Lenses Question. My Work Inside!?

Question

Question: A diverging lens has a focal length of -32 cm. An object is placed 19 cm in front of this lens. Calculate (a) the image distance and (b) the magnification. Is the image (c) real or virtual, (d) upright or inverted, and (e) enlarged or reduced in size?

My Approach: for (a) I got di = -11.9 cm, (b) m = 0.63, (c) virtual - di is negative, (d) upright - magnification is positive, (e) reduced - magnification is less than 1.

Tell me if you get the same numbers or different numbers. I would like to know if I am approaching this question without making any errors? Thanks!

Answer 1

Wiki, not always my favorite source, has a great section on optics. The appropriate page (ref.) gives the formulas for image size and magnification, which you seem to have followed. Define S as distance, X as size, f as focal length, 1 as object, 2 as image.
1/S1 + 1/S2 = 1/f, so S2 = 1/(1/f - 1/S1)
magnification M = X2/X1 = -S2/S1 = f/(f-S1)
So for (a), (b), I get S2 (or di) = -11.922, M = 0.627.
Agree on (c), (d) and (e).
For me the hard part is keeping straight the signs of S1 and S2. S1, the object distance, is always positive. S2 is only positive if it is on the opposite side of the lens. So when S2 is negative (because S1 < f), they are both on the same side, and you have a virtual image.