 All right, friends, so here is a question on ray optics. So ball is kept at a height h above the surface of a heavy transferent sphere made of material of refractive index mu. The radius of the sphere is r at t equal to 0. The ball is dropped to fall normally on the sphere. Find the speed of the image form as a function of time. Now, consider the image only by a single refraction. So let us try to visualize the situation first. So we have a transparent sphere like this, OK? This is the line along which you can say that the object is falling, OK? So this is the object which is at a height h above the surface of the transparent sphere, OK? This sphere has a radius of r, right? So at t equal to 0, the ball is dropped to fall normally on the sphere. So let's say that at any moment in time, its height from the sphere is y, OK? Let's say this object has reached here, OK? So this distance is y. So let's try to find out the location of the image as a function of y, OK? So what I'll do, I will use this formula, spherical interface formula mu 2 by v minus mu 1 by u is equal to mu 2 minus mu 1 by r, OK? So if I substitute the values over here, what I can get is, see, mu 2 is what? Refractive index of the sphere and mu 1 is the refractive index of object. So I can write mu 2 by mu 2 is basically given. Mu 2 is mu of the question. So I'll write mu 2 as mu by v minus mu 1 is 1 divided by u. What is u? U is negative of the y, isn't it? So this minus y is equal to mu minus 1 divided by r, where r here is what? Plus r, right? Because I'm going from here towards the center, which is along the direction of incident ray, isn't it? Now this particular relation, let's call this relation 1, gives you a relation between the object distance, which is y, and the image distance, which is v, OK? So if I differentiate this, let's say I differentiate it, I'll get mu divided by v square into what? dv by dt minus 1 by y square dy by dt, OK? This you will get as 0 because the right-hand side is constant, OK? So I will get the speed of the image, which is what we need to find out, that is dv by dt. So do not confuse dv by dt with rate of change of velocity, because here in optics we use v as the distance of the image, OK? So dv by dt will come out to be minus of v square by y square into dy by dt, OK? Now here I need to get the value of dy by dt over here, OK? That how will you get? dy by dt is nothing but the speed of the particles, right? So if the object has traveled a distance down by a value of, let's say, h minus y, so this is h minus y, I'll get the velocity of object or let me put it like this. I'll get dy by dt as under root of 2g h minus y, OK? And also since we need to find as a function of time, let us try to put dy by dt also as a function of time. So dy by dt will become nothing but g into t, isn't it? After time t, its velocity under the free fall becomes g into t. So this I can directly put over there, OK? And then I need to also write down the value of y and v over here as a function of time, OK? For that I will use this equation, which is spherical interface formula, OK? So if I put the values over there, I'll get mu 2 by v minus mu 1 divided by u. u is what? u is basically y only, OK? So y is a function of time. How will I write? I know that it will travel a distance of half a into t square. It will travel a distance of half a into t square in a time t, OK? So the value y will be equal to h minus half a t square, OK? So if I substitute here h minus half a t square, a is what? Acceleration due to gravity only. So I'll put here half gt square, OK? This is equal to mu 2 minus mu 1 divided by r. Now here mu 2 is what? Mu and mu 1 is 1, OK? So you will get the value of v as a function of time now, OK? So using these relations, let's say this is equation 2, this is 3, and this is equation number 4. So using 2, 3, and 4, you can get the value of the speed of the image as a function of time, OK? So like this, you have to deal with this particular question.