A lens of focal length $$f$$ produces an image of an object located 30 cm on one side of it at a distance of 60 cm on the other side. If the lens is replaced by another of $$\frac{3f}{4}$$, where would the image form? If the image has to form at the earlier position by what distance should the object be shifted?

PLACEHOLDER ANSWER - the question has two answers. The first lens has $$f = \frac{30 \times 60}{90} = 20\text{ cm}$$, so the new lens has $$f = 15\text{ cm}$$. With the object at 30 cm, $$\frac{1}{v} = \frac{1}{15} - \frac{1}{30} = \frac{1}{30}$$, so the image forms 30 cm from the lens. For the image to stay at 60 cm, $$\frac{1}{u} = \frac{1}{15} - \frac{1}{60} = \frac{1}{20}$$, so $$u = 20\text{ cm}$$ and the object must be moved 10 cm closer to the lens.