Welcome to the lecture module on optics in the last lecture, we discussed about refraction at a plane, interface at a plane interface, and we also saw the condition under which the total internal reflection takes place. Today we will discuss refraction at a spherical interface and then we will extend it to refraction by lenses, because lenses are widely used for various applications. So we will first discuss refraction at a spherical interface, followed by refraction by lenses, so refraction at a spherical interface. Spherical surface and by lenses so first here I am showing refraction at a spherical surface, so let me show the diagram first. This is the interface, a spherical interface between two media of refractive index n, one and n two.

This is medium one, this side and on the other side, medium 2, and in this case, I have considered n2 greater than n1. So here, o is a point object whose image is formed in medium 2.At a position I so there is a straight ray which is incident normally on the spherical interface, which passes undeviated and a ray which is coming at an arbitrary angle. Alpha a small angle. Alpha is refracted, because the dotted line here shows the normal to the interface. I am therefore the angle of incidence and because n2 is greater than n1 the ray bends towards the normal and therefore the ray bends towards the normal gear. It intersects the straight ray. At the point i and therefore i is the image point of this object. Now, this is the angle of incidence and, of course, I have shown a ray here.

A small fraction of light is also reflected because reflection is always present. However, first this fraction is small, typically four to five percent. If it is an air and glass interface, but this fraction can be minimized by coating this surface by what are called anti reflection coatings and therefore in the subsequent diagrams, we simply neglect this reflection and we are focusing only on the refracted ray here. So alpha, beta and gamma are the angles here, as shown in the figure. Alpha is the angle subtended with the axis and beta is the angle subtended here by the normal, with the axis and gamma is this is gamma, and this is r refracted angle r.

At the point of incidence m, this is the object distance from the surface here at the point p to the object position is object, distance U? we will look at the sign convention later, but right now U refers to the object. Distance and v refer to the image, distance and r capital r is the radius of curvature. Of this surface c is the Centre of curvature and r is the radius of curvature of the spherical surface? Now we assume a small aperture, the condition of small aperture this. I have already discussed in one of our earlier classes, so what it means, basically is so let me show it here so small aperture here refers to.

So when we have an optical system, it may have several components or several surfaces, but if this is the spherical surface by small aperture, what a mean is, if we put a block here, that is an opaque stop in front of this, with a small opening, a small aperture and the rays of light, which are entering through this aperture only are undergoing reflection or refraction or whatever. So we are considering a small aperture, which means the rays. So let me show you the different color so rays which are making.

So if a have, let us say a point, object here, o or a point source p here then rays which travel along the straight line rays, which make small angles here, will only be able to pass through this aperture. So small aperture means we are restricting to rays which are passing close to the axis and rays which make small angles only, and that is nothing but the paraxial approximation. So small aperture satisfies the paraxial approximation. This is what we had discussed so paraxial approximation, so rays, which are close to the axis approximation is valid. That means that i keep back the slide here, and that means that angle alpha the angle alpha here.

Actually, this m is very close to this, but only for clarity i have shown is a little away so that the angles are clearly visible. But the angle alpha is very small, because the point m is very close because we are assuming small aperture. Therefore, paraxial approximation is valid, which is which means point m is close to p. That means the angle alpha beta and gamma all the angles i and r, because if this point comes here, then the normal will be like this and i will be very small and then we have the approximation tan alpha nearly equal to sine alpha nearly equal to Alpha when alpha is very small, of course alpha is in radians tan beta is nearly equal to sine beta is nearly equal to beta etcetera.

So these things are valid, so this is of course I mentioned that the reflection of light can be minimized by using what are called anti reflection coatings. We will not discuss that here because to understand anti reflection coatings we need to know wave optics and therefore we will discuss this at a later stage. Now we come back to the problem, and here it is so refraction at a spherical surface. So let us first focus on the angles here, so what we see is the angle. I if you look at this triangle, o m and c o m c then alpha plus beta is equal to i, so i is equal to alpha plus beta.

Similarly, if we look at this angle m c - I this angle m this triangle here - triangle m c: I then we can see that r plus gamma is equal to beta. That is this angle of refraction r here and some angle here, gamma, which I have denoted gamma. So please see the difference between r and gamma r is written like this r, whereas in gamma we write like this and straight. This is gamma, and this is r, so i could have used some other symbol, but just i had used gamma alpha beta gamma come together, so i used alpha beta gamma, so the point is beta is equal to r plus gamma and therefore we are interested in I and r, because we want to apply Snell’s law and therefore, we write r is equal to beta minus gamma.

Then the second point, because of the paraxial approximation which we just discussed alpha nearly equal to tan alpha, is equal to in this diagram. If we see alpha tan, alpha is m d by o d m d by o d m d by o d, but because the point m is close to the point p here, that is, it is close to the axis whether we write op is nearly equal To od, because the point m is close to the axis and therefore this is nearly equal to m d divided by o p, that is, we approximate od by op. This is true for a paraxial approximation or when we consider small apertures. So exactly like that. For this angle, if you look at the triangle m d c, tan beta nearly equal to beta is equal to m d, divided by c d and as before, we are approximating c d by c p, because c p is exactly the radius of curvature. That is why we are a proc making this approximation and gamma is equal to tan gamma.

So if you look at the triangle here, m d, i then gamma equal to tan gamma is equal to m d by i d m d, divided by i d, but i p is the image distance. So we are approximating it by m d by i p and therefore the angle i is given by i is equal to alpha plus beta, which means alpha is here: m d by o p. Beta is m d by c p, so i is equal to m d by o p plus m d by c p and angle. R is equal to beta minus gamma Beta is here m d by c p: minus m d by i p m d by c p, minus m d by i p. So i have denoted these as equation Three and four now we apply Snell’s law because i we have r.

We have and therefore sin i by sine r, is equal to n two by n, one or none sin. I is equal to n two sine r, but again we know that the angles i and r are very small and therefore for small i and r. We can write sine, i nearly equal to i sine r, nearly equal to r, which means n one into i equal to n two into r. This is nearly equal, almost a very good approximation n one. I is equal to n two r. Now i and r are given here, therefore n one into i i from equation: three is equal to n two into r r from equation four, so let me call this as equation number six.

Now we continue further so here and therefore, if we, so let me keep this so that we focus on this from the pre page we have n1 into. I is equal to n2 into r and md is common throughout, so this m d goes off and therefore we are left with n one by o p plus n two by i p. So this part, that is, m d, has gone so n. Two by i p minus n. Two by i p. I am bringing to this side, so n 2 by i p, is equal to c p was common and therefore we take this term to the other side to make it n 2 minus n 1 by c p. Now here we have to look at the sign convention.

We are going to substitute for o p, i p and c p appropriately and what is the sign convention? So just let us very quickly recall the sign convention. It is almost the same as what we have in the case of mirrors, so we have a refracting surface here, and the point which is normal to the here to the axis here is the origin x. Equal to 0 x, equal to y, is equal to 0 and for light incidence from the left. We are considering light incident from the left uniformly. Therefore, x direction. Positive x direction is along this, so this is the positive direction which means from this point.

Whatever distance is on, the left is negative and whatever distance is on, the right is positive and therefore the object distance is the same diagram object, forming an image and for the objective distance, o p here will be equal to minus U because it is on the left of point p, whereas i p is the image distance, which is positive, c p, which is the radius of curvature, which is positive in case we had a if we had a concave surface like this. The object is here for this case. By chance, the image is also, on the left hand, side object is here so a ray which is bending here. The straight ray does not intersect with this, but they appear to come from a point.

I here and therefore the image virtual image is formed at the point i in any case in this poi. In this case, we see that the object distance is also on the left of this point x, equal to 0 y equal to 0 image Distance is also on the left, so it is minus v, and the radius of curvature is also on the left side, because this is the concave surface. Its center of curvature c is on the left side and therefore all of them are negative, whereas in this case we see that the object distance is negative, but these are positive. So this needs to be taken care when we substitute in the expression, because then only the result that we will get will remain valid, whether we take a concave surface or a convex surface.

Ok, so coming back, therefore applying the sign convention. Now we have come back here applying this is the equation. Applying the sign convention, Op is equal to minus U c, p is equal to r and the object image Distance is v positive, so we substitute here and 1 by minus. U, plus n 2 by v is equal to n 2 minus n n1 by r, or we can put it in the form n2 by v - minus n1 by: u is equal to n2 minus n1 by r. Now this is a very important equation in the sense that it gives a relation between object, distance and image distance in terms of the refractive index and radius of curvature of the spherical surface.

Given a spherical surface, it means the radius of curvature and refractive indice of the materials are given, then for any position of the object. It will tell us what is the position of the image. So if we take an example, it will become more clear. So let me take an example here, so let us take an example. A very quick example here. So here is a spherical surface and an object is at a distance of 100 centimeter. The radius of curvature of the spherical surface is given 25 centimeter here.

The material is given as glass with refractive index 1.5 and outside it is air with 1.0. So the question is: determine the position of the image when the point object is at a distance of 100 centimeter, 50 centimeter and 25 centimeter its a simple problem, basically substituting in the formula, because the single interface for which we have just now derived this formula. So let us quickly pick up that for 100 centimeter. Here, u is 100 centimeter and r is 25. Centimeter r is positive for this convex surface here and refractive indices are given. So if we substitute in the expression we get, this as v is equal to 150 150 centimeters real image in the glass medium.

It is positive, 150 centimeter, which means if this were 100 centimeter here, the image would have formed somewhere here at 150. Centimeter from the point p here, 150 centimeter, so that would be the position of the image so its a quick illustration of the application of this formula. If i pick up similarly, you can do for 50 centimeter, but let me take quickly the third one that is, for: u is equal to minus 25 centimeter substitute back in the formula straight forward, substitution and you get v, is equal to minus 75 centimeter minus 75 Centimeter, which means v is also on this side and that that is where we get a virtual image.

So the situation is what i had briefly explained earlier. So we have a spherical surface like this, and here is the axis, and in this case the object point is relatively close. Closer to the surface, the centre of curvature is somewhere here. The center of curvature is on this side, see here, but because and therefore the line joining the center of curvature. So let me take a ray like this one ray going like this. Normally incident will pass through the secondary. I am picking any arbitrary ray which is like this will. If i draw the center of curvature here so the normal to the surface, so let me use a different color, so this is the normal to the surface, which is the line joining the point of incidence to the center of curvature.

Then we see that this will of course bend so the ray will refract, so the ray refracts bends towards the normal. However, in this case, it is still diverging its not coming to this and intersecting with this ray, which means this appears to come from a point which is just extending this back to a point somewhere here. So this was the object distance, so this was o and therefore it forms a virtual image at this is the image distance. So This is the image distance and this is what we got answer as minus 75 centimeter. For the third case, when this was 25 centimeter, we got.

This was 25 centimeter, which means U is minus 25 centimeter. We got the image position as minus 75, which is a virtual object, virtual image, which is formed on the same side. That is why we have this situation, because the object is closer. If the object were a little bit further off, it would have been incident and it would have refracted and intersected with the axis somewhere here. So you would have got a positive, ah image distance in that case, so, depending on the position of the object, we will have the position of the image for the same spherical surface.

So that is why i have picked up those two simple examples. So let’s proceed Further - and we consider so here so this lets, ah after a single interface, now let’s go to refraction by a lens. So here we are considering refraction by a lens. Now let’s look at the refraction first and then come back to the lens. So here is the lens Is a biconvex lens. This is refracting surface one. This is refracting surface two. This is of refractive index n two. The lens medium is of refractive index n two, and in this case i have taken on both the sides n one.

It could be n one and n three on the other side also, but in a simple case we have considered outside. There is a particular medium and the lens is of a particular medium, usually a glass lens, and it has refracting surface one and two. There is an object here, so rays are emanated from the object and therefore i have shown three rays, one straight ray which passes through this along the axis. Now, what is the axis? We will see in a minute and then two other rays i have shown they are refracting. First, they undergo refraction here and then they undergo refraction. At the second surface, there are two refractions here so first refraction and second refraction to form the image. Now the lens is shown we have to remember that the lens has two spherical surfaces. We had discussed this in the first class that these two surfaces are part of spheres: two spheres here of radius of curvature, r, one and r, two first surface radius of curvature, r, one its center of curvature c.

One here - and the second surface, which is here, is part of this sphere here of radius of curvature r, two with the Centre of curvature here, the object point we have considered is here, and the image point is here. So there is the image point. The object point, and in this course, we particularly look at thin film, thin lenses, thin lenses, thin lenses means the separation, a b here, a to b this separation. If i call that as thickness t, this thickness is very small, it is a thin lens under this approximation, the distances o a is approximated to o p, provided this thickness is small, oa is nearly equal to op nearly equal to ob.

So this is an approximation which is followed in the case of thin lenses. That is why we are considering here thin lenses in this course. The axis is the line passing through the center of curvatures, the two Centre of curvatures line joining the two center of curvature C one and c two, so that is the axis. So this is the diagram which indicates the spherical surfaces surface one surface two and radius of curvature. Two and radius of curvature are one here, so let us go further and to determine the image to determine the image position of the image we treat the lens because it has two spherical surfaces.

We have seen refraction at a single spherical surface. Now we will treat individually each of the spherical surfaces and we will look at refraction by the lens as successive refractions by the two surfaces. So that is what i am going to show in the next slide here. So here i have drawn these diagrams in advance. So that they are relatively clear, so we can see object undergoing refraction here at the first surface and then undergoing a second refraction here at the second surface, to form the image point here. If this surface, the second surface was not there, the refracted ray would have traveled here somewhere here, and this is medium one, medium, two and medium one.

I have considered this n2 greater than n1 and this is the object distance, and this is the image distance with the appropriate sign conventions. When we go for the derivations, we will see that now. As i mentioned, we will treat this as you can see exactly this surface is shown here, and this surface is shown here, so we treat the net refraction here as successive case of successive refractions at interfaces 1 and interface 2. Why do we do that? Because we have already seen refraction at a single interface the case of refraction at a single interface first, medium of refractive index n1, second, medium of refractive index n 2 and radius of curvature r 1.

Here then, we have this equation: that refractive index of the second medium divided by the image distance here minus refractive index of the first medium by the object distance is equal to the refractive index, difference divided by radius of curvature of the spherical surface. Now the second refraction is as if it has nothing to do with this, because the ray has already refracted here. So the ray has refracted and it is proceeding when it encounters the second medium here and therefore, we show this as it is as if the entire medium on the left is of n2 and the medium on the right is of n1.

In other words, now this is the first medium, and this is the second medium and therefore we write the same equation for refraction at this interface as the. If the second interface were not there, the object would have formed a image here I 1 at this point, but because of the second refraction at the second interface, the actual image is formed here. Otherwise it would have formed at i 1 here it is in the same line as i 1. Therefore, as far as this interface is concerned, the ray is coming from here. There is no object there, but it is, This i1 acts as a virtual object.

The image i1 acts as a virtual object for the second interface and therefore the distance from here to i 1 is the object distance in this case and the distance to i is the image distance. So the object, distance image, distance and R2 is the radius of curvature. Therefore, the formula is refractive index of the second medium Second, medium is on this side, so this is now n. One refractive index of the second medium divided by the image distance refractive index, divided by the image distance image distance Is this one which is v so what is shown here v from the center to i.

This is v so refractive index of the second medium. Always we on the left is the first medium right is the second medium, so refractive index of the second medium divided by the image distance, minus refractive index of the first medium. First, medium is now this one, which is n two of refractive in x, n, two divided by the object Distance, object Distance is now v one here v, one object Distance is equal to refractive index of the second medium minus refractive index of the first medium divided by radius of curvature so equation One and two, this equation applicable to the first interface, this equation applicable to the second interface and therefore, if we now add 1 and 2, please see 1 and 2.

If we add this term is common and therefore this is with the negative sign. Therefore, this term cancels off and we will have n1 by v, plus n1 by v minus n1 By u is equal to so this. We can flip this. We can make n two minus n one with the negative sign, and that is what we get here. So let me show the in the next slide here, therefore, adding equations one and two, so let us focus here adding equations 1 and 2. We get n 1 by v - minus n 1 by - u is equal to n 2 minus n 1 into 1 by r 1 minus 1 by r. We can take the n 1 to the other side and we can write this as one by v minus one by: u is equal to n two by n, one that is dividing throughout by n one n, two by n one into one by R1 minus One by R2 note that what is there on the right hand, side is a constant.

This is a constant for a given length. A lens is given means the refractive, the radius of curvatures are fixed and the refractive index of the lens medium is fixed and, of course, depending on where you place, n1 is also fixed, and therefore this is a constant. This is the image distance. This is the object distance, so this also gives a relation between image, distance and object distance in terms of the parameters of the lens now for large distances. So let us look at this for large distances 1 by u, 10 to 0, when u large distances object distances when the object is at infinity, let us say that 1 by: u tends to 0. This means that we have 1 by v is equal to a constant. What is there on the right hand? Side is constant; it has nothing to do with the U, whatever be the position of U, it does not depend on the position of U, therefore, for large distances we have, one by v is equal to a constant which is independent of u. That is when the object is at a large distance.

It means the rays from the object are almost parallel to the axis, but they all concentrate, or they all converge to a point at a distance v, and that point is called the focus. The principal focus will discuss this in more detail in the next slide, so the when the image point is fixed for large values of u 1 by v is constant image. Point is fixed independent of u, and this is called the principal focus f. We will show this in a diagram. The corresponding image distance is called the focal length and therefore 1 by v is equal to 1 by f that constant, the constant on the right hand, side is denoted by 1/f is equal to n 2 minus n 1 minus 1 into this from 4 and 5.

We have 1 by v minus 1 by: u is equal to 1/f. Basically, what we have said is this is a constant which is equal to 1 by f now. What is that f? f is the focal length. F Is the focus where parallel rays, from a distant object, focus on to converge on to the point f? So this we i will illustrate so this is the important formula which is called the lens formula. Lens formula relates the object distance to image distance for any given lens of focal length f, which depends on the parameters of the length that is radius of curvature and the relative refractive index difference.

Now let us discuss this focal length a little bit more here, so here I am so we are. We will discuss the focal length that’s, a very important property of a given lens so focal length. So this is the lens formula with 1 by f is equal to, This is what we called as focal length and n 2 n 1. This is a biconvex lens r, one is greater than zero and R2 is less than zero, because R2 is having a center of curvature on this side. Therefore, R2 is less than zero.

For U tending to infinity this. What we discussed rays from the object are almost parallel to the axis and the object. The image distance v is equal to f, which is the focal length so the rays, the parallel rays, which come all of them converge to a point F because they are independent of the distance U all of them have the same image distance, which we call as the focal point. They converge to a point f and the distance between the lens and the focus. The principal focus is called the focal length now. This is true for a given lens and if we cannot that here we have taken n 2, for example, as glass and air, then we have a certain value for the focal length. But if we immerse the lens in a liquid, just as a case when the lens is immersed in a liquid of refractive index n l, then one by f l. f l is the focal length in the liquid is n2 by ni. Instead of n1, i have used n l that is refractive index of the liquid minus 1, Divided by this now note that n l is greater than air n 1. If it is air outside, it is one, but the liquid has a refractive index greater than hair, so n l is greater than n air. Therefore, the focal length in liquid is greater than the focal length in air, because n l is greater than one and therefore this difference is now smaller. Therefore, this quantity is smaller compared to the case of air. This is smaller means, one by f. L is smaller or f. L, that is the focal length in the liquid, is greater than the focal length in air. There are several applications where the lens is immersed in a liquid to have a different focal length or the effectively the focal length changes, and we know that the focal length in the liquid is larger than the focal length in air all right. So let us take. Ah, go further and see the lens makers formula i want to now discuss. This is a familiar or more common formula, because most of the common applications for common applications of a lens n one is equal to n air is equal to one when we use a lens.

Normally, the outside medium is air, except in special cases. When we have a liquid on the outside, so it is air and therefore refractive index is one and the refractive index of the lens is denoted by n, because there is only one other refractive index. So there is no point in writing N one and n two. So we n is the refractive index of the medium of the lens material of the lens and n2 is equal to n refractive index, and then we have 1. Over f is equal to n minus 1 into 1 by r1 minus 1 by r2. This is called lens makers formula because when one would make a lens for a particular application to obtain a required focal length f, the lens maker can choose a material and required values of the radius of curvature r1 and r2. r1 may be equal to r2 or may not be equal to r 2, but he can choose the radius of curvature to achieve the required focal length for a particular application. Hence this formula is traditionally called as the lens makers formula, although the general formula is what we had already seen that one by f. So this is the general formula. This is valid in for all refractive indices, but in the special case, when n one is air, then we use the lens makers formula, which is the simpler. Where n is the refractive index of the medium, the formula indicates to the choice of R1 and r. Two to obtain a desired focal length.

Now we proceed further for a symmetric, biconvex lens symmetric means. Both the radius of curvatures are the same, which is R1 is equal to R2. Of course, R2 is with a negative sign and therefore R1 is equal to minus R2 equal to r. So it is a symmetric biconvex lens. Then we substitute here in the formula we have one over f is equal to n minus one into one by r minus r naught, which is which is equal to two by r into n minus one. So note that n is the material of the lens which is greater than air n is greater than one. Therefore, f, the focal length is greater than zero, which is positive, so that is called a converging lens. A converging lens has a focal length which is positive, so we will see so what about diverging lengths. So let us see converging and diverging lenses, so here it is converging and diverging lenses for a symmetric, biconvex lens.

Just now, we showed that one by f is equal to two by r into n, minus one or f is equal to r by two into one by n minus one. So here is the converging lens that a biconvex lens, a symmetric by convex lens. It need not be symmetric, but the formula i have for a symmetric. I have considered as a special case when R1 is equal to R2, so we have f positive for a symmetric biconcave lens that is a biconcave. Here is a biconcave lens, so we can see that this is R1. The first surface second surface is R2 R1 has a radius of curvature on this side, so the center of curvature is here.

Therefore, radius of curvature is negative R2 has a radius of curvature on the other side and therefore this has a positive radius of curvature. So R1 is equal to minus r, both r 1 and r 2 magnitude is equal to r, because it is a symmetric lens, but r 1 is negative and r 2 is positive and therefore r 2 is equal to r gives f is equal to minus R by two, so this r is now a magnitude only so this r, because negative sign has been taken into account. So this is positive. Only f is equal to minus r by two into n minus one, because n is greater than 1, F is less than 0; in other words, the focal length is negative.

So we can see here that if we have a by concave lens, the focal length is on this side and therefore f is negative. F is positive in the case of a bi-convex lens, so in this case the rays diverge away as if they are coming from a point f. The principal focus is here on this side and therefore this is a diverging lens, whereas this is a converging lens. A biconvex lens is a converging lens, whereas a biconcave lens is a diverging lens. Note Something interesting that for n is equal to consider this, Consider the biconvex lens for n is equal to 1.5. F is equal to r n is equal to one point five. This is one point five minus one.

That is point five multiplied by two is one. Therefore, f focal length is equal to radius of curvature, whereas for n is equal to two focal length is equal to r by two. If you put n is equal to two here, then this this whole thing is one and therefore f is equal to r by t. So it clearly indicates that not only it depends on the radius of curvature the focal length, but also it depends on the refractive index of the material. So in one case the focal length is r. In the other case, focal length is r by two. This is like in the case of concave mirrors. We have seen earlier in the case of mirror that the focal length is r by two, but in the case of a lens focal length need not be r by two.

So in any problems do not jump to the conclusion that okay focal length is r by two that’s, not correct for the case of a lens. It depends on the refractive index of the medium and therefore you have to substitute in the formula. One by f is equal to ah n, two minus n one into one by r, one minus one by R2 and find out the focal length now several situations here, various situations which one encounters that R1 is greater than zero. There is a by convex lens r, one greater than zero R2 less than zero. The normal biconvex lens, which i have been discussing. There are lenses which are used for special purposes where both of them have a convex surface here, so R1 is greater than zero.

The radius of curvature is on this side and R2 also has a radius of curvature on this side. R1 may not be equal to R2, but both of them are convex surfaces and therefore R1 greater than zero R2 greater than zero. Both of them could be concave surfaces, in which case R1 is less than zero and R2 is also less than zero and we could also have plano convex lens or plano concave lens. This is a plano convex lens. Therefore, this is R1 is greater than zero. Radius of curvature is here, and this is a plane surface. therefore, the radius of curvature is infinity. R2 is infinity, but R1 is greater than zero. Now, Finally, in this situation, all the while we have been discussing the case when n2 is greater than n1, but what if n1, is greater than n2? If n1 is greater than n2, that is, if the outside medium as a refractive index greater than n2, the situation will change, a convex lens can become a diverging lens and a concave lens can become a converging lens earlier. I had shown that in diverging and converging lenses, a convex lens is a converging lens and a biconcave lens is a diverging lens converging and diverging lenses, but all the while we had assumed that refractive index of the lens is greater than that of the surroundings.

But in the reverse case, when the refractive index of the lens is smaller than that of the surroundings, it is possible that if this is immersed in a liquid of refractive index greater than that of glass, then it is possible to have this situation and in this Case the convex lens can become a diverging lens and a concave lens can become a converging lens right. What, if the next question? What if i have been considering light incident from the left side of the lens, with the radius of curvature, r, one and r, two? What, if light incident from the right side, so will it have the same focal length?

So, let’s see so here what if light, is incident from right side so now, r, 1 and r 2? This is the lens and we have so. Let me block this for a minute, so the case is here, light incident from here and getting focused at a point here. So this is f1. Ah, i had initially written f2. That is why it, but it is f1 and f1. So the question is: is this distance f1 is the same as f. F we had considered when light parallel. Light was incident from on this side and focus to a point to the principal focus here and we call this focal length as f. If light were to be incident from here, whether it will focus to a point here and whether the focal length on this side is the same as the focal length on that side now, light is incident on the surface with the radius of curvature r. Two, therefore, i can equivalently rotate this and put the lens like this, so that light is still incident on the left side, but now it is encountering R2 first R2 here. The light which was incident is encountering the surface here with radius of curvature r. Two so the same situation, so i have just flipped it and put it r, two first and r, one here and therefore one over f, now f1 one over f1. So this is not f2, it is f1.

So one over f1 is equal to n two minus by n one minus one, divided by one by r, two minus one by R1 earlier we had the formula one by r, one minus one by R2, but now because R2 has become r. One and R1 has become R2 in this case, because we have flipped the lens, so it is 1 by r 2 minus 1 by r1. So what is this? This is nothing but minus 1 by f and therefore mod f1. Whether in this case or in this case, the distance is the same, the focal length is the same. Whether light is incident from this side - or it is incident from this side - although r 1 and r 2 are different only therefore so long as n 1 is the same n 1 on this side and n 1 on this side is the same.

It is, it is worthwhile to check what happens if it is n, 1, n, n2 and n3, n1, n2 and n3. But right now i am considering the case where, on both the sides, we have n1, and this is n2 and so long as n1 is the same on both sides of the lens. Even though r1 is not equal to r2. Mod f1 is equal to mod f2. Mod f5 mod i have used because the focal length on this side is negative and focal length on this side is positive. If we look at it, but of course, when the light is coming from this side, this direction is positive. Therefore, the focal length f1 continues to be positive. This is not negative, but anyhow for this case we have shown, because we are all the while. We will consider light incident from here and therefore it will have a focal length f1 to this side and a focal length f2 to that side and f2 is positive and f1 is negative. Thus a lens has two principle foci. So let me discuss this. A little bit more so here the principal foci and focal length of a lens. So here is the lens light incident from here from the left side. All light rays are incident from the left side here, and it is focused to a point f2 here with a focal length. F, two, whereas f1 is the first principal focus rays which come from the first principle. Focus here, f1 will be rendered parallel, because if these, if light were to travel from here that is from right to left, then it would have focused to this principle.

Focus point f1, and this is what we saw in the earlier slide. The focal length is called f1 now in this case, light is travelling from here, but the parallel light is focused at the principal focus f2 and the focal length is f2, whereas light rays emanating from the first principle. Focus f1 will be rendered parallel. So f1 magnitude is equal to f2. So f1 is the first principal focus, because when we go from here, we encounter the first principle focus first surface first principle focus first focal length when we go further. We encounter the second surface. Second refracting surface. Second principle, focus and second focal length, so f, one is the first principal focus F, one is the first focal length. F, two is the second principal focus and f2 is the second focal length.

This one and f1 and f2 are equidistant from the lens, because we have just now showed that f1 is equal to f2 in magnitude. Therefore, the principle foci f1 and f2 are the principal foci which are equidistant from the lens.

Normally, when we refer to the focus of a lens, when we normally talk about a lens of focal length f, we are referring to the second focal length f2, because that is the one which we encounter later. It is beyond the lens, and it is the second focal length that we are referring to and the focus of the lens. Also we are referring to capital f2. That is the second principle focus, so here it is f2 and the focal length f2 then, what is the importance of1 because light is incident from here? So what is the importance of f1? As we can see, the importance is illustrated right here.

Any ray which is coming from f1, will be rendered parallel. So where do we need this? We need this in determining the images formed by lenses, so that will be the next topic. That is imaging formation of images by a lens, so formation of images by lx. So let me briefly discuss formation of images. We have discussed formation of images in detail in the case of mirrors, so now we are discussing formation of images by a lens laterally extended. I have already discussed formation of image of a point object, but now we are considering laterally extended object. That is a line. Object here of dimension a b, a b is the object.

F1 is the first principal focus. F2 is the second principal focus, so let us focus on the diagram here. So a parallel ray, which is coming from the object, passes through the second principle focus a ray which passes through the center of the lens here will pass undeviated and it will intersect the ray which is coming from the focus, and that will be the image point Of a so image point of a is marked as a’ or the image of the extended object. A b is a’ b’ here now, a third ray which is passing through the first principal focus, will be rendered parallel.

There are situations, many cases we are not able to get two. So these two rays, sometimes we are not able to draw - particularly in the case of ah by concave lenses, and then we have to use this fact that a ray which comes from the principal focus will be rendered parallel. A parallel ray parallel to the axis Will pass through the principal focus, but a ray which is passing or comes from the principal focus will be rendered parallel. The intersection gives us the position of the object. Now, Let us quickly see this, because we are familiar with the formation of images. So, look at the triangle: a b, p and triangle: a’ b’ p, so a b p here and a’ b’ b.

So this triangle and this triangle these are equivalent triangles, because this opposite angles are equal 90 degree. So all three angles are equal and therefore we have a b, by b p, a b by b p, which is actually tan Theta, a b by b p is equal to a’ b’ by b’ b, a’ b’ by p b,’ p b’. So this is tan, theta actually or a’ b’. So i am shifting this ah here, so a’ b’ by a b is equal to b’ b by b p now applying the sign convention. We know what it is, so we are interested in finding a’ b’ by a b, because we are interested in the lateral magnification, just as in the case of a mirror. We are interested in the lateral.

Magnification m is equal to size of the image by size of the object size of the image by size of the object. That is, we are interested in a’ by a’ b’ by a b. So a’ b’ by a b is equal to b,’ p by b p, so substituting. This is h, h’ as per sign convention. This is negative and this is positive. Any distance above this above the axis is positive and therefore we substitute for a’ b,’, minus h,’ and a b h is equal to v object, distance, which is positive and image distance, which is positive and object. Distance that is, b p, which is the object distance, is negative.

Minus U so that is what we have substituted here or m, is equal to h’ by h is equal to v by u now very quickly if we see the formation of image for the case of biconcave lens, so i am do not have to discuss, but you can see this very clearly that here is the object, a b. Now a parallel ray here incident will be diverging. It is a diverging lens appear to come from the first principal focus here: f 2 is coming from here, a ray which is which would have gone, which is which would have gone to this principle Focus here would be rendered parallel, because if a ray were to start from here, then it would have been rendered parallel, and that is why this ray would be rendered Parallel and array, which is passing through the center of the lens, will go undeviated all the three rays, 1.; 2.; 3.; Do not intersect anywhere here rays, coming from a do not intersect anywhere on the other side of the lens, but note that they appear to come from a point a here where they intersect they appear to come.

If we extend these backward, then they appear to come from a point, a’ and therefore a’ b’ is the image of a b due to a biconcave lens very quickly. If you look at the triangles a b p and a’ b’ p, they are equivalent triangles and therefore a’ b’ by a b is equal to b’ p by b p. That is h’ by h, a’ b. b’ is h, h,’ here, which is positive, it’s above the axis by h. This h is equal to minus v, the image distance and minus U so minus v by minus U, which is equal to v by u or the lateral magnification m, is equal to v by U, as before, as before means the same formula that we got in the case of a convex lens, because we have followed the sign convention in the next class. We will take up some examples and proceed on to the topic of power of a lens when a lens is converging or diverging. There is a power associated with what is the converging power and what is the diverging power this we will take up in the next lecture.

Thank You