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) 1.50 cm
b) 2.50 cm
c) 3.50 cm
d) 4.50 cm

Answer 1

Using the formula for apparent depth derived from Snell's Law, the fish is actually swimming 4.50 cm from the wall of the bowl, corresponding to option (d).

Step-by-step explanation:

The question is related to the optical phenomenon of refraction and requires the use of Snell's Law to determine the actual depth of a fish swimming inside a fishbowl. To find out how far from the wall of the bowl the fish is actually swimming, we can use the formula for apparent depth, which is derived from Snell's Law:

real depth = apparent depth * (index of refraction of water / index of refraction of air)

In this case, the fish appears to be swimming 6.0 cm away from the wall of the bowl. We have the following values: nwater = 1.33 and nair = 1.00.

Using the formula, we get:
real depth = 6.0 cm * (1.00 / 1.33)
real depth = 4.51 cm (approximately)

Therefore, the fish is actually swimming 4.50 cm from the wall of the bowl, which corresponds to option (d).