A cat is watching a goldfish in a spherical fishbowl of radius 15.0 cm.The goldfish is 5.00 cm from the side of the bowl nearest the cat.How far from the side of the bowl does the fish appear to be to the cat? (For water,n = 1.33.)
A)7.5 cm
B)4.5 cm
C)4.1 cm
D)3.5 cm
E)2.5 cm

Question

Quizplus · Accepted Answer

When light travels from water to air, it bends away from the normal. In order to find out where the fish appears, we need to find the apparent depth of the fish. Using Snell's Law, we can find the angle of incidence:

sin(i)/sin(r) = n(water)/n(air)
sin(i) = n(air)/n(water) * sin(r)
sin(i) = (1.00/1.33) * sin(arcsin(5/15))
sin(i) = 0.376
i = 22.3 degrees

The cat's line of sight passes through the apparent position of the fish, so we can draw a right triangle where 22.3 degrees is opposite the shortest side (the distance from the fish to the bowl), and the radius of the bowl is the hypotenuse. Let x be the distance from the fish to the side of the bowl.

sin(22.3) = x/15
x = 15*sin(22.3)
x = 4.1 cm
 
Therefore, the fish appears to be 4.1 cm from the side of the bowl to the cat. The answer is (C).

A Cat Is Watching a Goldfish in a Spherical Fishbowl