00:01
Hi, here in this given problem, focal length of the converging lens that is given as f is equal to 25 .0 centimeter and as it is a converging lens so as per sign convention the focal length will be positive.
00:35
Gap between the object and the screen means it will serve as the sum of magnitude of object distance do with the magnitude of image distance d .i.
01:03
And that is given as 125 cm.
01:10
So if we draw a rough diagram.
01:14
This is screen.
01:18
Here this is the object is assumed to be.
01:23
Then lens is anywhere between the object and the screen so that we should get a real image of this object over the screen.
01:33
So as do plus di here this is the object distance do and this is the image distance di, do plus di is 125 cm.
01:46
So, image distance in terms of object distance is 125 minus do.
01:54
So using thin lens formula 1 by di minus 1 by do is equal to 1 by f.
02:12
So plugging in all the known values here, for di this is 125 minus do minus d .o...