00:01
So we have a lens l1 and the focal length of this length of this lens is negative 3 .00 meters.
00:11
And we have an object, do, which is 2 meters away.
00:19
So we need to find the location of the image.
00:24
So we have the thin lens formula 1 over do plus 1 over d .i is equal to 1 over f.
00:32
We're trying to figure out 1 over di.
00:35
So 1 over 2 plus 1 over di is equal to 1 over negative 3.
00:42
And di will be equal to solving this equation for di gives the i is equal to negative 1 .2 meters.
00:52
So what that means is that the image will be 1 .2 meters away from the first lens.
00:59
So this distance is negative 1 .2 meters.
01:04
Now for part b, there's another lens that comes into play.
01:07
First, before we go there, let's find the magnification.
01:10
Magnification is negative d .i.
01:13
Over d .o., which is negative of negative 1 .2 divided by 2.
01:20
So the magnification for part a is positive 0 .6...