 Friends, so those who are online, what we are doing right now is, now we are in middle of the chapter. So, this is a class that is going on, so those who are online are actually seeing a live class. We are in between a chapter, this chapter name is Ray Optics. And till now what we have done is, we have talked about reflection, we have discussed about refraction. When it comes to reflection, we have discussed about plane mirrors, spherical mirrors. And why we are discussing all this? Because these things will deflect the ray that is coming towards the observer. And if rays get deflected, then you may not be able to see the object where it is. You will be able to see only the image, you will feel as if rays are coming from somewhere else, if rays gets deflected. So, we learned about the laws of reflection. We applied laws of reflection in plane mirror to identify the location of image. Then we also discussed the spherical mirror, how it creates the image. So, after that we had discussed about refraction. In refraction, we first discussed about interfaces, how interface affects the deflection and how it can create the image. So, not only how much ray will deflect depends on the refractive index that is mu 1 and mu 2. So, light if it changes the medium, it will be always in some, already will be in some medium, isn't it? And then it will change the medium, it will go to next medium, fine. So, what happens is that this deflection not only gets affected because of the refractive index, but also it get affected because of the geometry of the interface itself, fine. So, interface could be horizontal, could be flat, right. So, we discussed about how image gets formed from the flat surface and this flat interface is very common, like water in a bucket is a flat surface, water in a swimming pool is again a flat surface, right. So, we discussed that and then we also talked about how spherical surface will create the image, are you getting it? So, we had in fact a formula for a spherical surface, what is that formula? We had mu 2 by V minus mu 1 by R, this is the formula which can be used for every image formation in any interface. And what is the interface? It is a boundary between two refractive indexes, getting it. In fact, this is what we call mother of formulae's because even for planar surface, you can use it, R for the planar surface will be infinite. So, you can use this formula and what is mu 1? So, you can use this formula and what is mu 1? So, you can use this formula and what is mu 1? So, you can use this formula and what is mu 1? Refractive index, there is no first or second medium. You will not be able to identify based on the question, what is first medium? What is the second medium? But then you can say that medium of the object, fine. So, mu 1 is medium of the object and mu 2 is the other medium, fine. So, like that we will be using, we had actually used this formula and solved many numericals on it, fine. So, after this what we did? We started talking about lens, right. So, this is convex lens, fine. So, we discussed about convex lens, right. And any lens, our first assumption was what? That it is a thin lens, okay. Thickness of the lens is negligible, fine. And another assumption is both sides of the lens have the same refractive index, okay. Once you assume these two things, then you can derive some formulas which will be called as lens maker formula or lens formula. And how you derive this formula? Using this interface formula itself, okay. So, we had done this. So, we had this interface, sorry, you have this first interface which has this refractive index and that refractive index. Then you had that interface which separates this one and that one, okay. So, first formula which we have derived for lens was lens maker formula, okay. What are the lens maker formula? 1 by F is equal to mu 2 minus mu 1 divided by mu 1 multiplied by 1 by R1 minus 1 by R2, right. What is mu 2 and mu 1? What is mu 1? Reflective index of? Reflective index of the surrounding, okay. Mu 2 is refractive index of the lens, fine. Now, if you see lens maker formula for fixed R1 and R2 and for fixed refractive index of the lens, even if you change mu 1, the focal length can change. So, if you change the refractive index of the medium also, then also focal length will change, okay. In fact, you can change the sign of the focal length also by changing the mediums, sorry, the surroundings refractive index, fine. So, we had discussed this also last class. What else we discussed? We discussed about the lens formula, right. What are the lens formula? 1 by V minus 1 by U equals to 1 by F. This is true for thin lenses and for the scenario where both sides have the same refractive index, fine. Now, using this formula, you can directly get the location of the image. But if you use this formula, even this formula is valid here, but you have to use it twice. You have this interface, first interface and then second interface, okay. And if you already know that both sides have the same refractive index, you can directly use this formula, all right. So, we discussed all this and then after that what we did? We combined lens with mirrors and saw a few numericals on it, fine. So, it need not be that only one lens will be present all the time. There can be multiple lenses together, which all of them together forming some sort of image, okay. So, this is what, you know, is the brief summary of what we did till now. So, you know, I just wanted you to recall all that, recollect all that, okay. Fine. Next topic, which we are going to take up in this class is prism, fine. Write down prism in some sort of experiment or something. 10th grade, right. There you have used prism to do some minimum deviation experiment. So, no. So, what we did in that experiment, what was the purpose of that? The purpose was to find the refractive index of the medium, okay. So, prism offers a very simple experiment using which you can find the refractive index of the material. That is why we are studying prism, fine. So, experimentally, it becomes very difficult to find out what is the angle of incidence and what is the angle of reflection. Because you will not be able to see a ray going like this and then you draw normal and find the angle. You will not be able to do that, okay. So, you should, I think you are not able to find the angle of incidence or refraction directly, okay. But the Snell's law relates refractive index with angle of incidence and angle of refraction, okay. But you cannot measure it directly, experimentally. But if you somehow relate angle of incidence and angle of refraction with the angles which you can measure easily, then you will be able to find a refractive index. You can write I in terms of angles which you can measure directly. Are you getting it? And similarly, angle of refraction also, if you can somehow write in terms of angles which you can directly measure, then you will be able to find refractive index, okay. So, prism offers a very simple experiment to do all that, okay. Let us see what is that experiment. Do you know how prism looks by the way? Prism is, suppose I take a book like this and this is prism, okay. So, from front, it looks like a triangle. Are you getting it? So, I'll just draw the front view because everything is symmetrical on the lateral side. So, I don't need to show what is happening. There is no variation on this inclined face, fine. So, draw this prism. All of you who are live also, please take a notebook and draw it on your notebook. Suppose this is the instant ray that is coming. Can you draw its path? How it will travel and come out from this side? You have to show the normal and everything. Only 5 people. So, total we have 5 plus you guys. Anyways, so first step is what? To see where it goes. Draw the normal, right? So, this is the normal suppose. Now, lightly bend away or towards the normal? Why? Because you are assuming that, suppose this is refractive index mu1. Mu2 is more than mu1. That is fine. I mean, it's a reasonable assumption. So, you should know what you are doing. So, mu2 is greater than mu1, you are assuming. So, what happens to this light? Then what will happen? It will bend away from the normal. But first step is to draw the normal itself, right? So, that is normal. So, bending away from the normal means that angle of refraction will be more than this angle. So, it will bend away. So, this ray of light, this one. This is the emergent ray. Right now, nomenclature. This is incident one, which is coming from this side, left hand side. Incident ray. This is angle of incidence. Right this as I. Okay. Let's name this as R1. This angle as R2. And that angle as E. Fine. Okay. Now, what we discussed initially, we wanted to find out the relation between angle of incidence refraction and the angles which I can easily measure. Okay. So, first tell me this. If I rather than taking like this, if I take prism like that, will that affect anything? So, it doesn't matter. So, prism's identity is not what is happening. The identity of this angle. Because when you do experiment, the refraction happens from this phase and that phase. So, you call this as prism angle. Let's call it as angle A. There is another angle which you can track. That angle is, can you show me where is the angle of deviation? What is deviation? This is the original path. What is the path now? This is a path. So, how much is a change? This angle is a change. Right. Let us say this angle is delta. Fine. Experimentally, it is seen that delta is in those two pins. Okay. When you track delta, you'll see that delta will keep on, you know, keep on decreasing. But after some point, it starts to increase if you change the angle of incidence. Fine. So, you can actually pin point. Where is the location or what is, what, where is the situation? What is the situation in which deviation angle is minimum? Because initially we will keep on decreasing. And then after which it will start to rise. So, the point where it just starts to rise, that is the point where minimum deviation happens. Are you getting it? Fine. So, we will assume here is that we can measure A and delta. The relation between refraction angle, incident angle, this angle with respect to A and delta. Relation between R1, R2 and A. There is a relation. Geometrically, you can find out. Do it. What is the relation? Is it equal to R1 plus R2? Are you getting it? You can see here that there is a quadrilateral here. This, this angle is 90 degree. This one. And this is also 90 degree. So, these two sum is 180 degree. So, these two sum should be also. And this angle is what? 180 minus R1 plus R2. So, this is 180 minus R1 plus R2 plus this angle should be. Okay. So, write down 180 minus R1 plus R2 plus A is equal to 180. Get R1 plus R2 equals to A. So, this is our first relation. We got a relation which connects angle of refraction. Are you getting it? Now, can you find out angle of deviation delta in terms of I, R1, R2 and E? All four angles will feature now. This is what you are getting? How you got that? Exactly. So, this angle delta is an exterior angle. These two. Fine. So, this angle is what? This one is I minus R1. And this angle is I minus R2. So, delta is this angle plus that angle. So, fine. Now, this will become I plus E minus R1 plus R2. Right? R1 plus R2 is what? A. So, I can write this I plus E minus A. Okay. This is delta. Right? So, we have a second. Fine. Now, let us try to rearrange the terms in delta. So, delta will be equal to minus of 2I into E plus 2 times I into E minus A. So, delta will become this constant. But if you see this bracket term will be minimum. This is a square term. The minimum value of a square term can be 0. It cannot be negative. Right? And it will become 0 when? When I is equal to E. Fine. So, I can say that delta direct derivation that this is what this necessary condition. But this is just to show that, you know, when angle of incidence becomes angle of emergence, then delta will, experimentally, this is true also. I is equal to E 1 and R2. Now, you can hear. Are you able to hear? Hello. Check your net speed, internet speed. For that internet speed. Now, what we have found? Okay. So, you get 2R1 is equal to A. So, R1 will be equal to A by 2. Okay. Now, I must be audible. Fine. Refractive index is more for higher frequency. All right? But then in our numericals, it will be written that for this medium refractive index is 1.5. So, till now, we have always used refractive index as a fixed number. We never thought about that. What is the wavelength of the ray that is hitting the interface? Right? So, we will continue doing that only. This is just for your information. There are not many numericals on it. Unless you write J advance and probably some paragraph question may come. But that is like far-petched.