00:01
Okay, in this problem we're asked a lot of questions concerning a concave mirror with a focal length of f.
00:06
And we have an object located some distance greater than f, which will, of course, result in a real image.
00:13
So to start us off, we're asked to prove that the magnification, m, is equal to f over f minus d not.
00:19
We're then asked to sketch what m looks like for varying values of d0.
00:23
And we're asked to find the point where m equals 1, where the image height and object height will be identical.
00:29
So there's no magnification at all.
00:31
And we're asked to find some general region where m is much greater than one resulting in a big amount of magnification.
00:37
So to start us off, we're going to be making use of the definition of magnification.
00:42
M is equal to the ratio of heights, but it's also equal to the negative ratio of distances from the image and object respectively.
00:52
With that said, we can write the image distance as negative m -d -0.
01:00
We can now combine this with the ideal lens equation, 1 over f is equal to 1 over di plus 1 over d1, or d .i.
01:12
1 over d0, o is equal to 1 over d .i for object distance and image distance...