00:01
In this problem we have an object that is in front of a conversion lens at 16 ceres so we're gonna put the object right here and the problem says that the focal length of this first lens is 12 centimeters so we're gonna put 12 here and 12 here also the problem says that exists a second length at 20 centimeters from the first lens, so this is 20 cem mirrors.
00:42
And the focal length of this second length is minus 10 ceres, negative, because it's a diversion length.
00:52
So this is 10.
00:55
And the problem asks to compute where the final image is produced because of this system.
01:03
Okay, in order to compute where the final image is, we first need to compute where the image produced by the first length is.
01:20
For that, we are going to use the lens equation here for the first length.
01:29
So for the first lens, we can compute where the image is, 1 over q, 1 is equal to 1 over f1, so this is 1 over 12, minus 1 over p1, p1 is 16.
01:50
It's the object distance, so it's 1 over 16.
01:57
So from here we can compute the value of q1, and this is 48 centimeters, right? so the image produced just by the first lens is at 48 centimeters from the center of the first length.
02:20
So it will be maybe at this point over here.
02:25
Look, yeah, because here we have the distance between the lenses is 20, so more 28 centimeters, it's maybe here, right? and we can see that the image produced by the first length is over there using rays.
02:51
So let's see how to compute that.
02:59
Well, first we have array that comes from the image and pass through the focal point of the first length.
03:09
Something like this.
03:10
Okay, something like this, and emerge parallel to the principal axis.
03:20
So it's something like this.
03:27
Okay, yeah, this should be parallel.
03:31
And a second ray is a ray that comes parallel from the axis, something like this, and then pass through the focal point for this lens, that is this point in red here.
03:51
Okay, it's something like this, maybe.
03:57
Okay, maybe like this.
04:01
And this is the location of the image produced by the first lens, this one, right? okay.
04:11
So once we have computed this image, we must remember that this image is, it becomes an object for the second lens...