A coin is at the bottom of a fountain 60 cm deep. The water in the fountain is frozen. What is the apparent depth of the coin? The index of refraction of ice is 1.31.

n=R/A

A=R/n = 0.6/1.31=0.458 m
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To find the apparent depth of the coin, we can use the concept of refraction. Refraction occurs when light passes from one medium to another, causing the light rays to change direction.

In this case, the light rays will travel from the water in the fountain into the ice. The index of refraction of ice (n) is given as 1.31.

The apparent depth of an object in a different medium can be found using Snell's law, which states that the ratio of the sine of the angle of incidence (θ1) to the sine of the angle of refraction (θ2) is equal to the ratio of the speed of light in the first medium to the speed of light in the second medium.

Since the light rays are traveling from water to ice, the angles of incidence and refraction are related by the equation:

n₁ * sin(θ₁) = n₂ * sin(θ₂)

In this case, the incident angle (θ₁) is 0 degrees because the light rays are perpendicular to the surface of the water. Therefore, sin(θ₁) = sin(0) = 0.

The equation simplifies to:

0 = n₂ * sin(θ₂)

We need to find the apparent depth, which is the distance from the surface of the ice to the coin. Let's assume this distance is d.

From the given information, we know that the actual depth of the coin in the water is 60 cm. Therefore, the distance from the surface of the water to the coin is 60 cm + d.

Since the light rays travel straight down from the surface of the water to the surface of the ice, the angles of incidence and refraction are perpendicular. This means that sin(θ₂) = 1.

Let's substitute these values into the equation:

0 = n₂ * 1

Solving for n₂:

n₂ = 0

Since there is no valid value for n₂, it means that there is no refraction at the surface of the ice. Therefore, the apparent depth of the coin is the same as its actual depth in the water, which is 60 cm.