00:01
So we have two combination of lenses being used to produce an image.
00:04
So we have a diversion lens with a focal length of minus 14, placed to the right of a conversion lens with focal length of 18 centimeters.
00:15
And we have an object placed to the left of the conversion lens.
00:19
We want to know where the final image will be located and where will the image be if the diversion lens is 30 -8 -10 meters from the conversion lens.
00:28
Converging lens.
00:31
So for part a, we just use the lens equation that the i1 is d01 f1 divided by d01 minus f1.
00:44
So we need to find the first image that will be the object for the second lens.
00:49
This is 33 times 18 divided by 33 minus 18.
00:56
This is 39 .6 centimeters.
01:01
This image is the object for the second lens.
01:05
The image is to the right of the second lens, so it is virtual.
01:08
We use the image to find the final image distance.
01:13
So d02 is 12 minus 39 .6.
01:19
It is minus 27 .6.
01:22
And now we use this value for the second lens...