Question

A fishbowl has a radius of curvature of 7.0 cm. a beta fish is swimming inside the bowl. as you stand outside of the bowl, the fish appears to be swimming in the water 6.0 cm away from the wall of the bowl. how far from the wall of the bowl is the fish actually swimming? the index of refraction of water is 1.33 and the index of refraction of air is 1.00. assume the fishbowl is very thin.

a. 11.1 cm from the bowl
b. 0.1 cm from the bowl
c. 10.3 cm from the bowl
d. 15.1 cm from the bowl

Answer 1

The correct answer is c. 10.3 cm from the bowl.

How we can solve this problem?

Snell's Law: When light travels from one medium (air) to another (water), its path bends due to the difference in refractive indices. Snell's Law describes this relationship:

n air * sin(θ air) = n water * sin(θwater)

where:

n air and n water are the refractive indices of air and water, respectively

θ air and θ water are the angles of incidence and refraction

The radius of curvature (r) of the bowl relates to the object and image distances (d o and d i) by the equation:

1/d o + 1/d i = 1/f

where f is the focal length of the bowl, which can be approximated as f = r/2 for a thin lens.

d o = d i * (1 + (n air - n water) / n water) / (2 - (n air - n water) / n water)

d o = 6.0 cm * (1 + (1.00 - 1.33) / 1.33) / (2 - (1.00 - 1.33) / 1.33) ≈ 10.3 cm