A converging lens and a diverging lens, separated by a distance of 30 cm, are used in combination. The converging lens has a focal length of 15 cm. The diverging lens is of unknown focal length. An object is placed 19 cm in front of the converging lens; the final image is virtual and is formed 12 cm before the diverging lens. What is the focal length of the diverging lens?
I know for sure I'm suppose to use the equations:
1/f1 = 1/p1 + 1/q1
1/f2 = 1/p2 + 1/q2
p2 = s - q1
The Attempt at a Solution
Here's what I tried to do. I understand that I'm suppose to do one lens at a time:
1/q1 = 1/f1 - 1/p1 = 1/15cm - 1/19cm = 71.25cm
p2 = s - q1 = 30cm - 71.25cm = -41.25cm
f2 = 1/q2 + 1/p2 = 16.923cm
I tried changing signs of each number part, but I'm still not getting the right answer. I tried reading class notes and textbook and I still can't figure out what I did wrong, or if there's something I'm missing
There's another similar problem to this I'm having difficulty with (which adds height), but I think I'll understand that too once I get this.
Any help would be greatly appreciated. :)