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## Homework Statement

A small fish, four feet below the surface of Lake Lansing is viewed through a simple thin converging lens with focal length 30 feet. If the lens is 2 feet above the water surface, where is the image of the fish seen by the observer? Assume the fish lies on the optical axis of the lens and that n

_{air}=1, n

_{water}=1.33.

## Homework Equations

[tex]\frac{1}{f}[/tex] = [tex]\frac{1}{s_{0}}[/tex] + [tex]\frac{1}{s_{i}}[/tex]

and

[tex]\frac{1}{f}[/tex] = [tex]\left(n-1\right)[/tex][tex]\left[\frac{1}{2R}\right][/tex]

where R

_{2}=-R

_{1}

## The Attempt at a Solution

This problem is counter intuitive and I'm not sure how to solve it. The focal length of the lens is 30 feet... so the fish is before the focal length. Once the rays refract from the sunlight into the water, the light rays refract towards the normal, which in my mind would cause the light rays to converge further away... So I'm not sure how the water and converging lens would allow someone to see something closer than the focal length when in a isotropic medium an object before the lens would create a virtual image...