 Let us take up the next question, first job is fitter, so we are actually doing one of few of the most difficult questions from this chapter okay, so you can see that difficulty level of the most difficult question is not very high, so it is a very boring chapter doing properly, in fact this entire chapter is all about sign convention, once you go to sign convention right, you will be able to solve any question, you have a concave convex lens made up of glass of refractive text 1.5 and has surfaces of radii 20 centimeter and 60 centimeters okay, fine initially you had only one lens, this was not there, initially you had only one lens okay, so if you place an object 80 centimeter to the left of the lens as in the object is here 80 centimeter fine, you need to locate the image of this object, so basically your observer is here, this is your observer, this is your object, light on the object should reach the observer, this lens is not there right now okay, find the location of image okay, but what you will do here, how to solve this question, this is a lens okay, it is a thin lens, so our user tendency is to use a lens formula 1 by v minus 1 by u equals to 1 by f right, but you do not know what is f, for that we have to use, let us make a formula okay, so do not find out f and then substitute here 1 by f, find 1 by f only and then substitute here, that is what I am trying to say, unless actually you divide it and you will find out f and then you will even finding 1 by f and putting it here, anybody want to answer, anyone? Sir you have 2.8, 2.8 okay, let us put this our solve this, 1 by f will be equal to mu 2 minus mu 1 divided by mu 1, 1 by r 1 minus 1 by r 2, so when I substitute here 1.5 minus 1 divided by 1, 1 by 20 minus 1 by 16, how much is this, 1 by f is 1 by 60, 1 by 16 okay, so this nice number comes, so whether you substitute f or 1 by f, but at time what will happen, you get 1 by f suppose you know, suppose you get 0.345, so do not invert it and then put here, you directly cut 1 by f, then substitute 1 by f here, so 1 by v minus what is u, minus 80, so this is minus 80 and this is 1 by 16, so v comes out to be 240 centimeter, this is what it is okay, now what you are doing here, you are putting another lens, a similar lens you are putting here at a distance of 160 centimeter okay, and your observer is now here looking at things, where is the English, anyone got, how much, 48 from the second end, what you are getting, what you are getting, what you are getting, what you are getting, you are signing the mention, if I do one thing draw a red icon, it will really help you, if you draw a red icon, take a point object, do not take an extended object, from here the image of that will be 240 centimeter, so it will try to form an image somewhere over here, whose distance is 240, fine, but it is not able to do that, so what will happen, light further converts, because this converging lens will convert like this, so finally it will be here somewhere, this is initial image I1 and that is I2, so I2 is where the actual image gets formed, I1 is the first image, this I1 is the object of this lens, and now it is converging lens, because so far light comes out to be positive, it converges a parallel phase forward, that is why it is converging, what you got, 240 by, that is approximately how much, no, anybody else, any other answer, 18, no, see 18 is anywhere 80 centimeter from here, isn't it, this is what, 80 centimeter only, I2 has to be less than 80, because it further converges, same, identical, what you are getting, 80, you are also getting 20, okay, so you can talk among yourself, see where, what you are getting, 48, should I do it now, lens converges what, 1 by v minus 1 by u equals to 1 by s, now v is what we need to find out, what is u, u is what for this, 80 or minus 80, so this is plus 80, why, because object is that side, okay, f is what, f is 60 or you guys did, you were doing better in addition, so v is, is, what I am doing, what is written is, find a thin lens of focal length plus 12 centimeter, okay, this is the focal length given, now this lens is immersed in water of refractive index 4 by 3, what is a new focal length, how much it is, focal length in air is given, focal length of this lens in air is given, we need to have it in a gentle water, so if I change the refractive index of medium, will focal length change, extra change, it depends on the medium's refractive index also, mu 1 is medium's refractive index, we have 1 by f equals to mu 2 minus mu 1 by mu 1, 1 by r 1 minus 1 by r 2, so we have to use this, so in case number one, it is air, the the surrounding. So, 1 by f which is 12 is equal to mu 2 is let us say mu minus 1 divided by mu sorry divided by 1 only 1 by r 1 minus 1 by r 2 ok. We need to find out what is f in case the refractive measure of medium becomes 4 by 3 mu is 1.5 ok. So, it is not given, but it should have been. So, it will take mu as 1.5. So, when you substitute here 1.5. So, if you mistake like that. So, how it will not happen. No, perfect sir I do not have sense like I thought you have to find the u 2 and then substitute into the thing. No, you have this 1 by r 1 minus 1 by r 2 is also variable fine you have f also variable you have only 2 equations you cannot make another variable. I took that 1 by r 1 minus 1 by r 2. So, divide once you divide these will go away. Sir like I should be given the refractive measure of material should be given. So, it is 1.5 fine. So, this completes the function of the which these with lenses ok.