00:01
In this problem, we have a microscope that is formed by two converging lenses.
00:08
So we call these lenses the objective and the ips.
00:13
And the problem says that the focal length of these lenses is 1 over 18, where 18 is the refracting power.
00:24
So remember that the focal length is just the inverse of the refractive power.
00:31
Also the problem says that the distance between these two lenses is 28 meters or 0 .28 meters.
00:39
And in the first part of the problem, we want to compute what is the tithe length.
00:45
So this l here.
00:47
So remember that in a microscope, this tithe length is just the distance between the two focal points of the lenses.
01:00
So this is the secondary focal point of the objective here, and this is the primary focal point of the ips.
01:10
So as we have the values of the focal length of the lenses and also this distance, we can easily compute l.
01:21
So l is just .28 minus 1 over 18, minus 1 .18 minus 1.
01:31
Over 18 from the figure and all of this is just 0 .17 mirrors.
01:42
In the second part of the problem we want to compute what is the angular, total angular magnification.
01:52
So remember that for a microscope this total angular magnification is minus l.
01:59
This l is the tube length over the focal length of the objective times the near distance over the focal length of the ips.
02:13
And remember that this near distance is 0 .25 meters...