Refraction at a curved surface: A swimming pool is filled to a depth of 2.0 m. How deep does the pool appear to be from above the water, which has an index of refraction of 1.33?

Question

Quizplus · Accepted Answer

When light travels from one medium to another, its speed changes, and it bends as a result. This bending of light is called refraction. The amount of bending of light depends on the difference between the indices of refraction of the two media.

In this problem, light travels from water (with an index of refraction of 1.33) to air (with an index of refraction of 1.00). As a result, the light bends away from the normal as it exits the water.

To determine how deep the pool appears to be from above the water, we need to take into account the apparent depth due to refraction. The apparent depth is the distance from the surface of the water to the point where the light appears to come from.

We can use the formula for the apparent depth, which is:

(1) 1/f = (n2/n1 - 1) (1/r1 - 1/r2)

where f is the focal length, n1 is the index of refraction of the medium the light is coming from (in this case, air), n2 is the index of refraction of the medium the light is going into (in this case, water), and r1 and r2 are the radii of curvature of the two surfaces (in this case, we assume the surfaces are flat, so the radii are infinite).

Solving for f, we get:

(2) f = r2 / (1 - n1/n2)

In this case, we want to find the apparent depth, which is f + d, where d is the actual depth of the pool (2.0 m). So we have:

(3) apparent depth = f + d

Substituting (1) and (2) into (3), we get:

(4) apparent depth = r2 / (1 - n1/n2) + d

Plugging in the given values, we have:

n1 = 1.00
n2 = 1.33
d = 2.0 m
r2 = infinity (since we assume the surface is flat)

So (4) simplifies to:

(5) apparent depth = 2.0 m + infinity

Since infinity is undefined, we need to approximate the result. In general, when light travels from a medium with a higher index of refraction to one with a lower index of refraction, the apparent depth is less than the actual depth. In other words, the pool will appear shallower than it actually is.

The difference in depth is given by:

(6) h = d - apparent depth

At the surface of the water, where the angle of incidence is 0 degrees, the light does not bend, so the apparent depth is equal to the actual depth. As the angle of incidence increases, the light bends more, and the apparent depth decreases.

At the critical angle, which is the angle of incidence where the refracted light travels along the surface of the water and the angle of refraction is 90 degrees, the apparent depth is zero. Beyond the critical angle, the refracted light is totally reflected back into the medium with the higher index of refraction (in this case, water).

For water and air, the critical angle is approximately 49 degrees. We can use Snell's law to find the critical angle:

n1 sin(theta1) = n2 sin(theta2)

where theta1 is the angle of incidence, theta2 is the angle of refraction, and n1 and n2 are the indices of refraction of the two media.

At the critical angle, theta2 is 90 degrees, so sin(theta2) = 1. Plugging in the values for water and air, we get:

1.00 sin(theta1) = 1.33 sin(90)

sin(theta1) = 1.33 / 1.00

sin(theta1) = 1.33

This is impossible, since the sine of an angle cannot be greater than 1. Therefore, there is no critical angle for light passing from water to air, and all light that strikes the surface of the water at any angle is refracted into the air.

Since we know that the apparent depth is less than the actual depth, and we also know that the difference in depth is small (less than 20%), we can use a simple approximation:

(7) apparent depth = d / n2

Plugging in the values for water and air, we get:

(8) apparent depth = 2.0 m / 1.33

(9) apparent depth = 1.50 m

Therefore, the pool appears to be 1.50 m deep from above the water. The best choice is A, 1.5 m.

Refraction at a Curved Surface: a Swimming Pool Is Filled