00:01
Part a of our question says to use the thin lens equation to derive an expression for q in terms of f and p.
00:08
Okay, so we are going to start with the thin lens equation, which says that 1 over p, object distance, plus 1 over q, the image distance, is equal to 1 over the focal length, f.
00:24
So, we're going to isolate q, 1 over q, is equal to 1 over f minus 1 over p.
00:40
Solving for q here, we find that q is equal to f times p divided by p minus f.
00:54
You can box that in as our solution for part a.
00:59
Part b says, prove that for a real object in a diverging lens, the image must always be virtual.
01:05
So the hint then is to set f equal to minus f and show that q must be less than zero under the given circumstances.
01:13
Okay.
01:15
So for part b, we're just going to simply replace f with minus f.
01:21
So now we have minus fp divided by f plus p.
01:32
Or in other words, we can type this out, we can say that thus a diverging lens will always form a virtual image of real objects...