Question

A small fish, four feet below the surface of Lake in viewed through a simple thin converging lens with focal length 30 feet. If the lens is 2 feet above the water surface (Fig. (1)), 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 = 1.0, n_water = 1.33.

• A. 2 feet below water surface
• B. 4 feet below water surface
• C. 6 feet below water surface
• D. 8 feet below water surface

Answer 1

Answer: (B)

4 feet below water surface

Answer 2

An object at P in water appears to be at P' as seen by an observer in air, as Fig. (2) shows.

Fig. (2)

The paraxial light emitted by P is refracted at the water surface, for which

1.33 sin i₁ = sin i₂

As i₁, i₂ are very small, we have the approximation 1.33 i₁ = i₂. Also,

i₂ = a », i₁ = b » .

Hence, we have

$\overline { \mathrm { OP } }$= 3 ft.

Let the distance between the apparent location of the fish and the center of the lens be u, then

u = 2 + = 5 ft.

From =+, we have

v = – 6 ft.

Therefore, the image of the fish is still where the fish is, four feet below the water surface.