00:01
So we need to solve this problem in numerous parts.
00:05
So we'll start by first analyzing the image formed by objective.
00:17
So we have the distance of the object from the lens is 114 meters.
00:23
We know that the focal length is 1 .50 meters and we will try to find the position of the image.
00:32
So using the thin lens formula, 1 over do plus 1 over d .i is equal to to 1 over f we have 1 over 114 plus 1 over di is equal to 1 over 1 .5 or 1 over di is equal to 1 over 1 .5 minus 1 over 1114 so that gives us di is equal to 1 .52 meters so we can now find the magnification of the objective which is negative the i over d o so negative 1 .52 divided by 114 gives us the magnification of the objective as negative 0 .0133 so now that we know the magnification of the objective we will go ahead and find the discuss the image formed by the i piece okay, so we know that do, the distance of the object, would be equal to l minus 1 .52, where l is the distance between the two lenses, and that is actually equal to the sum of the two focal lens.
02:00
So this would be fo plus f .e minus 1 .52.
02:05
The fo is 1 .5.
02:08
The f .e is 0 .07 ,000, the f .e is 0 .07 minus 1 .5 .5.
02:12
So the object distance for the objective comes out to be 0 .05 meters.
02:19
And we already said that the focal length is 0 .07 meters.
02:26
So we'll start by finding out the location of the image.
02:30
So 1 over do plus 1 over d .i is equal to 1 over f.
02:35
1 over 0 .05 plus 1 over d .i is equal to 1 over 0 .07.
02:45
Or 1 over the i is equal to 1 over 0 .07 minus 1 over 0 .05 so that gives us the i is equal to negative 0 .18 meters so we can now find the magnification of the i piece which is negative di over d o so negative of negative 0 .18 divided by 0 .05 so that comes out to be negative, that comes out to be positive 3 .6.
03:26
So now we know the magnification of the objective.
03:29
We also know the magnification of the i piece.
03:33
So the overall magnification, m would be equal to m .o.
03:42
Times m .e.
03:44
So we had m .o.
03:46
Is equal to negative 0 .133.
03:48
And then times that with 3 .6.
03:53
So we get the overall magnification is equal to negative 0 .048.
04:01
So this is the linear magnification.
04:04
Now let's go ahead and find the angular magnification.
04:09
So first thing for angular magnification that we need is the angle form by the object on the unedited...