00:03
This problem is about a galilean telescope.
00:07
This telescope uses a converging lens for the objective and a divergent lens for the ips.
00:18
The problem says that these two lenses are 32 ceres apart and that the focal length of the objective is 36 ceres.
00:29
Also, the problem says that the primary focal point of the objective is at the same place as the secondary focal point of the ips.
00:46
So it's this point here.
00:50
And with this information, we want to compute in the first part of the problem, what is the focal length of the ips.
00:59
So from the figure we see that this distance here is 36 ceres, is the distance between the primary focal point of the objective with the objective.
01:14
And also we have that this is 32 ceres, so from here we can say that this distance here is just four ceres and this represents a focal length of the ips.
01:34
This is just four centimeters, but we must remember that for a diverging lens, the focal length is negative, so this is minus four.
01:44
In the next part of the problem, we want to compute at what distance from the ips the final image is produced, so we can call it just qe.
01:56
Qe is the image distance from the ips, and we're gonna take just just the absolute value as we only want to get what this distance is.
02:07
So for the ips we can use the focal, sorry, the lens equation.
02:14
So the lens equation is this.
02:25
And this object distance for the ips is this distance here.
02:36
And this distance is for cere mirrors, but as this object is proposed, on the opposite side from the original object.
02:46
This is a negative object distance.
02:50
This is for the ips.
02:52
So if we replace this value here in this equation, this is minus 4 zero mirrors.
03:00
The focal length is also minus 4 3 meters.
03:03
So these two numbers are the same.
03:05
We get that 1 over qe is just 0.
03:11
So from here we can get two values.
03:12
From qe.
03:16
So this is infinity or this also can be minus infinity.
03:23
But in a telescope we want that the final image will be produced on the same side of the original object.
03:32
So the original object is in this region.
03:35
We also expect that the final image will be produced in this region.
03:40
This region is a negative image distance for the ips as it is on the same high of the original object.
03:51
So that's why we are going to take this value here.
03:57
But for the answer of the problem, of this part of the problem, we just want the absolute value of that.
04:04
So it's no problem, this is just infinity.
04:07
We just want to know what that distance is.
04:13
So the distance is just the absolute value.
04:15
This is different from the image distance.
04:18
Okay, in the next part of the problem, we want to know if this image is virtual or it's real, but as we already know that this number here is negative.
04:34
This is because the image is virtual.
04:39
So remember, when the final image distance is negative, it's virtual, and if it's positive, it is real.
04:47
In order to see if this image is inverted or upright, we're going to just compute the total transversal magnification.
04:58
This is not the same as the angular magnification.
05:03
The angular modification is other concept.
05:07
This is just the magnification from each lens.
05:12
And we multiply this value.
05:14
So for the first lens, for the objective, this is minus qo from objective over bo.
05:23
Minus qe for ibis over be...