Refraction (magnification) - very basic

1. Dec 10, 2006

Question:

A lens produces a real image that is twice as large as the object and is located 15 cm from the lens. Find (a) the object distance and (b) the focal length of the
lens.

$$m = \frac{-i}{p}$$

So the image is twice as large so obviously $$m = 2$$. The lens produces a real image that is 15cm from the lens. Thus, $$i > 0$$ RIGHT??? So we are simply left with,

$$p = \frac{-15cm}{2}$$. How can $$p$$ (the objects distance) be negative? This doesn't make any sense to me. I can resolve this physically but not in the math... I would like to know why.

This problem should be trivial.

2. Dec 10, 2006

freethinker

The issue here is that there are bunch of easy-to-forget sign conventions that go along with the optics formulas. In particular: we are dealing here with a REAL image. Real images are ALWAYS inverted. Because the real image is inverted, it's magnification is technically -2, not +2. This will solve your sign problem and make the object distance positive.

3. Dec 11, 2006