, this is an electric field which has some component, for example, y cap into e y. If I want to write the electric field e of this vector here, then y cap y is the unit vector in the direction y here. So, that is y- and this is z, y cap e y plus z, cap e z, where e z is the z component, each one of them is oscillating, one is oscillating like this, and the other one is oscillating like this. Therefore, every component, which is shown here, whether it is this component or this component, any randomly oriented components can be resolved along the x direction and y direction, and the net unpolarized light can be equivalently represented here in the form of one y component and one z component, this is an equivalent representation, but the electric field is varying randomly in different directions. So, that is what is explained here again that the electric field, vectors of the randomly oriented polarizations, are resolved into their components along y and z directions in the equivalent representation.

Here, unpolarized light comprises of two equal components. Both the components are equal in magnitude because there are random polarizations in all directions and therefore, on an average we will have equal. Both the components are equal in two orthogonal directions comprises of equal components of the electric field, light field of light in y and z directions. So therefore, now onwards we will represent unpolarized in light in this fashion, so the representation of polarized light.

So, in this here I am showing the representation summary of the discussions that we had so far so representation of polarized light, assuming the following coordinate system here with the x axis as the direction of propagation so y is here Z being into the board, and therefore we have y polarized light means is represented like this. That is, this is the direction of propagation and the electric field is oscillating in the direction y. Similarly, z polarized light direction of propagation x and the electric field is perpendicular to the plane of the paper.

Here that is why it is shown as a dot and unpolarized light will have both y component and the z component, and therefore unpolarized wave is represented like this. In two d, what we have shown is in two dimensions. If you look at the cross section from this direction, if you look in the x direction, then you will see the cross section is a x y x y z plane. So, in the y z plane, we see that so this is x coming out of the paper and we have y polarized light will look like this and z. Polarized light will look like this. It is horizontal - and this is vertical and an unpolarized wave can be represented by two arrows which are at this double-sided arrow. All the arrows, as you can see, are double sided arrows. The double siding comes because electric field once become positive. Other times become negative. That is why it is always represented by double sided arrows, and this is the representation of polarized light, including unpolarized or randomly polarized way. Now, the next question would be how to obtain polarized light, how to obtain polarized light. The answer is straight forward. The answer is here by passing unpolarized light through a polarizer. Now, there are several types of polarizers: a polarizer is a device, an instrument, or a component which polarizes. That means. If you launch a particular state of polarization, then it will. It can change the state of polarization to something else or it can polarize. It can polarize an unpolarized light. That is, if you launch an unpolarized light, then the output of the polarizer will be plane polarized light. There are several types of polarizers based on different working principles. The simplest, least expensive, and most widely used are sheet polarizers or a polaroid sheet. These are simple sheets. I do not have a sheet here right now, but these are small sheets which are widely used in laboratories.

Now let us discuss a little bit about this simple polarizer tolerance sheet, so polaroid sheet or sheet polarizer. So, what I have shown here is a sheet where I have shown some molecules, so the polaroid comprises of a sheet of certain long chain. Polymeric molecules are polymers. Polymers are long chain molecules comprising of a few atoms or molecules. So, these are long chains, so long chain, polymeric molecules which are almost aligned like a wire grid. So, you can see the polymer molecules are all aligned here. In this case there are techniques of making it aligned. So, a polaroid sheet comprises of long chain polymeric molecules which are almost aligned like a wire grid wire grid, is what you see here the grid here. So, it is a grid, so it is aligned all of them in the form of as if it is a wire grid.

Now, the polarization component, which is parallel to the long chain, suffers loss. Now, if light is an incident like this unpolarized light is incident like this. So, that is what is shown here: unpolarized, comprising of two orthogonal components. Here, we have resolved one of the component along the molecule and another component perpendicular to the molecule perpendicular to these chains long chains, the polarization component which is parallel to the long chain, suffers loss or attenuation or it undergoes loss, whereas the component, which is perpendicular here, hardly suffers any loss, which means, if you incident polarized light here, then this component will pass through with very little loss, but the other component will be highly absorbed or highly attenuated. So, on the other side of the sheet, what you will get is a polarized output. So, in this figure, the vertical component in the figure is attenuated, that is, is absorbed.

The horizontal component passes through the sheet with little loss and therefore the horizontal axis is called the pass axis of the polarizer pass axis of the polarizer refers to that axis, which allows polarization of light to pass through. So, in this case, the path axis is horizontal. Let me repeat again that the vertical polarization here undergoes loss, but the horizontal polarization passes through the sheet and therefore the horizontal axis here is called the pass axis. Let us make it clearer. Let us look at the diagram again in a unique way, so unpolarized light passing through a polarizer here is a polaroid sheet and unpolarized light is incident. As usual. We have resolved it into two components, one component parallel to the path axis and another component perpendicular to the path axis. The component, which is parallel to the path axis passes through the sheet. The perpendicular component is blocked and therefore we get 50% of the light. Passing through because, as we have already discussed, unpolarized light can be thought of as comprising of two components: one vertical component and another horizontal component, each of them 50% strength, equal strength and therefore fifty percent of the electric field is attenuated and fifty percent passes, which means, if I had the input intensity I0, then what we will have on the other side. Is I0 by two, because 50% light is blocked by the polarizer, but on the other side we get a plane, polarized light, so plane, polarized light on the other side, with the plane of polarization parallel to the path axis of the polarizer. So, this is the way it works, so I0/2, is the output intensity. Of course, we have neglected absorption of the vertical component.

There is a little bit of absorption even for the vertical component, even though it is passing through the axis, but if otherwise, in practice, it is slightly less than zero by two. But we neglect the absorption and say that if I0 is the input, then I 0/2, is the output of the on the other side.

Now, what if we rotate the polarizer? So instead of this, we rotate the unpolarized light that is coming here. The polarizer we are rotating, that is the path axis we are rotating. So, what would happen? Because, if the path axis, for example, if the path axis is like this, then we can always resolve this randomly polarized light into one component, parallel to the path axis and the other component perpendicular to the path axis as before the component. Parallel to the path axis. Will be on the other side, but the perpendicular component will be blocked. However, now the output state of polarization. So, let me show this here. If I consider the path axis to be like this at an angle, then what we do is the light which is coming here will be resolved. One like this component and the other component will be perpendicular to this. The perpendicular component would be blocked and then, on the other side we will have light, which is polarized like this. So, if I rotate the polarizer or if I rotate the polarizer, then the plane of polarization at the output also rotates earlier. We had the polarizer pass axis like this, so at the output we had polarization coming like this output, polarization vertically polarized.

Now, we have rotated the path axis. What is shown here is the path axis. Then the plane of polarization will rotate, but 50% of light will still come on the other side. So, if we had I0 here, we still have I0 by two independents of the rotation of the path axis. What it means is, if you launch unpolarized light passing through a polarizer, then if you rotate, what if we rotate the polarizer, if you rotate the polarizer about an axis, obviously the path axis rotates, but there is no change in the intensity of light at the output.

We have already answered this question why does unpolarized light pass through a polarizer? This is one of the ways. One of the ways is to use a polarizer sheet or a polaroid sheet, but there is another important technique that is polarization by reflection. So, let us look at the second technique, which is polarization by reflection of light. Now, let us first look at this recall reflection of light at a plane, interface in ray optics earlier. We discussed reflection of light at an interface and Snell’s law. So, here we had discussed in terms of ray here. The ray represents the direction of propagation of the wave. The wave is the electric and magnetic fields are perpendicular to the direction of propagation, and this is the direction of propagation. So, the wave is incident here, the wave now as per Snell’s law. We know that sin I/sin r, is this. I is the angle of incidence between the normal and the direction of incidence. Here, I and r is the angle of refraction. These are two different media of refractive index n1 and n2, and this is the interface we are looking at reflection of light at an interface now, Snell’s law says that sin I/sign r, is n two by n1, which is also written as n. Two one now let us start with here a small angle, the first one here, the black one.

So, I and then this is r. This is the transmitted ray. This is the reflected ray the reflected wave or ray here has the same angle. I so reflected angle is equal to angle of incidence. If I increase the angle further, we look at the blue line here. Then this is reflected here and the transmitted ray or transmitted wave is here. If I increase further, then to an angle I designate, as Ib, we will know that b is standing for Brewster Ib, then the reflected ray is here again. The reflected ray makes an angle Ib. I am recalling from the ray optics here and then this is rb, but an important observation which we did not discuss. This angle, Ib, is an angle between the reflected and transmitted rays of 90°. Here, it is 90° and therefore we can write. Rb is 90-Ib, the rb is 90-Ib and therefore from the figure you can see clearly here that this is Ib.

Therefore, this is 90-Ib and if this is rb, then rb will be equal to 90-Ib from the figure and therefore sine Ib by sine rb is equal to sine ib by sine 90-Ib, which is cos Ib, which is equal to Tan Ib and therefore Tan Ib=n21, and this is called the Brewster’s law, and the angle of incident is Ib is known as the Brewster angle.

What is special about booster angle? This is ok, this comes from ray optics also, and we know that this is the Brewster angle and at which the refractive index, for example. If this was air refractive index 1 - and this is some glass refractive index say 1.5 - then we know that tan ib will be equal to n21, which is equal to n2. That is the refractive index of glass. So, we can determine the refractive index of glass. If we know the Brewster’s angle but how to find what is special about the Brewster angle, let us discuss that reflection of unpolarized light at the Brewster angle in the earlier diagram. I did not show anything about the polarization now I am showing here in this diagram polarization of the light.

So, let us look at the diagram carefully. There is unpolarized light which is incident at the Brewster angle. So, one component is: we have resolved unpolarized light. One component perpendicular to the plane of incidence that is out of the paper here and one component in the plane of incidence plane of incidence contains the normal, the normal and the ray which is incident and the reflected. So, this is the plane of incident and we have one component in the plane of incident, and one component coming out of the paper that is perpendicular to the plane of what is observed is when light is incident at the Brewster angle.

The reflected light here does not contain the component in the plane of incident. It contains only one component, which is perpendicular to the plane of incident, which means it is completely polarized, whereas the transmitted light contains both components. The vertical component here and the in-plane component, and therefore it is called partially polarized light because it contains both, but the reflected light is perfectly plane polarized. This is the example, of course, about the Brewster angle, but what is important for us is reflection. Light does not have an in-plane component. The answer y is a little beyond the scope of our discussion, but for the sake of completeness, I will briefly explain why the reflected light does not have the transmitted component.

So, let me explain here so here is the interface and light are incident like this, and this is the transmitted light reflected light and we have the transmitted light when light is incident on any material or propagating in any medium. So, this is a medium n one, and this is a medium n two, so light is an electromagnetic wave. Therefore, it comprises of electric fields varying like this. The light is assumed that light is incident on a medium here. The electric field is positive here, and electric field is negative, so electric field positive negative when light enters the medium it the due to the electric field, the medium comprises of atoms. It is a matter which is made up of atoms, the or and molecules, and if I look at individual atoms or individual molecules, then the center, when there is no electric field, the Centre of positive charge and Centre of negative charge coincide at a point and the atom is neutral now when there is an electric field, for example, if you place this in an electric field, you apply an electric field.

Keeping this just a hypothetical situation, place one atom between two plates and apply an electric field, then the positive. So, if you apply an electric field, that means, if you apply positive here and negative here, then the electric field is in this direction and the negative charge moves to the other side and positive charge moves to the towards the second electrode. So, these are what I have shown are two electrodes and there is an atom comprising of electrons and the positively charged nucleus. And then the charges separate because of the applied electric field and this such an entity here is called a dipole, because now it is an entity which has a negative charge here and a positive charge here. So, it is so. I am showing it like this. It is a dipole now this is a study dc field. If I apply suppose i reverse the field, then I will have the positive coming this side and negative charges moving to the other side, I reverse again, then I have positive negative here and positive here and so on. Therefore, when a time varying electric field is incident, so if you have an electric field which is changing with time as it propagates, so it is positive negative then this is equivalent to changing the field.

Positive, negative, positive, interchanging, the positive negative. So, in the medium, this electric field of the light, the varying electric field of the light, induces what are called dipoles or induced dipoles, so induced dipoles. This is beyond the scope of our discussion, but just for the sake of completeness, let me just describe it very briefly and the induced dipoles, if i show the induced dipole here, that is with the charge like this and changing at a later time like this, because the electric field is varying at a different time and so on. So, this is at time t1. This is at time, t2 and time t3 and so on it changing such a dipole emits. So, this rapidly changing positive, negative, negative, positive, positive negative leads to emission of electromagnetic wave, so emission of e m wave of the same frequency e m wave of the same frequency same frequency.

The important point is if the dipole is here now, this is at different times, but it is the same dipole if the dipole is here becoming plus minus, minus, plus, plus minus and so on, then this gives out radiation. So let me show the different color, so this gives out radiation in the transverse direction, so the electromagnetic wave it emits electromagnetic wave in all directions. So, what I have shown are the field lines. These are a matter of details, but the important point that we need to know is there is no. The fields are propagating in the transverse direction. There is no field propagating along the axis of the dipole, there is no electric field variation or the electric field. Variation is of the same frequency as that of light. There is no field along the dipole now.

How is this related to the problem at hand? So let me put back the slide here when electric field variation is incident here or let me take a different slide. Let me draw that again one last time, because this is not so we have the electric field variations in one case, the electric field is varying like this. In the other case, the electric field is varying in a direction perpendicular. Therefore, the medium here. This is one medium of refractive index n1, another medium of refractive index n2, when the transmitted wave.

So, this is the transmitted wave this polarization at which is incident on the medium, so we have seen that at Brewster angle, this is 90°. This is 19. This is the incident light. This is the reflected light and this is the transmitted light at Brewster angle. This is 90° angle between the reflected and the transmitted light, and therefore this variation here, the dipole which is oscillating like this recall again, the earlier when the dipole is oscillating like this.

There is no radiation or no electromagnetic wave propagates in an along the axis of the dipole, exactly like that when the dipole is oscillating because of this field. When the dipole is oscillating like this, then there cannot be any radiation in this direction in the direction here. There is no radiation in this direction because it’s along the axis, because this angle is 90°.

However, the dipole which is oscillating like this does give wave propagating in this direction and therefore this polarization is reflected back, but the other polarization is transmitted. This polarization can only be transmitted. There is no reflection this wherever you have not followed. It does not matter because it is a slightly advanced concept, but the important point that we need to know is this: that the reflected light contains component only perpendicular to the plane of oscillation. One additional point which i want to discuss before I take up the last topic in this: is the electric field and intensity electric field and intensity of light, consider light propagating this direction and it’s polarized in this direction. That is y polarized wave. So, this is why Y polarized. This is x is the direction of propagation.

The electric field can be written as e is equal to y cap y cap is a unit vector in the y direction. Sometimes we denote it as j. I j k instead of x cap y cap z, cap into e0, is the amplitude and . This is called the phase term phase term, and this is the amplitude. We are recalling what we have studied under electromagnetic amplitude and fist. If you need to know the intensity, then you take mod e square, gives you the intensity, which in this case, will be equal to the whole square now is angular frequency of Light this is a new is very, very big number.

Therefore, so nu is of the order of ten to the power of fourteen or ten to the power of fifteen hertz for light, and therefore this is an extremely rapidly varying function and therefore the mod square gives you an average. You have to take an average of this and this will be equal to mod e 0 square. Into average. Of this is half that is sin k. Omega k is x, minus omega t if you take the mod square and take the time average. This comes out to be half. We have discussed this in the case of interference in the case of intensity measurement. If we take an x polarized wave, then we represent e is equal to z, cap z, polarized wave.

That is the horizontal polarization. So, we have this horizontal polarization and this the direction of propagation is x, z, cap into e, zero sin, k, x, minus omega t and again the intensity in this case will be equal to mod e zero square into half as before. If we consider light at an angle here, light polarized at an angle now like this, it has two components and therefore the electric field can be rep. This is x direction. The electric field e, is represented by y cap into e x e zero cos theta. The amplitude is e zero. Therefore, this comprehenses of one y component and one z component and therefore the electric field can be represented by y cap e, zero cos, theta, plus z, cap e, zero sin, theta e, zero sin theta.

What is theta is the angle here theta angle, between y, so we can see this more carefully. Let me draw it more carefully so here, so this is the y direction. This is the electric field and therefore this angle is and therefore, it has one component along y along y, which is if this is e0. Then this is e0 cos . The other component, which is along z, will be e0, 90- cos and therefore this is e0 sin . So that is what I have shown. It has one component that is this component and these two components. So, this component is e0 cos and . If you take mod square, that is intensity, is equal to mod e square, will again come out to be e zero square into cos square theta, plus sin square theta.

So, we have this into. This is amplitude variation into sine kx-. This phase term is always there. What I have written here is the amplitude, because now it is at an angle, theta, and therefore we get e zero square into half as before. We get e zero square into half. What is the implication, what it means is the intensity of light intensity of light does not depend on the state of polarization whether it is y, when light is passing through a medium, not passing through a polarizer passing through a medium. The intensity does not depend on the state of polarization whether it is this or this or this all of them give the same e0 square by half and therefore, in the remaining discussion, the phase term simply gives you a factor half. Otherwise, there is no change.

Phase term simply gives you a factor half and therefore, in the remaining discussion in problem in problems, we can drop the phase term. We can and discuss only the amplitude variation in determining the intensity when we determine relative intensity. That is input by output or output by input when we want to calculate the half factor will get cancelled and therefore, we can simply look at the amplitude variation.

Why I discuss this will become clear when I take up the following problem now, let us take up the problem of propagation of a linearly polarized light through a polarizer. So, what is shown here is plane polarized, light at an angle incident at an angle to the path axis. So, the path axis is here along y, the incident plane polarized light the it’s a linearly polarized light is making an angle theta with the path axis. Then the path axis will allow only one component of this, so this is what we have discussed here: e one.

The electric field here is - a component here - so I discussed this just now that the component along y is e, zero cos theta and the component along that is E0 sin . Therefore, the electric field can be written as a y component plus z component. We have dropped the phase term, as I discussed. We do not, we are. We have dropped the phase term, which is common everywhere, and therefore the path axis is along y, which means the y component will be allowed to pass through, but the z component will be completely blocked by this polarizer and therefore the electric field.

Here E2 will comprise of y cap e0 cos . That is the first component. Only the second component is along z. Component is along z-axis. Therefore, it is blocked, so we have y cap E0 cos . Therefore, the intensity here will be , which is equal to .

What about the original intensity? The input intensity? I is mod of E12 e. One is the electric field here, which is , which is simply I1= E0 square intensity here at the input is E0 square intensity at the output after the polarizer is E0 square cos2 and therefore, I two the intensity at the output is equal to intensity at the input into cos square theta.

This important relation is called malus law, malus law, where theta is the angle between the pass axis and the polarization the plane of polarization of the input light.

Now, with this, we now take up the second problem, namely unpolarized light passing through two crossed polarizers now unpolarized light passing through two crossed polarizers. So, the diagram shows here there is unpolarized light, which is incident on the first polarizer, which has a pass axis like this, and the second one has a pass axis perpendicular to this. Therefore, such an arrangement is called crossed. Polarizers crossed means crossed polarizer means that polarize the at pass axis are perpendicular to each other.

The two polarizers have pass axis perpendicular to each other, and if light is unpolarized, light is incident at the input. Then I1=I0. The I1 is here I2 is here, and I3 is the intensity here, the intensity at the input. If it is, I0, then after passing through the polarizer, the intensity will be I0/2. We have already discussed that unpolarized light passing through a polarizer, 50 percent of the intensity will be lost, so the output intensity.

Here, if I designate it as I2, then it is I0/2 now when light continues here. This is a polarized light y polarized light, but the path axis here is along z. It is perpendicular and therefore this polarization will be completely attenuated or absorbed or blocked by this polarizer, and we get no light.

What if we rotate any one of the polarizers, so let’s keep this constant and what, if we rotate this, of course, as we rotate, then, when the pass axis changes and finally, when it passes axis, becomes parallel to y, we get full light coming through this when it is perpendicular to y then no light when it is perpendicular parallel to y.

Then all the light passes through the second polarizer as well. What if the path axis makes an angle, theta that’s what we want to know what, if we rotate any one of the polar, whether it could be this or that, but it is easier to imagine rotating the second polarizer.

Now, if we introduce a third polarizer, what would be the output? Let me discuss this problem so introducing a third polarizer. Please see the diagram unpolarized light incident on the first polarizer, whose path axis is along the y direction. The third polarizer, the second polarizer, as I showed in the earlier problem, there was no third polarizer here. We call this as the first and the second. This is the third polarizer that we have introduced when this polarizer was not there. We had pass axis like this path. Axis perpendicular and therefore output is 0. There is no output, no light, because we are passing through two crossed polarizers.

Now, if I introduce a second, a third polarizer, one polarizer, two and third polarizer with path axis at an angle with the y axis, then let us look at the output intensity. Let us estimate the output intensity. We start with the input the intensity. I1=I0. The intensity I two is fifty percent, because this is a polarizer. So now we have only the vertical component that is y polarized light with the intensity. I0/2. If this passes through a polarizer, which makes an angle theta with the plane of polarization plane of polarization, is here y and therefore there is an angle . Therefore, the intensity here must be. input intensity into . That is the mallow’s law. Malus law says that if the angle is theta, then we have. I two intensity here will be. I1 cos2 and the same thing we apply here the intensity.

Is I0 by two? Therefore, the intensity here. Is I0 by two into cos square theta? Now we again apply malus law. Now the polarization makes an angle theta with the y axis, because beyond this, the polarization is tilted making an angle theta with the y axis. When we come here, the angle between the plane of polarization and the path axis.

Is this one or this one, which is 90-, 90- and therefore, the intensity? Here I4 on the other side will be equal to so let me write so it will be equal to . I3 is the intensity here. Intensity here will be equal to , so that’s what I have written here that I4 the intensity will be equal to. this. , and that is is sin and therefore, this is, which is I0 by two into sine two theta by two and which is equal to I0/8 into sine square. Two theta is the angle between the pass axis and the plane of polarization of the third polarizer, which is introduced between the first and second crossed polarizers. Between the two crossed polarizers. We had introduced a third polarizer which now brings a finite intensity at the output. Before we introduce the third polarizer, there was no output.

Now when we introduce the third polarizer between the two, we get a finite output. I4 is maximum when =45°. That is when the angle here of the introduced polarizer makes 45° to the y axis. Then we have maximum light coming out at the output, which is equal to I0/8. That is 1 8 of the intensity, so I4 is maximum and I4=0 when , which means when. We have this, which means the third polarizer and second polarizer are crossed. Therefore, output is 0. If , which means this rotates like this and then this polarizer and the third polarizer are at 90-degree.

Although this is parallel to this and again, the output is 0. Output is 0 here itself and therefore output here is 0. That is what the mathematics shows here, that I4=0, when theta is equal to 0 and = 90°, there can be several numerical based on the physics, the simple calculations that i have discussed here. You can have different polarizers at different angles: two polarizers three polarizers and so on the picture. If it is clear, then all these numerical can be worked out.

So here I stop the discussion on polarization wave optics and the optics module. Thank you.