Hello! Welcome to this lecture module on Optics. In the last lecture we discussed about reflection by spherical mirrors and, in particular, we discussed the formation of images by a spherical mirror. We also derived the mirror equation, which tells us precisely the position of the image when the position of the object is given. So today we will move on, and we will discuss refraction of light. Refer Slide Time 01:32 To begin with, we will first discuss refraction at a plane, interface, refraction of light refraction of light from a transparent, medium, a transparent medium. It has been historically known for a long time that, when light is incident on a transparent, medium such as glass or surface of water, then a part of the beam is reflected back, and a part of the beam is transmitted into the medium light incident. On a reflecting surface, here it is a transparent, dielectric, transparent medium, a part of the light is reflected. We know that this satisfies the law of reflection, that the incidence angle is equal to the emergent angle and a part of the beam is refracted or transmitted into the medium. This was also known. It was also observed that the angle of refraction was not equal to the angle of incidence, while the angle of emergence here - that is the reflected angle, the angle subtended by the reflected b was equal to the angle of incidence. The refracted angle here, that is the angle subtended by the refracted beam or the transmitted beam into the medium, was not equal to the angle of incidence. This was historically known for a long time. This is partial reflection and partial transmission of a light beam. A light beam incident, a part of it is reflected and a part of it is transmitted. So, this is called partial reflection.

Refer Slide Time  03:26 Now, if we consider an air water interface in particular, because water air interface is very commonly encountered in day to day life that a light beam incident on a water surface, a part of it is reflected back and a part of it is transmitted into the medium. As long as in the year 140 AD, the Greek Physicist Ptolemy, had tabulated the angle of incidence and the angle of refraction for different angles of incidence: , etcetra. He had measured the angle of refraction and tabulated and given as a table. If this is the angle of incidence, then this will be the angle of refraction, but he did not know any further about the relation between them. However, in 1621 W. Snell, based on his experimental observations, found that ; for a given medium for refraction at an interface. This is subsequently known as the Snell’s Law. The Snell’s Law is now given by ; a constant when we have an interface between a medium 1 and medium 2 of refractive index and then, if is the angle of incidence and is the angle of refraction, then , where is called the refractive index of the second medium with respect to the refractive index of the first medium, that is, Refer Slide Time  04:54 All subsequent discussions is the angle of incident where the ray is incident, we call that as medium number 1 and there could be 2, 3, 4 and so on. So the refractive index of the second medium, with respect to the refractive index of the first medium, One note that and This has some important applications; we will see this. The medium with higher refractive index higher of the two is called the denser medium, and the medium with lower refractive index is called the rarer medium. This denser medium density has nothing to do with the mass density, which is . So, this has nothing to do with that. This is denser here; refers to a higher refractive index of the medium and rarer refers to a lower refractive index. It is a relative term. It’s not any absolute value, it is just relative to the other one. Denser relative to the first medium here, so rarer medium is a medium with refractive index lower than the second medium or the other medium. Refer Slide Time 07:23 Now let us see in particular refraction when light is entering from a rarer medium to a denser medium and when it is entering from a denser medium to a rarer medium. Note that angle of incidence , angle of refraction , here I have considered air-glass interface just as an example. Air has refractive index 1, and glass has 1.5, so the light is entering from the rarer medium to the denser medium. So, , because

In other words, the ray bends towards the normal. This is the normal here, normal to the interface. This is the interface between the two media, this is glass, and this is air. This is called the interface, and this line here is the normal to the interface and the angle, with the normal to the interface is the angle of refraction here, and the angle of refraction in this case. When light enters from rarer to denser light bends towards the normal and when it is the reverse, that is when it enters from a denser medium to a rarer medium, so, for example, glass and air. This is the interface then, In other words, the ray bends away from the normal. So, the ray here, the transmitted ray or the refracted ray bends away from the norm, bends away from its original direction. The dotted line here is the original direction, so it bends away from the normal, whereas here the original direction is here, light bends towards the normal. The ray here, we are referring to the refracted ray.

Refer Slide Time 09:46Let’s, take now the refraction through a glass slab. That is, now we will encounter two interfaces. Earlier, we had discussed refraction at a single interface. Now, we are discussing refraction at two interfaces. That is, what happens if we consider a glass slab here, a rectangular glass slab. Then it has one interface here and the second interface here. At the first interface is between air and glass. At the second interface, it is glass to air, so consider incidence of a ray here, making an angle , then at the first interface, the ray bends towards the normal and therefore here, is the refracted angle. I have now used the notation when there are more than one interfaces or several interfaces. It is convenient to use and so on, rather than and , because there will be more and the same would become. for the next interface and so on. Therefore, it is convenient to useand so on. So, I have used here, and is the angle of refraction. Earlier, this was and , but now we have a second interface where this ray, that is, the refracted ray at the second interface subtends, an angle , which is the angle of incidence. Now, as far as this interface is concerned, and then is the angle of refraction at the second interface. Now applying the Snell’s Law at interface 1 and 2. We have noted with the subscript, so is and is , is the thickness of the glass slab. is the lateral shift. We will talk about in a minute. At the second interface, is now the angle of incidence. So, Now there is the third medium here that is air and therefore So this simply gives You can multiply the two equations and you see that and cancels, and cancels here and one, and therefore What it means is when the ray passes through the glass block, the ray which is coming out makes the same angle which is equal to , which means, as far as the direction of the transmitted ray is concerned there is no deviation. However, there is a lateral shift. As you can see here, there is a lateral shift. So, no deviation but a lateral shift of the ray and this lateral shift depends on the thickness of the glass block, as we will see later now.

Refer Slide Time 13:35 I extend it further, so we have now considered two interfaces but suppose I have now several interfaces. So, we look at refraction through a multi layered structure. Now there are four layers, 1, 2, 3, 4 and, of course outside here is air and outside here too. This is a stack, comprising of four layers of different refractive indices. They are , and they are different. So, this is a stack comprising of four layers and therefore there are six refractive indices, one from outside here and one here. Now, if we apply the Snell’s Law, it has to be applied at each interface. Depending on whether it is going from rarer to denser or denser to rarer, the ray will bend away or towards the normal. We can see here, for example, the ray here is bending away from the normal and again away from the normal. I have just taken some refractive indices, but we have not given any values here. If we apply the Snell’s Law at each interface, so the first interface or we can cross multiply this ; this is a more convenient form of Snell’s Law. If we apply to the second interface here, then it gives us The angles are indicated here: is the angle of refraction here, which becomes the angle of incidence when the second interface is considered. is the angle of refraction here, which becomes the angle of incidence for this interface and so on. So we have

here is the angle of refraction, which becomes the angle of incidence here for this interface and here is the final angle of refraction. If these are all equal, it simply means If the first and the last medium are the same in this case, for example air. This was a stack of four layers on both sides. There is air, a ray of light was incident here, and the ray is emerging from here. If the first and last medium are the same, it may not be same; it could be water in another problem. This could be water, for example, but if they are the same then , which implies there is no deviation, that is, the final emergence angle does not depend on the thickness and refractive index of the layers. Then what is the need? This is quite interesting to see that the angle of deviation, there is no deviation which is independent of the refractive index and thickness of the layers. Then what is the need to use such a multi layered structure? There are large number of applications of multi layered structures in optics. To understand this, ray optics will not be able to help us. To understand and design these applications, we will have to go to wave optics.

Refer SlideTime 17:30 However, let us see something so, we will come back to that a little later and here we summarize. Therefore, the laws of refraction:

the incident ray, the reflected ray and the transmitted ray or the refracted ray - lie in a plane perpendicular to the interface. So we can see here. This is the incident ray. There is a reflected ray and there is a refracted ray, refracted or transmitted, because the energy is partially transmitted into the medium and partially reflected from here, and therefore all of them lie in one plane, perpendicular to the interface. The Snell’s Law, which is , which is more conveniently written as or could be the angle of incidence.

Refer Slide Time 18:48 Now we will see we will discuss some natural consequences of refraction of light, so some natural consequences of refraction of light. The first one that I have shown here is apparent depth is illustrated here if we have a beaker of water or a container which has water, and if there is a coin at the bottom, I have taken a point P. Maybe a point source here at the bottom, then it appears as if the depth where the point source is present is smaller compared to the actual depth. That is illustrated here why this is happening, so you are observing from here. Try to show I and here is a point P. It could be a point source. So therefore, light is emitted, it could be an object. We know that when we discuss the image of an object, then also we consider rays coming from a point on the object, and this could be a point source as well so point source rays, which come out from here are refracted at the interface. This, for example, is water, and this is air, so it bends away from the normal at each of these interfaces. We can mark the interfaces here. So, this is the interface and if you draw the normal here, then light bends away from the normal, because refractive index here is smaller than refractive index here and it bends in this direction. Therefore, it is a diverging beam. What you have here is a diverging beam, but all of them, if you extrapolate here, then they appear to come from a point , which is different from the actual point P. In other words, if we look from here, it appears as if the apparent depth, the depth of the point , that is from the surface to the point , here is . I have denoted it by the apparent depth. Of course, the point P is at the bottom of the container. The actual depth is d and is the apparent depth. So, the apparent depth in this case is smaller compared to the actual depth. We can quantitatively determine how much it is smaller.

Refer Slide Time 21:39 So let’s just continue here and see this now. The equivalent problem, as shown here. The point P was here: a ray of light was at an angle. here it is angle of incidence at this interface, comes out here with an angle of refraction , and if I extrapolate it here, that is, if you were observing from here, the ray appears to come from the point , and this angle here is , because the angle of refraction here is and therefore this angle is angle of incidence is here, and this angle is also because these are two parallel lines, so this is a ray which is normally incident which transmits, through the medium of course, it’s also partial transmission. Everywhere it is partial transmission, so is the apparent and depth d is the actual depth. Now for small angles , these are actually small angles, because we are looking from here, so I can show the , which is here so the is here actually so this is the which is observing the point P. Therefore, the rays which enter, are those which make very small angles. The ray which would come here would enter the eye. A ray which is coming here is any ray which is making a larger angle does not enter eye. So, all the rays which enter your eye here are the ones which make a very small angle. Therefore, the approximation is very much valid for small angles and . If is small is also small, although is a little larger than , but it is still quite small. Therefore we can write ; for small angles From and similarly Therefore, by Snell’s Law. ; if we divide one by the other, it is which is nothing but . So, . is always the second medium and is the first medium by notation. Usually, is air, and is from where we observe air, as in our problem, this is the liquid, where the point was at the bottom of the container, and here it is air apparent depth is equal to actual n2 is equal to one, therefore actual depth. If we take apparent depth, actual depth comes here. The as and is the . So If we consider glass or water, then we see that the apparent depth is smaller compared to the actual depth. Everyone can observe this. In practice, we will take up some numerical to illustrate this point further.

Refer Slide Time 25:46 As a second example, we look at the inclination of the setting sun. Inclination of the setting sun, apparent inclination, or apparent position of the setting sun. So, the diagram here is not to scale it’s just a schematic, because what I have shown is the earth has an atmosphere around it. Up to few hundred kilometers, earth has an atmosphere. This is, of course, several thousand kilometers. Therefore, this diagram is not to scale. So, earth surrounded by an atmosphere and beyond that, of course, it is free space, and the stars and sun are in free space which are far away from the earth. So, this distance is, of course, much larger compared to the thickness here or the width here of the atmosphere so it’s not to scale, but just a schematic illustration. What is being illustrated is the following: There is an observer here on the surface of the earth. The size of the observer is, of course, negligible compared to the dimension again, not to scale so the observer here looks at the sun. If this is the horizon, then the observer is looking at the sun. The apparent position of the sun is here, he is looking at the sun, which is above the horizon, but the actual fact is that the sun is below the horizon, because when the when the sun is here, I consider a typical ray one ray just to Illustrate the ray which is coming from the sun. There are, of course, large number of rays, a bunch of rays which are coming but array which is coming like this when it enters the atmosphere, free space, or vacuum, has refractive index n is equal to 1. Exactly 1.0 and atmosphere here, which has air and other gases, has refractive index slightly more than 1. Maybe 1.00 something, but it is slightly more than one and therefore the ray is entering from a rarer medium to a denser medium and it continuously bends towards the normal. We can stratify this, for example, we can see that this is when it is entering. So, this is the atmosphere, so the ray is entering like this ray is entering here. There is a higher refractive index. Therefore, if I stratify this, that is, if I consider this into a large number of layers, then the ray which is incident here bends towards the normal, so it bends towards normal it bends towards the normal it bends towards the normal. It slowly, slowly, why slowly? Because the refractive index here is 1 and maybe here it is 1.0078 or something like that. But it is slightly larger and therefore the ray is continuously bending towards the normal. And therefore, we see that, however, the observer who is here when it reaches the observer, then the ray is entering his eye like this. So, it appears to him as if the ray is coming from some point here. If this was the position of the actual sun, then the sun appears to him as if it is coming from a point which is here, whereas the horizon is here, so this is the horizon, so that is the diagram. Let me put the pre drawn neat diagram here, so the ray we can see here starts bending and continuously towards the observer, and the observer finds it as if it is above the horizon. This is known as inclination. The apparent inclination of the setting sun. This is an illustration which shows that, although the refractive index is very small because the length of the atmosphere is of the order of hundred kilometer or two hundred kilometers, then it starts over that period. It bends significantly and there is a significant difference between the actual position of the sun and the apparent position of the sun. We have these two natural examples which I have illustrated. We will take up some numericals to illustrate refraction of light and let me take up some examples. With these examples, we have got a better appreciation of the refraction of light, and there are numerous problems which are possible, particularly the multi layers are very important, and

Refer Slide Time 31:09 Let us take some examples now to illustrate what we have studied. Here, a narrow beam of light travels from medium 1 through three layers of different, transparent media into medium five, as shown in the figure, so see the figure. A narrow beam of light travels from medium 1 here through medium 2, medium 3, medium 4, into medium 5, as shown in the figure rank the media in ascending order of their refractive indices. We have to find out, which is the medium with the least refractive index to which is the maximum one, and therefore we have to ask we have to rank them or list them in ascending order from the lowest to the highest refractive index. The data - the figure shows angles here - 45, 30, 40°, 50° and 35° here. So, how do we go about this? We use the Snell’s Law. We use the Snell’s Law in the form that ; for any given medium. We apply it to this and find out which one would be the lowest refractive index. Using Snell’s Law ; so a medium which has the largest angle here, which means will be largest. ; therefore, when the angle here that is , this becomes here this is, but here so when the angle is the largest, the refractive index must be the smallest because increases with and therefore medium 4 must be having the smallest refractive index. The smallest one is medium 4, the largest angle. So medium 4will have the smallest refractive index. Let me rank it in the ascending order of refractory index. So this will be: medium 4 The next angle that we see is 45 ° here. The next largest angle is 45 Therefore, medium 1 will have the next higher refractive index. Medium 1 Then we have 40, so Medium 3 Then after 40 we have35. Medium 5 And finally, the smallest angle that we have here is for Medium 2; and therefore, this will have the largest refractive index. We have now ranked the various media here in ascending order of refractive index medium, two, where it makes the smallest angle will have the largest refractive index and medium 4, where it makes the largest angle here. This is the rarest medium among all of them. You can see that it is the rarest medium. That is why it goes well above the well beyond the normal bending away from the normal making a large angle, 50 ° here so medium 4>1>3>5>2 As I mentioned that when we have several media, then it is easier for us to write the Snell’s Law in the form of ; for every one of the media. This is like a quiz question. Very quickly we can identify without doing any mathematics. Just looking at the angle we can identify, which are the ah media that have largest refractive index.

Refer Slide Time 35:44 Let me take another example here, so let us go to the next example. A glass beaker of height 10 centimeter contains water of refractive index 1.33, up to 4-centimeter height from the bottom, and then a transparent oil n equal to one point three, one above water up to the top edge of the beaker. This here I have tried to draw the diagram. There is a glass beaker here. The first of total height, 10 centimeter and the first 4 centimeter is filled with water and the next 6 centimeter is filled with that transparent oil of refractive index When viewed from above that from the top, when viewed from the top, what would be the apparent depth of a small coin located at the bottom of the beaker? There is a small coin placed here at the bottom of the beaker. What would be the apparent depth when viewed from above so this is actual depth, is 10 centimeters, but will it appear as 10 centimeter deep or will the apparent depth is smaller or larger than the actual depth? That is the question: to determine the apparent depth of the bottom of the beaker. A small coin is placed, or it could be a point source. It could be a point P at the bottom of the beaker, but basically to estimate the apparent depth of the beaker. Let us comprehend this problem a little bit more carefully. Here is the beaker. Let me draw that again so beaker and there is water up to a certain level, so 4 centimeters of water. This is 4 centimeters, and this is 6 centimeters. So, this is of refractive index 1.31. This is 1.33, slightly different refractive indices and viewed from above, which means you are viewing from the top here so viewing from here. That means the eye is here. I am showing eye a little bigger, just for convenience. The point to see is when you view this: a bunch of rays enter the eye. There is a small cone, so a cone over which rays enter the eye so rays enter over a small cone coming from the bottom. If I have a point P at the bottom, or there is a point P here - a point source P, then a bunch of rays which come out will enter the eye over a small cone. This angle, here is very small. This angle of the cone is very small. However, this, as we will see, will lead to an apparent depth and we are asked to determine the apparent depth of the coin in this mixture of in the in the liquid in the beaker containing two different liquids. So here I have drawn a neater figure to illustrate the problem more carefully.

Refer Slide Time 39:15 Here is the solution, so the first medium of refractive index , height . I have not put any numbers. We are handling it in general, analytically and second medium is of refractive index and of height and the third medium which is here outside, which is the air in this case. Let me call it as, and here is the eye. We are viewing from here; the eye is here. In the last diagram that I had made, I showed that in practice this angle is very small when you are viewing from the top, but it is not necessary that I should be viewing from the top. Eye could be viewing from an angle, so even then a small cone of rays would pass through this. So, eye may be watching from here, so my eye could be here, so this is viewing from above, but this is viewing at an angle. This angle, eye may be watching at an angle of 40°, for example, so both cases are keeping both cases in mind. Here I have tried to analyze this problem, so here it is from the point P. The point source here, ray which is coming here, is incident at an angle. The refracted angle in medium two is and the refracted angle in medium three is. See the diagram, if you are watching from here, if the eye is here so you may not be able to see, let me draw an eye there. Here is the eye then. The eye is watching this point observing this point, which is coming at an angle theta three from the geometry. What we can see is if this angle is , this angle is. I have marked this distance as and therefore from the geometry, we can see that are the thickness of this water column here and h is the apparent position of the point object P. There is a point object here, but eye sees it as if the point object is located here. In other words, is the apparent depth in this problem. is the apparent depth, so we have to determine what is the apparent depth? The actual depth is, of course, , the total height from surface to the bottom, . But the apparent depth is . Looking at the geometry,. OR Now from the geometry, we can also see , because this is parallel to the normal here. This is also a normal and therefore and therefore . However, here is equal to this. Height is Therefore Therefore, Similarly, this is , and this is . Therefore Therefore, Note that we have not made any approximation here, there is no approximation involved and therefore this is valid for any angle , at which the observer is looking at the coin. Therefore, if we know that the observer is looking at an angle , I can calculate by using the Snell’s Law, because the refractive index of the media are given Therefore I can calculate . If I know I can calculate and therefore I know , and therefore . The apparent depth For any can be determined using Snell’s Law. However, in this problem it is said viewing from above viewings from above means, we are viewing right from above, which means, as I indicated in the problem, a small cone here. So, we are looking at from above, which means a small cone of angles which we are considering, which enters the eye and therefore viewing from above implies the angles are all small angles and therefore This approximation is very much valid for all small angles and if we apply that and substitute here that instead of , if you substitute and , then we will get We can simply substitute that and see so We know that and therefore and Therefore, this expression here is Because our surface here is air, we are looking through the liquid medium here. Therefore, it is understood that it is air here. Therefore, . With we have

Refer Slide Time 46:48 We are asked to find out the depth. So, what is the answer? The apparent depth and , and So, There are several variations of the problem, which is possible, so can be, may be larger than , may be larger than and so on. Several combinations are possible.

Refer Slide Time 50:08 There is also an interesting extension, so here it is. Let me just discuss an interesting extension, a particular. So here is a point P. A point P could be a point source p, which is observed by an observer here, so the actual height that he would see, or the point P is located at some depth here, so some depth . Now we introduce a block. So, if someone introduces a glass block of certain thickness , and refractive index , so, this was this - was here now a block is introduced. A thick glass slab is introduced of thickness and refractive index . So where will the point P now appear to be or what is the shift? This is going to shift so it may be shifting here, or it may be shifted here because of the introduction of this, because there is a refraction which is taking place and therefore, we are asked to find out determine the shift in the. This is the object now. This point is the object, so I can also call this as , so this may be shifted to a point . I am just showing it here, . Determine the shift. The shift is this. So, what is the shift? I can call this as or or . Determine the shift, determine the apparent shift apparent shift. The object is, of course there, but it is appearing so determine the apparent shift of the object shift of the object it just occurred to me. I had not drawn any pre drawn any diagram, so we can just determine the shift. So let us take this as .

Refer Slide Time 54:00 Now, if I draw a line here, we can simply extend the earlier problem, so let me draw a fresh so the object is here . We have introduced a glass block here of certain thickness. Let us say this is this was the original position and therefore let me call this, so this is let this be . Let this distance be , because I am interested only in determining the shift. So, shift and . Let this be , and this is the shift to be determined. This is to be detailed. I can think of just extending the previous problem where we had apparent depth, so and . So if I call this as , so this new position as , we see earlier, we had actual height and as the apparent depth. I am now calling it as and . This height does not matter because it remains the same. There is no change taking place and therefore, if I determine that will give me the shift. The shift is simply To determine , what we had done was earlier, if you remember that we had thickness of refractive index and the remaining part here that is . The total length is . So, I can write this as I am just extending the same result: we know this refractive index is 1 here, because this is air. This is air. This is where a glass slab has been introduced. A glass slab of thickness has been introduced. Therefore, I have simply applied the earlier result, we said that the apparent depth So in this case , is this thickness and the remaining length which is . I had called the total length as . Therefore, , which has which is of refractive index one, because that is air, and therefore . So, the shift is remarkably interesting, indeed. See that, it does not matter where you introduce the slab. The slab could be introduced here. The slab could be introduced here. Anywhere, the slab could be introduced. The shift simply depends on the thickness of the slab and the refractive index of the slab, so it’s just a variation of the other problem which I had discussed earlier.

Refer Slide Time 56:45 There are several possibilities, and I would suggest that you work out several problems to get a better feel of this subject. A second part, which I would also recommend you take it as an exercise, is to determine the lateral shift. I had not worked it out here, so we have a ray of light incident like this: it bends towards the normal. If this refractive index is greater than the refractive index here, and then it again bends away from the normal, if the media are the same, the angle that it subtends here is the same as the angle it subtends here, which means we said there is no deviation, is the thickness of the glass slab and is the refractive index of the medium. So, what we have got? We have already discussed this, that we have lateral shift. What we get is a lateral shift here so which I think I denoted, as . Find an expression lateral shift. We can extend this problem by taking two slabs, so the like, we had done for the apparent depth for the lateral shift. Also suppose you had two slabs one after another, so this is of thickness , and this is of thickness and refractive index and and outside, of course it is , or air. You can say this is 1, and this is 1 which means again, as we have already discussed, there is no deviation, however, the lateral shift. So let me show the lateral shift here. This is bending towards the normal. Maybe it will bend a little bit away, but finally, it will come out in such a way that it will be parallel to this. In other words, there will be no deviation. We have discussed this several times, only there will be a lateral shift Now, the lateral shift will depend on and , and and . Determine the lateral shift. Follow a procedure which is almost like what I have outlined here and determine an expression for the lateral shift. So, there can be several similar problems.

Thank you.