Question

A catfish is 1.5 m below the surface of a smooth lake.

Answer 1

Answer:

A. The diameter of the circle would be 3.4 m

B. As the fish descends the diameter would increase.

Explanation:

Part A

The critical angle can be obtained with the expression below;

Sin θ = 1/n

given n is the refractive index = 1.333

sinθ = 1/1.333

θ = $sin^{-1}$ ( 1/1.333)

θ = 48.75$3^{0}$

Therefore the critical angle would be θ = 48.75$3^{0}$

from the diagram attached the sine of the critical angle can be evaluated thus;

Remember tan θ = Opposite / adjacent

From our diagram the radius r is the opposite, the adjacent is the 1.5 m depth. Therefore the tan of the critical angle would be;

tan θ = r/1.5

Tan 48.75$3^{0}$ = r/ 1.5

1.1404 = r / 1.5

r = 1.1404 x 1.5

r = 1.7106 m

Since the radius is 1.7106 m the diameter would be;

D = r x 2

D = 1.7106  x 2

D = 3.4212 m

D ≈ 3.4 m

Therefore the diameter of the circle would be 3.4 m

Part B

Since the fish is descending at a constant critical angle and same refractive index. There would be an increase in the adjacent resulting in a corresponding increase in the opposite which represents the diameter.

Therefore as the fish descends the diameter would increase.