 starting our first lecture on Geometrical Optics. So, what is the first thing that comes to your mind when we talk of well Geometrical Optics? I suppose the first thing that comes to your mind is you know the passage of light through lenses, how curved lenses that is a convex lens or a concave lens or maybe reflection from curved mirrors. So, but one has to realize that this is rather pretty old subject, this is rather a pretty old subject in physics. We have been doing things for hundreds of years. In fact, it is the use of these lenses and mirrors which had a huge role in the development of astronomy. For example, Galileo was the one who used just two lenses, he just used two lenses and then put it in a cylindrical tube and he watched the boons of Jupiter and he made the certain astronomical observations. So, in a sense well this subject is it is one of the primary things which was responsible for advancements in astronomy and not only in astronomy, I mean not only in physics for that matter, I mean if you think of the simple magnifying glass to look at small things. I mean I by bet the modern biology owes its origin to the observation of small things. So, the study of geometrical optics had rather far reaching consequences not only in the development of optics, but also in other branches of physics and also in other branches of science. So, now that we know that it is an important thing to study, although it is an old subject, what is the area of applicability? So, rather at what wavelength range range should we apply geometrical optics? Now, why do I say that? Because of the back of your mind, you also have things like diffraction, interference which have been explained so well with the electromagnetic waves. So, what is the area of applicability here? For that, let us draw a simple diagram and define our area of applicability. Let us say we have a small slit or a small hole and we are going to put some source in front of that, some source of light in front of that. Now, we put a screen a little distance away from this hole or from this slit. Now, what do we expect to observe? Well, I mean we have a reasonably big hole here. So, we can draw these two lines, we will come to this point that we are drawing lines here and then we observe the image of this slit or this hole on the screen. Now, which is kept at a certain distance away from this slit. Now, if you observe you have sharp boundaries for this image and you just observe only one image of this hole, then I am sure you are going to say that we are in this regime of geometric optics. Now, why? I mean look at the other look at the other extreme. I mean if your hole is small enough, you will have if your hole is small enough, let us say let me draw another picture. If your hole is small enough you can have and then you put a source of light here, I call this some source. You are going to have alternate regions of light and dark regions. Now, that is diffraction. So, you have alternate regions of light and dark regions here. But the field of geometrical optics would not be too concerned with this diffraction, but would be talking of things in which when you have a slit, the image of the slit is very well defined on the screen. Now, when does that happen? That happens if the size of the slit that you have taken, let us say the size of the slit is D. Now, that is much greater than the wavelength of light that you have used. So, this is one important criterion for the application of geometric optics. And well, if it is actually if it is not, if the slit dimension is some way it is very near or comparable to the dimensions of the wavelength, you will as I said you will have to deal with things like diffraction. So, this is one thing about the sizes. The second thing is a bit involved, but then it is another conceptual thing. It is that the light or the energy of the photon in this case should be lesser than the energy sensitivity of the instruments that you are using. So, it should not interact with the instruments. So, basically you are not in the quantum region, quantum regime here. So, we are still in this classical regime and we will be using what we call as these lines or these rays of light. Let me draw it once again. So, these are the rays of light by which you are going to get some sharp images on the screen. Now, what are these rays of light? These rays of lights are nothing but infinitesimally thin. So, these will be using the rays of light. So, they are the infinitesimally thin beams of light. Now, for that reason geometrical optics is also called ray optics. So, that is what we are going to study. We are going to study ray optics or geometrical optics here. Now, let us do one more thing. Now that we know that we are going to deal with rays. So, the next question to ask is what path do these rays take? So, the question that we ask ourselves is what path does the ray of light take? Now, this was something which bothered many people and quite early. And apparently people had lots of ideas where they said that it will follow the least distance between two parts. Well, if it is the same medium, that is fine. I mean the least distance is fine. But that may not be the case in which you have in the media of different refractive indices in between. Now, I will come to that again. Now, apparently there is a principle which tells you what path the ray of light will take. Well, it is called Fermat's principle or in its original form it is also called the principle of least time. So, we will be talking of the Fermat's principle or the principle of least time. Well, it simply states that well the ray of light will follow that path from one point to another in which I have point A and point B. Well, you can take whatever refractive index in between. Well, it will follow from point A to point B the path in which the time taken will be the least, will be the minimum one. So, that was Fermat's principle in its original form. So, that is what Fermat said that light is going to travel from A and B and the path it is going to take and it is going to take is that for which the time is going to be minimum. So, instead of this path or the other path the one from A to B will be the one. The one the light will take and the ray of light will take the other will be the one in which the time taken is the least. Now, apparently the whole of geometrical optics can be derived in principle from this one simple concept. So, that is how important it is. Well, in its this is the statement in its original form it is the principle of least time, but I will come a little later on some modifications, but let me just tell you 2 or 3 consequences of Fermat's principle. So, let us just do one or 2 consequences of Fermat's principle consequences of Fermat's principle. Well, the first is that if from point A to point B let us say light is traveling and it has taken the path of least time. So, point A to point B suppose this is the path and this is the path of least time from point A to point B. Well, if I start from point B to point A the light will actually take the same path back. So, if it was a least time on one way that will be on the other way also it will be the least time compared to other paths. Now, this is called the principle of reciprocity. So, it retraces its path back. So, this is the principle of reciprocity. Second thing is about well I mean what comes to my mind is I mean I started this talk on you but to talk by talking of some lenses. So, let us draw a lens let us draw a simple convex lens and on the axis. So, light let us say I have some point on the axis on the left hand side here. So, this is my convex lens and then let us say light from this point truly comes and then it comes here and then I have well comes on the other side. So, this is the object distance and this is the let us say the image distance which I denote by V. Now, there are well there are infinite paths of light let us just take two or three of them. There is one path which goes entirely through air. So, the one up at this point and then comes and comes back here. There is one which goes partly through air which is actually along the axis and then there is one path which goes through the lens and then comes through and then comes travels in air to reach the image point. Now, if you look at these two points the length of these two points are different the path lengths are different. The actual length from on the axis that is the least compared to the length when it goes to the tip of the lens and then comes to the image point. Now, however, by the principle of least time you can argue that the time taken by these two light rays but this one the one in which I have drawn the one goes to the tip and then comes to the image point and the one in which it passes through the which is along the axis of the lens. They both of them take the same time otherwise you are not going to have an image on the other side. So, of course, I can also derive the simple laws of reflection and refraction using Fermat's principle. But before that I wish to remind you once again about well maybe I have not said this before is about the statement of Fermat's principle. Now, we do not say that the time taken is minimum what we say is that the light will follow that path. So, from A and B so, Fermat's principle the way we say it now Fermat's principle is that light will follow that path will follow that path for which the time taken is the time taken is actually an extremum is actually an extremum compared to the nearby paths. Extremum compared to what? Compared to nearby. Now, when I say an extremum I do not actually specify whether the time is actually minimum or maximum it is an extremum. So, pictorically or diagrammatically how do I write that I mean I have point A and point B let us say point A and point B now there are actually infinite number of paths which connect point A and point B. Now, what it tells what Fermat's principle tells is that light will follow that path for which the time taken is actually an extremum compared to these paths. Compared to for example, it will be it will follow the path in which I have shown by a solid line as opposed to and that will be an extremum path compared to let us say the ones which I show by the dashed lines. So, that is how we know Fermat's principle now. Now, this is actually a far reaching consequence of the principle of least action or again it is the Hamilton's principle in mechanics. Now, let us as I said we can derive the laws of reflection and refraction from Fermat's principle. Let us see how we can do that and so that will give us some more confidence about the applicability of this theory of this principle. So, we want to check what happens during reflection draw a figure first. So, here we have a mirror a plane mirror let us say and say we have a point A and a point B. So, let us say we have a point A and a point B. Now, the thing that we want to do here is that we want a ray of light to come and hit the mirror reflection and go to B. Now, which direction should it go? Well common sense and what we have learnt in school is that well it should go in a such a way that the angle of reflection should be equal to the angle of incidence rather should be equal to the angle of reflection. Now, why is that? Because you see from going from path going from point A to B via the mirror I can actually have an infinite number of paths. Now, why should light choose the one in which the angle of incidence is equal to the angle of reflection? Let us find out using Fermat's principle. So, let us draw a normal at the point of incidence and consider that this point A is at an height h1 from the mirror point B is at an height h2 and the distance on the mirror you know the distance of the mirror is l let us say and let us also think that the distance at which the light strikes the mirror that is x. So, let me see have I drawn all the distances? Yes, I have taken the distance in the mirror the distance where it hits the light hits the mirror is incident on the mirror. So, I have not drawn the angle of incidence let me call this I and the other one here is the angle of reflection let me call this R. So, and this is the normal. So, the one this dotted line is normal and I bet you are going to say that what will be this angle? This angle is also R and this angle is I and these are all right angle triangles. So, let me I have not labelled these points height of this is P let me call the point at which the light is incident as O and let me also call this height. So, B this is Q. So, P Q is l P O is x some distance and the height from the mirror is h1 here and the height from the mirror height of the mirror of B is h2. Now, the thing now the question that I am trying to address here is I mean light comes and hits the mirror at a distance x here well it could have hit a little bit I mean if it hits at x what happens to the angles this is one thing or to put it in another way round I mean why should it hit only at x it can hit at the other places and come to B. Now, which is the one according to Fermat's principle should be the path of the light here. So, let us do this what is the time it takes for light to go from. So, time to travel let us say to travel the distance A O B this distance A O B let me say let me call that time t and the velocity of light in this medium. So, let me say this medium itself is air. So, velocity of light let us say the speed of light is c. So, what is the distance A O well the distance A O is nothing but h1 square plus x square pi by c that is the time it takes to travel the distance A O and the distance O B. So, that will be square root of h2 square plus l minus x square divided by c. Now, what do I do now? Now, I say that the time will be and compared to different paths compared to its neighbouring or nearby paths. So, this is going to be I am going to have this is going to be a next step. So, that is what I have. So, I have let me let me do this dT dx. So, this is equal to what is dT dx? I am going to taken 1 by c here that is the speed of light that is a constant thing I am going to differentiate the 1 in the square root. So, I will have a 2x at the top and then if I differentiate the entire thing. So, I am going to have a 2 times 2T over h1 square plus x1 square x square in the denominator. What about the 1 on the on the for the for the part O B? So, this is twice l minus x. So, I am going to differentiate this point h2 square plus l minus x whole square and the square root of that. So, that is equal to twice of h2 square plus l minus x whole square. So, I have not taken a minus sign out. So, because l minus x, okay fine I put a minus sign here and then there is a 1 by c. So, when I say it is an extremum I am going to put this equal to 0. So, which immediately tells me that x by root over h1 plus x square is equal to l1 minus x h2 square plus l minus x root of that. So, what does this tell us? So, it tells us that if I want the time to be an extremum, okay, compared to all different parts light can travel. Now, if this is the path light should travel, okay, I better have x by well I should have put an h1 square plus x square root over of that. That should be equal to l minus x h2 square plus l minus x whole square square root of that. Now, what does this mean? Let us look at this figure. Let us look at this triangle APO, okay. Now, what is x? What is x divided by h1 square by x1 square by x square, okay. So, let us look at triangle APO, okay. Let us look at the figure triangle APO. Now, in triangle APO, what is the sign of the angle i? Well, it just happens to be x by x by the distance h1 square by x square, okay. Now, compare that in triangle BOQ, what is the sign of the angle r? Okay. Now, these are what? So, sign of the angle r is l minus x divided by h2 square plus l minus x whole square, take the square root of that in the denominator. So, which tells you immediately here that light will travel that path in which I am going to have just the, you know, in this same medium, the sign of this angle i is equal to the sign of this angle r or in other words, I need this angle i to be equal to the angle r, which is actually the familiar laws of the reflection that you have, okay. So, well, to think of this as a mechanics problem is also quite interesting. Actually, if you, if you, if you give, if you, if you set a problem like this, that I want to run from point A to let us say point B by touching some wall, okay, which is PQ. Now, which path should I take so that the time taken will be released here, okay. Well, actually we will be taking that path for which, you know, the point at which you hit the wall and then you draw a normal to that, then you find an angle, okay. That should be equal and you still call that an angle of incidence, okay, should be equal to the, the angle that you make with the normal when you run away from the wall to reach point B, okay. So, this again will actually follow from the principle of least action in mechanics. But right now we are doing geometrical optics, we are dealing with light, rays of light. So, and we have just now verified again the laws of reflection using Fermat's principle, okay. Now, with this basic verification, how about doing it for the, for refraction, okay. So, you can always say that, okay, fine, I have done this for a medium in which the refractive index is the same, right. So, your reflection is in this, in this region in which, you know, in the region A, O, B, that is in a region where the refractive index is the same, okay. So, let us find out what it is going to be when you have two different media. Let us draw a figure once again. So, okay, so we have two media, okay, which is infinite on both sides, let us say. So, in one side, the refractive index is N1 and on the other side, the refractive index is N2, okay. Now, you have, you are at point A, let us say and you wish to go to point, some other point, okay. Now, what is the point, what is the deduction that you need to follow, okay. In case you want to go to point A to point B, is it, is it the straight line? I should draw the point B here, is it the straight line? Well, that is the distance, that is the least distance. But then remember that the speed of light is different in these two media, okay. So, the time taken is necessarily not the same. If you go, it is definitely not the same when you, if you follow a straight line or let us say you travel this much in one media and then the other here. Or let us say you travel this much in media, in medium one and then the one here. Well, it is more like you, you, you want, let us say N1. So, this is region 1, let us say this is land and then N2, this is some water, okay. You want to go from point A to point B, okay. What do you do? Do you jump straight, you go straight into, do you go straight to the shore where water is and then swim all the way to B, okay. Or do we go, do you want to minimize your swimming because you know swimming, that is, that is a bit tougher than just running on land, okay, because of resistance. So, do you maximize your path on land, okay. Let us say you come here and then you swim a little bit in the sea or in water. Let us not bring in sea or anything else in water. So, which path should you take? Well, the problem is the same. So, let us redraw this figure once again and label it. So, let us say you need this to be the path and this is how you wish to go. Okay, you are in point A and this is the point you are incident on the interface of both the media. So, you want to go to point B here. Let us say again I call this distance, let me call this distance, well, this distance h1, this point as p, this point of contact within these two media where light comes and hits as o and then B and this to be h2 and then I call this pq. So, this point is q, okay. Now, I again take this distance. So, this distance is fixed, this distance pq, this distance pq, this is let us say let us call that as l, distance l. Let me call po. So, that is x, okay. So, that is variable. Where do I want? Where do I want to hit the surface here? Do I want to hit it here, here, here or here so that the time it takes compared to all these paths to go from point A to point B is an continuum. Well, let us do that. So, definitely what is point oq? So, this is l minus x, okay. So, what is the time it takes to travel this distance AO, okay. So, the time again to travel AOB, this distance AOB. So, this is let us say capital T. So, what is this distance by the way? So, this distance is AO in medium 1, okay, where I have the refractive index N1. Now, how is the velocity or the speed of light related with the refractive index? Well, if the speed of light in vacuum, air in vacuum, let us say is C and then I divide that in a medium where the speed of light is V, then the refractive index is N, right. So, what do I get? So, the velocity of light in a certain medium which has a refractive index N, which has a refractive index N is nothing but C by N, where C is the velocity of light or the speed of light in vacuum and then N is the one in N is the refractive index here, okay. So, the time it takes to go to do the distance AO, what is that? Well, let us find the distance. So, the distance is so this is x, okay. So, this is nothing but again H1 square by x square. Now, you need to divide this by the velocity in medium 1 or the speed in medium 1 which is C by N1 because you are in medium 1. Now, we need to add again, so this is the time it takes for light to traverse this distance AO, okay. Now, what about this distance OB? Well, that is H2 square plus L minus x whole square divided by C by N2, remember the speed of light in medium 2, I call that velocity V2 and that is again dependent on the refractive index, the C is the speed of light in free space or the speed of light in vacuum itself, right. Now, we need to take the next term of this. So, which means we simply take dt dx is equal to 0, so which tells us that well C and N and all these things are constants here. So, we need to differentiate this entire expression with respect to x. So, what is that? So, that is N1 by C, I am going to take a 2x upstairs and so and then I differentiate that that will be, so that is H1 square plus x square. So, differentiate 2x square, so I get a 2x here and then again H1 square by x square. So, this half by this entire thing to depend 1 minus half, so that is okay. So, square comes down and then I have again N2 by C, the other one, I get okay, so l minus 1, l minus x, so I will have a minus sign here and so I get a 2 of l minus x okay and again H2 square plus l minus x whole square, okay. So, what is the point? Well, what is the time like we will take once it goes from point A to B and via this, from one medium to another. So, we will be calculating the time and the time it takes to travel from A to A, A OB, we have calculated that. Now, we are estimizing, we are doing dt dx is equal to 0, we are trying to find out what is the extremum of that. We have found a condition okay. Now, what is that condition once again? We have just found the condition that N1 over C, remember N1 is the refractive index in medium 1, so times x by root over H1 square plus x square and that should be equal to N2 over C root over N2 over C and then l minus x root over H2 square plus x square okay. Now, what was the thing that we were doing? We were looking at refraction okay, from medium 1 to medium 2 okay. Now, this angle was i, let us say that is the angle of incidence and then this was height H1, so this was x, so this was A, it is at B and then the angle of refraction was R okay. So, this distance was H2 and so this was P and this was Q, so an OQ, this was l minus x fine, definitely this angle, that is angle OBQ, that is also R okay. Now, you have got a relation regarding some the refractive index times something times x, x by something some distance okay. So, let us find what that angle is, so by the way, so this angle is the angle of incidence, the angle PAO okay, that angle is also i right. So, in triangle APO what do I have sin i to be, I have sin i to be x by the distance AO which is nothing but root over H1 square plus x square and in triangle BOQ what is the sin of angle R, well sin of angle R nothing but l minus x divided by root over of H2 square plus x square. Therefore, when we come back to our old equation which we have got by taking the time to be by making the time to be an extremum okay is if we substituted back into our old equation, when we were this equation here, this one, what we get is n1 sin of i is equal to the is equal to this value n2 sin of R okay, where i is the angle of incidence, that is the angle with which the light makes or the ray of light makes with the normal when it strikes the boundary of these two media and then R is the angle, that is the angle in which light emerges in medium 2 okay. Now, this you know everybody knows this as Snell's law in optics. Now, you see that we have simply obtained the Snell's law in optics using Fermat's principle and remember this mechanics problem I was telling you. So, if you wish to go from point A to point B, let us say and then the region 1 was land and region 2 was water okay. So, what is the route you need to take? I mean you better take the route A or B okay. That is also the thing you can work out using the principle of least action of the in mechanics right. So, to summarize what we have done today. So, we have talked of the beginning of geometrical optics, we have talked of the fundamental guiding principle in geometrical optics which is Fermat's principle which actually tells you how the path of array of light okay. So, we have done that for reflection from a plane mirror and we have also seen the path of light, the ray of light will take when it goes from one medium to another okay and in doing so what we have also found is that we are right because we have reproduced the familiar laws of reflection and the familiar laws of refraction here fine that is Snell's law. So, thank you very much.