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.

 

[tomcue]

I would first address the assumptions made in the problem. The given information states that the fish is viewed through a simple thin converging lens, which implies that the lens is thin enough that we can use the thin lens equation. The problem also assumes that the fish is on the optical axis of the lens, which simplifies the problem and allows us to use the thin lens equation without any additional calculations.

Using the thin lens equation, we can calculate the image distance (s_i) by setting the object distance (s_0) to 4 feet (since the fish is 4 feet below the surface of the lake) and the focal length (f) to 30 feet. This gives us an image distance of -120 feet, meaning the image of the fish will be 120 feet behind the lens.

However, we need to take into account the refraction of light at the air-water interface. The problem states that the refractive index of air (n_air) is 1 and the refractive index of water (n_water) is 1.33. Using Snell's law, we can calculate the angle of refraction at the air-water interface. This angle can then be used to calculate the apparent depth of the fish, taking into account the refraction of light through the water.

Using the given information, we can then calculate the final position of the image of the fish relative to the observer. This will give us the location of the image of the fish as seen by the observer.

In conclusion, the image of the fish will appear to be 120 feet behind the lens, but due to the refraction of light through the water, the apparent depth of the fish will be different, resulting in a final image location that is closer to the observer. This counterintuitive result can be explained by the principles of refraction and the assumptions made in the problem.