 All right friends so here is another question in front of your screen this is again from Ray Optics chapter okay so a paper weight in the form of hemisphere of radius three centimeters is used to hold down a printed page an observer looks at the page vertically through the paper weight at what height above the page will the printed letter near the center appears to the observer okay now since it is a paper weight and you are holding it holding the papers so definitely you'll put the you know hemisphere like this only okay so hemisphere is placed like this okay and you have an object observer is looking at the page vertically down right so basically just imagine that there is something written over here so this is the object all right you need to find out what is the height of this printed letter near the center okay so basically the observer is here and it is looking at the object and we need to find out where is the image of this object okay so if you draw the ray diagram let's say one ray goes like this okay so if one ray goes like that then how this ray will you know refract okay so let us try to find out the location of the image for this object okay so we have a spherical interface here okay we can say that refractive index of this glass let us say that is 1.5 and defective index of air we can take it as 1 okay so object is in this refractive index so if I use this formula mu 2 by v minus mu 1 by u is equal to mu 2 minus mu 1 by r you know you can see that the object refractive index is 1.5 right so mu 1 is 1.5 and mu 2 is 1 over here so 1 divided by v minus 1.5 divided by what now this is the spherical interface I am talking about and this is this point is optical center okay so the incident rays are traveling like this so from optical center I am going against the incident light isn't it so u is negative of the r which is minus 3 centimeter okay this is equal to 1 minus 1.5 divided by r now tell me what is r r again in order to measure r I have to travel to the center of curvature which is again I am traveling against the incident ray okay so even r you have to put it as minus 3 okay so let's try to solve this so 1 by v is equal to minus of 1.5 by 3 you know and this becomes minus of 0.5 divided by minus 3 okay so this will come out to be minus 1 by 2 and plus 1 by 6 okay so 1 by v will come out to be equal to 6 and this will be minus 3 plus 1.