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Ch. 32 - Light: Reflection and Refraction
Giancoli Douglas - Physics for Scientists and Engineers 5th edition
Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook
Chapter 31, Problem 71

We wish to determine the depth of a swimming pool filled with water by measuring the width (x = 5.20m) and then noting that the bottom edge of the pool is just visible at an angle of 13.0° above the horizontal as shown in Fig. 32–61. Calculate the depth of the pool.


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Identify the key concept: This problem involves refraction of light at the water-air interface. The apparent position of the bottom of the pool is affected by the bending of light due to the change in medium. We will use Snell's Law and trigonometry to solve for the depth of the pool.
Apply Snell's Law at the water-air interface: \( n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \), where \( n_1 \) is the refractive index of water (approximately 1.33), \( \theta_1 \) is the angle of incidence (unknown), \( n_2 \) is the refractive index of air (approximately 1.00), and \( \theta_2 \) is the angle of refraction (13.0°). Rearrange to solve for \( \sin(\theta_1) \): \( \sin(\theta_1) = \frac{n_2 \sin(\theta_2)}{n_1} \).
Determine \( \theta_1 \): Use the inverse sine function to calculate \( \theta_1 \) from \( \sin(\theta_1) \). This angle will help us understand the path of the light ray inside the water.
Use trigonometry to relate the depth of the pool to the width and the angle \( \theta_1 \): Inside the water, the light ray travels at an angle \( \theta_1 \) relative to the vertical. The depth \( d \) of the pool can be found using the tangent function: \( \tan(\theta_1) = \frac{x}{d} \), where \( x = 5.20 \text{ m} \) is the width of the pool. Rearrange to solve for \( d \): \( d = \frac{x}{\tan(\theta_1)} \).
Substitute the known values into the equation for \( d \): Use the calculated value of \( \theta_1 \) and the given width \( x \) to find the depth of the pool. Ensure all trigonometric calculations are performed in degrees unless otherwise specified.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Refraction of Light

Refraction is the bending of light as it passes from one medium to another, such as from air into water. This phenomenon occurs due to the change in the speed of light in different materials. In this scenario, the angle at which the bottom of the pool is visible is influenced by the refraction of light, which must be accounted for when calculating the depth.
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Index of Refraction

Snell's Law

Snell's Law describes the relationship between the angles of incidence and refraction when light passes through different media. It is mathematically expressed as n1 * sin(θ1) = n2 * sin(θ2), where n is the refractive index and θ is the angle. This law is essential for determining the angle at which light travels from the water to the observer's eye, which is crucial for calculating the pool's depth.
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Trigonometric Relationships

Trigonometric relationships involve the ratios of the sides of a right triangle to its angles. In this problem, the width of the pool and the angle of visibility create a right triangle, where the depth can be calculated using the tangent function. Understanding these relationships allows for the application of trigonometry to find the depth based on the given width and angle.
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Related Practice
Textbook Question

Two plane mirrors are facing each other 2.2 m apart as in Fig. 32–60. You stand 1.5 m away from one of these mirrors and look into it. You will see multiple images of yourself. (a) How far away from you are the first three images of yourself in the mirror in front of you? (b) Are these first three images facing toward you or away from you?


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Textbook Question

(c) Determine the magnification of a plane mirror in this same limit.

(d) Are your results in parts (b) and (c) consistent with the discussion of Section 32–2 on plane mirrors?

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Textbook Question

A 1.80-m-tall person stands 4.20 m from a convex mirror and notices that he looks precisely half as tall as he does in a plane mirror placed at the same distance. What is the radius of curvature of the convex mirror? (Assume that θ ≈ θ .) [Hint: The viewing angle is half.]

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Textbook Question

The critical angle of a certain piece of plastic in air is θC = 35.8°. What is the critical angle of the same plastic if it is immersed in water?

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Textbook Question

The label on a laser says it produces light of wavelength 670 nm. The laser beam passes through a block of plastic for which n = 1.57. What is the wavelength of the light inside the plastic?

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Textbook Question

When light passes through a prism, the angle that the refracted ray makes relative to the incident ray is called the deviation angle δ, Fig. 32–64. Show that this angle is a minimum when the ray passes through the prism symmetrically, perpendicular to the bisector of the apex angle Φ, and show that the minimum deviation angle, δm, is related to the prism’s index of refraction n by


n=sin12(ϕ+δm)sinϕ/2.n = \(\frac{\sin \frac{1}{2}\)(\(\phi\) + \(\delta\)_m)}{\(\sin\) \(\phi\)/2}.


[Hint: For θ in radians, (d/dθ)(sin1θ)=1/1θ2(d/d\(\theta\)) (\(\sin\)^{-1}\(\theta\)) = 1/\(\sqrt{1 - \theta^2}\).]

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