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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 77

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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Identify the relationship between the wavelength of light in a vacuum and its wavelength in a medium. The wavelength in a medium is given by the formula: λmedium = λvacuum / n, where λvacuum is the wavelength in a vacuum, and n is the refractive index of the medium.
Substitute the given values into the formula. The wavelength in a vacuum is λvacuum = 670 nm, and the refractive index of the plastic is n = 1.57.
Perform the division to calculate the wavelength in the plastic: λmedium = 670 / 1.57. This will give the wavelength of the light inside the plastic in nanometers (nm).
Understand the physical meaning: The wavelength of light decreases when it enters a medium with a refractive index greater than 1 because the speed of light is reduced in the medium, but the frequency remains constant.
Ensure the units are consistent throughout the calculation. Since the input wavelength is in nanometers, the result will also be in nanometers.

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Key Concepts

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

Wavelength

Wavelength is the distance between successive peaks of a wave, typically measured in nanometers (nm) for light. It determines the color of visible light and is inversely related to frequency. In this context, the wavelength of the laser light in air is given as 670 nm.
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Index of Refraction

The index of refraction (n) is a dimensionless number that describes how light propagates through a medium. It is defined as the ratio of the speed of light in a vacuum to the speed of light in the medium. For the plastic in this question, n = 1.57 indicates that light travels slower in plastic than in air.
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Index of Refraction

Wavelength in Different Media

When light enters a medium with a different index of refraction, its wavelength changes while its frequency remains constant. The relationship is given by the formula: λ' = λ/n, where λ' is the wavelength in the medium, λ is the wavelength in vacuum or air, and n is the index of refraction. This concept is essential for calculating the new wavelength of the laser light inside the plastic.
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Related Practice
Textbook Question

Light from a laser (in air) strikes the exact center of one face of a solid glass cube (n = 1.40) at an angle θ relative to the normal. The refracted beam travels inside the glass until it strikes an adjacent face of the cube. The original angle of incidence θ is such that no light exits the cube where the beam strikes the second face. What is the maximum value θ can have?

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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

A triangular prism made of crown glass (n = 1.52) with base angles of 26.0° is surrounded by air. If parallel rays are incident normally on its base as shown in Fig. 32–66, what is the angle Φ between the two emerging rays?

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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

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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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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