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Chapter 32: Problem 46

You are submerged in a swimming pool. What is the maximum angle at which you can see light coming from above the pool surface? That is, what is the angle for total internal reflection from water into air?

Short Answer

Expert verified
Answer: The maximum angle at which light can be seen from above the surface of a swimming pool is approximately 48.8掳, which is the angle for total internal reflection from water to air.

Step by step solution

01

Find the refractive indices of the mediums

Given that we are considering light traveling from water (medium 1) to air (medium 2), we need to find the refractive indices of these two mediums. The refractive index of water is approximately 1.33 and that of air is 1.00.
02

Understand the Snell's Law formula

Snell's law relates the angle of incidence, angle of refraction, and the refractive indices of the mediums involved. It is given by the formula: n_1 * sin(theta_1) = n_2 * sin(theta_2) Here, n_1 and n_2 are the refractive indices of medium 1 and medium 2, respectively, while theta_1 is the angle of incidence and theta_2 is the angle of refraction.
03

Set up the formula for critical angle

At the critical angle, the angle of refraction (theta_2) is 90 degrees. Plugging this into Snell's law and rearranging for the angle of incidence (theta_1), we get: sin(theta_1) = n_2 * sin(90掳) / n_1
04

Apply values and calculate critical angle

Substitute the refractive indices of water and air found in Step 1: sin(theta_1) = 1.00 * sin(90掳) / 1.33 The sine of 90 degrees is 1, so the equation simplifies to: sin(theta_1) = 1 / 1.33 Now, find the inverse sine to get the critical angle: theta_1 = arcsin(1 / 1.33) 鈮� 48.8掳
05

Interpret the results

The maximum angle at which you can see light coming from above the pool surface is equal to the critical angle, which we calculated to be approximately 48.8掳. This is the angle for total internal reflection from water to air in a swimming pool, and represents the threshold at which light entering from above the surface can no longer be seen by an observer submerged in the water.

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Most popular questions from this chapter

An object is located at a distance of \(100 . \mathrm{cm}\) from a concave mirror of focal length \(20.0 \mathrm{~cm}\). Another concave mirror of focal length \(5.00 \mathrm{~cm}\) is located \(20.0 \mathrm{~cm}\) in front of the first concave mirror. The reflecting sides of the two mirrors face each other. What is the location of the final image formed by the two mirrors and the total magnification by the combination?

You are under water in a pond and look up at the smooth surface of the water, noticing the sun in the sky. Is the sun in fact higher in the sky than it appears to you while under water, or is it lower?

Use Fermat's Principle to derive the law of reflection.

Why does refraction happen? That is, what is the physical reason a wave moves in a new medium with a different velocity than it did in the original medium?

A \(45^{\circ}-45^{\circ}-90^{\circ}\) triangular prism can be used to reverse a light beam: The light enters perpendicular to the hypotenuse of the prism, reflects off each leg, and emerges perpendicular to the hypotenuse again. The surfaces of the prism are not silvered. If the prism is made of glass with in dex of refraction \(n_{\text {glass }}=1.520\) and the prism is surrounded by air, the light beam will be reflected with a minimum loss of intensity (there are reflection losses as the light enters and leaves the prism). a) Will this work if the prism is under water, which has index of refraction \(n_{\mathrm{H}_{2} \mathrm{O}}=1.333 ?\) b) Such prisms are used, in preference to mirrors, to bend the optical path in quality binoculars. Why?
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