A bright object is fixed at a distance of $$100\text{ cm}$$ from a wall. A convex lens placed between the object and the wall at position A forms a sharp image of the object magnified 4 times on the wall. Find the focal length of the lens. The lens is moved to a new position B to obtain sharp image again. Find the distance between positions A and B.

PLACEHOLDER ANSWER - the question has two answers. With magnification 4, $$v = 4u$$ and $$u + v = 100$$, so $$u = 20\text{ cm}$$ and $$v = 80\text{ cm}$$. Then $$f = \frac{uv}{u + v} = \frac{1600}{100} = 16\text{ cm}$$. The second sharp image is obtained when object and image distances interchange ($$u = 80$$, $$v = 20$$), so the lens moves $$80 - 20 = 60\text{ cm}$$.