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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°, 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°) / 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°) / 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) ≈ 48.8°
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°. 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

Convex mirrors are often used in side view mirrors on cars. Many such mirrors display the warning "Objects in mirror are closer than they appear." Assume a convex mirror has a radius of curvature of \(14.0 \mathrm{~m}\) and that there is a car that is \(11.0 \mathrm{~m}\) behind the mirror. For a flat mirror, the image distance would be \(11.0 \mathrm{~m}\) and the magnification would be 1\. Find the image distance and magnification for this mirror.

Answer as true or false with an explanation for the following: The wavelength of He-Ne laser light in water is less than its wavelength in the air. (The refractive index of water is \(1.33 .\)

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What kinds of images, virtual or real, are formed by a converging mirror when the object is placed a distance away from the mirror that is a) beyond the center of curvature of the mirror, b) between the center of curvature and half the center of curvature, and c) closer than half of the center of curvature.

A solar furnace uses a large parabolic mirror (mirrors several stories high have been constructed) to focus the light of the Sun to heat a target. A large solar furnace can melt metals. Is it possible to attain temperatures exceeding \(6000 \mathrm{~K}\) (the temperature of the photo sphere of the Sun) in a solar furnace? How, or why not?
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