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Chapter 34: Q 51 (page 992)

A 4.0-m-wide swimming pool is filled to the top. The bottom of the pool becomes completely shaded in the afternoon when the sun is 20掳 above the horizon. How deep is the pool?

Short Answer

Expert verified

The depth of the pool is 4.0 m

Step by step solution

01

Step 1. Given information is :Width of the swimming pool = 4.0 mBottom of the pool becomes completely shaded in the afternoon when sun is 20o above the horizonRefractive index of air n1 = 1Refractive index of water n2 = 1.33

We need to find the depth of the pool

02

Step 2. Using Snell's Law

As the Bottom of the pool becomes completely shaded in the afternoon when sun is 20o above the horizon, the ray refracting from the top edge of the pool do not reach the bottom of the pool after refraction as shown below :

Using snell's Law,

n1sin胃1=n2sin胃2sin胃2=n1n2sin胃1胃2=sin-1n1n2sin胃1胃2=sin-11.001.33sin70掳胃2=44.95掳

Let d be the depth of the pool. Then,

tan胃2=4.0mdd=4.0mtan44.95掳=4.0m

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Most popular questions from this chapter

A light beam passing from medium 2 to medium 1 is refracted as shown in FIGURE Q34.4. Is n1 larger than n2, is n1smaller than n2, or is there not enough information to tell? Explain

Shows a light ray that travels from point A to point B. The ray crosses the boundary at position x, making angles 胃1and 胃2in the two media. Suppose that you did not know Snell鈥檚 law.

A. Write an expression for the time t it takes the light ray to travel from A to B. Your expression should be in terms of the distances a, b, and w; the variable x; and the indices of refraction n1 and n2

B. The time depends on x. There鈥檚 one value of x for which the light travels from A to B in the shortest possible time. We鈥檒l call it xmin. Write an expression (but don鈥檛 try to solve it!) from which xmincould be found.

C. Now, by using the geometry of the figure, derive Snell鈥檚 law from your answer to part b.

You鈥檝e proven that Snell鈥檚 law is equivalent to the statement that 鈥渓ight traveling between two points follows the path that requires the shortest time.鈥� This interesting way of thinking about refraction is called Fermat鈥檚 principle.

A 20-cm-tall object is 40cmin front of a converging lens that has a20cm focal length.
a Use ray tracing to find the position and height of the image. To do this accurately, use a ruler or paper with a grid. Determine the image distance and image height by making measurements on your diagram.
bCalculate the image position and height. Compare with your ray-tracing answers in part a.

Find the focal length of the glass lens in FIGURE EX34.25.34.25.

Shown from above in FIGURE P34.54 is one corner of a rectangular box filled with water. A laser beam starts 10cmfrom side A of the container and enters the water at position x. You can ignore the thin walls of the container.

a. If x=15cm, does the laser beam refract back into the air through side B or reflect from side B back into the water? Determine the angle of refraction or reflection.

b. Repeat part a for x=25cm.

c. Find the minimum value of x for which the laser beam passes through side B and emerges into the air.
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