How Does Refraction Affect Image Location in a Thin Lens Observation?

[guilesar]

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

\frac{1}{f} = \frac{1}{s_{0}} + \frac{1}{s_{i}}

and

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

where R₂=-R₁

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...

 

[Delphi51]

I don't know how to apply the lens formula to this situation.
If I were doing the problem, I would go back to basics and sketch some rays. If the middle of the fish is on the optic axis then the nose of the fish will be a bit off the axis and you can draw a ray straight up out of the water to the lens and it will bend to the focal point above the lens. Ignoring the water/air surface for a sec, a second ray from the nose through the center of the lens would not bend and the intersection of the two rays would give the position of the image - looks like deeper than the actual location of the fish. But that ray hits the air surface at an angle greater than zero, so it will bend. One could calculate the bending and then try to deal with the change in the bending at the lens due to it not going through the center any more, or perhaps take a different ray that does go through the center but at a different angle due to the bending at the water/air surface. Either way, if you can compute the trajectory of two rays from the fish through the lens, you can then find the intersection of those rays and you'll have the location of the image.