 So, next we will take up convex mirror. So, convex mirrors look like this. This is the principle axis. Where is the focus? Focus is here and this is r and object will be there. This side will be the object. Now, there is no f or r here. So, there cannot be any cases here. There is just one case, object wherever you keep. So, put an object somewhere and draw its image. Where will be its image? One ray that goes towards the r. What happens to that ray? This will get reflected back because this is along the normal and one ray that goes like this. What happened to this one? This is I. It will reflect off like this. This angle will be also I. So, where is the image? Will these two ray ever meet this one and this one? No, but they will appear to meet here. So, this is the image. It is real or virtual? It is virtual image. Write down characteristics of image, characteristics of image, virtual, diminished, erect. So, if you have been given two virtual images, how will you identify whether this virtual image is from concave or convex? Concave, it will be magnified. Convex, it will be diminished. So, if I tell you that the size has gone down and it is a virtual image, you do not need to know that. I mean, it should not be mentioned that whether it is convex or concave. You should be able to identify. One more thing you might have noticed is that whenever the real image gets formed, of course, this mirror will never form a real image. So, whenever real image gets formed, it is inverted. Suppose it is given that the image is put on the screen. So, it should not be given that it is having a real image. If it is a real image, it must be concave mirror. These all things should be there in your mind. Now, although we have done almost every case by ray diagram, but it becomes very difficult to exactly locate the location of the image by drawing these plots, making ray diagrams and things like that. We will not be able to exactly pinpoint where is the image. So, we should have some sort of formula. We should have some sort of formula which will output as what is the distance of image. And what all thing we should already know? We will know that where I have keeping my object that is easy to measure, is not it? And I should also know the characteristic of mirror. I should know the focal length of the mirror. So, depending on characteristic of mirror and the distance of objects from the mirror, I should be able to find out the image distance. There should be some formula because only these two things, image will depend on characteristic of the mirror and where is the object. So, now we are getting into the algebra of the chapter where we will introduce some sort of formula. And this formula which you already know, it is a mirror formula. Draw neatly a concave mirror, slightly bigger will be better. This is the principal axis, this is F, this is R. Now, here is the object, draw its ray diagram. This is a ray diagram. You will see that the image will lie over here. Now, use the naming convention of the points like this. Let us say this is M, this is N, this is O, this is P, this is Q, R. This one was F. So, this is the naming convention. Now, I want to find out the distance of, let us say this is T, this point. So, I want to find out what is the distance of T p. I want to know what is T p. What is given? N p. N p is known, what else is known? Some property of the mirror is also known. What is the property of the mirror? Focal length. So, focal length is also known. So, P f is also known. I want T p in terms of N p and P f. Are you getting it? Now, look at it carefully. T p, N p and P f, what are they? They are the distances, horizontal distances along the principal axis. So, you need to bring in these horizontal distances in the equation somehow. Now, one easy way to do that is to use similar triangles. Then you can use ratio of sides. So, when you use ratio of sides, one of the side you should take as horizontal. Otherwise, horizontal side will not come in picture. Other side, do not take this slant one. Other side, take vertical. Why? Because this distance, if you drop perpendicular like this and drop perpendicular from there, these two distances, this distance will be equal to that distance and this distance will be equal to this distance. So, can you judiciously find out what are the two similar triangles where you can involve horizontal distances and vertical distances? Think over it. First, tell me the similar triangles which you want to involve. Is triangle M, N, F similar to triangle O, P, F, yes or no? And triangle, do it yourself. It will be better. T f s is similar to triangle. No, these two are not similar. P f q, O f p. From first condition, what equation will you get? M and F similar to triangle P f q. What can you write? M n divided by N f. These are two sides of this triangle. This should be equal to what? Do yourself. Try to come up with, after this, what should we do? Is left hand side for both? Same or not? M n is equal to O p and P q is equal to T s. So, left hand sides are same. So, I can equate right hand side. So, N f by P f is equal to P f by T f. So, P f square is equal to N f multiplied by T f. I can further write this down as P f square is equal to, what is N f? Is N f known? No. N p P f is known. So, what can you write about? N f. N f is equal to N p minus P f. What is T f? T f is what? T p minus P f. T p is what you have to find out. So, it should come in the equation. Otherwise, how will you find out? T p minus P f. Is this thing clear? Now, look at it. At times, what will happen if the object is between F and P? Image will go that side. So, T will go there. Are you getting it? So, if T goes there, the T f will be equal to what? T f will then become equal to F p plus T f. Are you getting it? So, this thing will not be valid then. But if I introduce some sort of sign convention and I say that this side, the sign is something and that side something else, then automatically sign switching will happen. So, this minus will become plus. Are you getting it? So, right now, we are introducing sign convention before we proceed further. So, write down sign convention. We want to make this formula more generic. It should be applicable for every case. Right now, it is applicable only for the case where object and image both are in this side. Write down sign convention. First, write down horizontal distances. All the horizontal distances are measured along the principal axis. Horizontal horizontal distances are measured along the principal axis. From point, from pole, all the horizontal distances are measured along the horizontal, sorry, along the principal axis from pole. From pole, you measure the distances. Next point. If starting from pole, you go against the direction of incident light, the distances will be negative. The distances will be negative. If you start from pole and go in the direction of incident light, then that distance will be positive. This is about horizontal distances. Now, write down vertical distances about vertical distances. Vertical distances are measured perpendicular to the principal axis. If one side is positive, the other side will be negative. If one side is positive, the other side will be negative. If one side is positive, the other side is negative. Usually, what we do? We take upper one as positive, but upper one is not easy to define. You cannot say that what is upper and what is down. But, the way you draw it in your notebook, you can define upper is what or lower is what. So, usually, you take upper as positive. So, this is the sign convention which we are going to follow. Now, tell me pf. Let us say focal length is f. So, pf is what? Plus f or minus f? Minus f. So, this will be minus f whole square y minus f. I am starting from pole along the principal. I am going against the direction of incident ray or not? Incident ray should come from the object. I am going against that. So, minus f square. What is np? np. Suppose, image is v and object is u. np is what? u or minus u. No, you always start from p. So, again minus u. What about pf? It is negative. What is tp? tp. It is an image. So, minus u find minus of minus f. So, f square is equal to f minus u into f minus v. So, you can further simplify this and write down f square is equal to f square minus uf minus vf plus uv. So, you can cut f square like this. So, you will get uf plus vf is equal to uv. Now, I can say that this is the mirror formula, but I can make it more concise. How I can make it more concise? By dividing everything by uvf. If you divide everything by uvf, you will get 1 by v plus 1 by u is equal to 1 by f. So, this is the mirror formula. It is valid for all the mirrors for plain mirror also. For plain mirror, what is focal length? Infinity. When you put f is equal to infinity, this term becomes 0. So, you get v is equal to minus u. That is how it should be. So, this is the mirror formula derivation. There are approximation here. There are few approximations. You have taken aperture less and you have taken r is equal to f by 2. So, all that approximation we have taken here. Any doubt on this? This is about horizontal distances. What about vertical distance? Vertical distance will tell you how much image has magnified. Let us see that. In order to do that, you should make a construction in the ray diagram. Draw a light like this, light ray. This will reflect off and will again meet at the same point, is not it? Like this. So, this is suppose height of object and this suppose is height of image. So, if you look at it carefully, triangle m and p is similar to triangle p, p, s. So, I can say that m, n is equal to m, n is equal to p by n p is equal to T s by T p. Are you getting it? m n is what? m n is u sign convention. If you are taking m n as plus h naught, you are saying that upward is positive. So, m n is plus h naught, n p is what? Minus u. T s minus h i, T p minus v. So, h i by h o is equal to minus v by u. This is what the magnification is. So, we have magnification formula. So, we have talked about horizontal distances and now vertical distances. So, I can exactly find out what, you know how the image will be and where it will be.