00:01
We're told that we have an object 20 centimeters to the left of a diverging lens.
00:05
So p here is 20 centimeters.
00:08
And the focal length f is equal to negative 32 centimeters.
00:12
Okay.
00:13
And it wants to determine a, the location, and b, the magnification of the image.
00:17
And then for part c, it wants us to construct a ray diagram that's consistent with the results from part a and part b.
00:23
So for part a, we want to find where the image is located.
00:27
So we're going to use the equation that says 1 over the object distance plus 1 over the image distance is equal to 1 over the focal length.
00:36
So we can solve for q, which is the image distance.
00:39
That's what we're asked to find for part a.
00:41
We find that q is equal to the focal length, f, times the object distance, p, divided by p minus f.
00:51
So plugging the values for p and the values for f, we find that this is equal to minus 12 .3 centimeters.
00:58
And this minus sign indicates it's going to be to the left of the.
01:01
The lens.
01:02
So it's going to be on the same side as the object.
01:05
We can box it in as our solution for part a.
01:09
Part b wants us to find the magnification, and this can be done using the equation for magnification, m, which says that it's equal to minus q over p.
01:21
So plugging in negative 12 .3 for q and 20 for p, we find that m is equal to 0 .615.
01:31
So it's a little more than half the size of the original object.
01:38
And then for the last part, part c, it wants us to construct the ray diagram.
01:43
Okay.
01:44
So to do this, let's first draw in green here, the lens.
01:50
It's divergent.
01:51
So it looks something like this.
01:55
I apologize about these crude drawings here.
01:57
We'll do the best we can.
01:59
Okay.
01:59
And then we'll draw something.
02:02
This is the line here...