00:01
So as you can see from the question that we are given some information here.
00:07
We are given the two lenses here, lens one and lens two.
00:14
Lens one is having focal length which they are calling is f1.
00:18
Lens 2 is having a focal length which they are calling as f2.
00:21
The value of f2 is given as 200 units and the value of f1 is given as 100 units.
00:29
Okay and we are given that there is a distance between these two lenses and the distance is given as 590 units so another thing is there is an object lying here on the principal axis which is at a distance of 150 meters 150 units so now we have to find the position of the image position of the final image formed by this system of lens, and that location of the final image we have to find with respect to the lens too.
01:11
So now let's do that.
01:12
We know that if there is an object, definitely, there would be a corresponding image that we are going to see somewhere here, and that will act as an object for the second lens.
01:26
So if i call this distance as, let's say, v1, we can get the value of v1.
01:32
1 out we know that 1 by b minus 1 by u is always equal to 1 by f so 1 by b i'm calling as b1 minus 1 by u u is 150 so it would be minus 150 because lying on the left side of the lens so this would be minus of 150 and the focal length we know here it is 100 so 1 by hundreds solving this we can get the value of v1 which is coming out to be 300 units so if this value b1 is 300 units and this entire value is 590 then this value we can see very easily this value would be equal to 590 590 590 which will give us the value 290 so this is coming out to be 290 now, this will serve as an object for the second lens.
02:34
For the second lens, we can apply the formula.
02:37
Let us do that.
02:39
We know that 1 by b.
02:41
I am repeating the formula once again, minus 1 by u is equal to 1 by f.
02:46
Now 1 by b, b is the final image...