00:01
A coin is located 20 centimeters to the left of a converging lens with a vocal length of 16 centimeters.
00:07
A second identical lens is placed to the right of the first lens in such a way that the image formed by the combination will have the same size and orientation as the original coin.
00:16
Find the separation between the lenses.
00:19
Okay, so the magnification, let's just start it off, is negative d .i.
00:25
Over d .o.
00:25
So the magnification of the final lens is going to be the two magnifications multiplied together.
00:36
We also have our thin lens equation 1 over f is equal to 1 over do plus 1 over d .i.
00:43
We were given that the focal length of the first lens is 16 centimeters, and it's an identical lens for the second lens.
00:53
And the distance of the object in the first is 20 centimeters.
01:02
So using this equation in only our comma 1 values, so f1 in d .o .1, we are solving for d .i .1...