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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
All textbooksGiancoli Douglas 5th editionCh. 32 - Light: Reflection and RefractionProblem 62
Chapter 31, Problem 62

(III) A beam of light enters the end of an optic fiber as shown in Fig. 32–59. (a) Show that we can guarantee total internal reflection at the side surface of the material (at point A), if the index of refraction is greater than about 1.42. In other words, regardless of the angle α , the light beam reflects back into the material at point A, assuming air outside. (b) What if the fiber is immersed in water?

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Step 1: Understand the concept of total internal reflection. Total internal reflection occurs when light travels from a medium with a higher index of refraction to a medium with a lower index of refraction, and the angle of incidence exceeds the critical angle. The critical angle can be calculated using the formula: θc=sin⁻1(n2n1), where n1 is the index of refraction of the denser medium and n2 is the index of refraction of the less dense medium.
Step 2: For part (a), assume the optic fiber is surrounded by air. The index of refraction of air is approximately 1. To guarantee total internal reflection at the side surface (point A), the critical angle must be less than 90 degrees. Using the formula for the critical angle, substitute n2=1 and solve for n1. This will show that the index of refraction of the fiber must be greater than approximately 1.42.
Step 3: For part (b), consider the case where the optic fiber is immersed in water. The index of refraction of water is approximately 1.33. Repeat the calculation for the critical angle using n2=1.33. Determine the minimum index of refraction of the fiber required to ensure total internal reflection at point A.
Step 4: Analyze the results. In part (a), the fiber's index of refraction must be greater than 1.42 to guarantee total internal reflection when surrounded by air. In part (b), the fiber's index of refraction must be greater than a different value (calculated in step 3) to guarantee total internal reflection when immersed in water.
Step 5: Conclude that the surrounding medium affects the conditions for total internal reflection. The higher the index of refraction of the surrounding medium, the higher the required index of refraction of the fiber to ensure total internal reflection. This is a direct consequence of the relationship between the critical angle and the indices of refraction.

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Key Concepts

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

Total Internal Reflection

Total internal reflection occurs when a light ray traveling in a denser medium hits a boundary with a less dense medium at an angle greater than the critical angle. This phenomenon ensures that the light is completely reflected back into the denser medium rather than refracting out. The critical angle depends on the indices of refraction of the two media involved, and for total internal reflection to occur, the index of refraction of the denser medium must be higher than that of the less dense medium.
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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. A higher index of refraction indicates that light travels slower in that medium. For total internal reflection to occur, the index of refraction of the fiber must be greater than that of the surrounding medium, such as air or water.
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Critical Angle

The critical angle is the minimum angle of incidence at which total internal reflection occurs when light travels from a denser medium to a less dense medium. It can be calculated using Snell's Law, which relates the angles of incidence and refraction to the indices of refraction of the two media. If the angle of incidence exceeds the critical angle, all the light is reflected back into the denser medium, which is essential for the functioning of optical fibers.
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Related Practice
Textbook Question

A fish is swimming in water inside a thin spherical glass bowl of uniform thickness. Assuming the radius of curvature of the bowl is 32.0 cm, locate the image of the fish if the fish is located: (a) at the center of the bowl; (b) 20.0 cm from the side of the bowl between the observer and the center of the bowl. Assume the fish is small.

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Textbook Question

(II) A ray of light, after entering a light fiber, reflects at an angle of 14.5° with the long axis of the fiber, as in Fig. 32–57. Calculate the distance along the axis of the fiber that the light ray travels between successive reflections off the sides of the fiber. Assume that the fiber has an index of refraction of 1.55 and is 1.60 x 10-4 m in diameter.

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Textbook Question

The critical angle for a certain liquid–air surface is 52.6°. What is the index of refraction of the liquid?

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Textbook Question

Two identical concave mirrors are set facing each other 1.0 m apart. A small lightbulb is placed halfway between the mirrors. A small piece of paper placed just to the left of the bulb prevents light from the bulb from directly shining on the left mirror, but light reflected from the right mirror still reaches the left mirror. A good image of the bulb appears on the left side of the piece of paper. What is the focal length of the mirrors?

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 beam of light is emitted 7.7 cm beneath the surface of a liquid and strikes the surface 7.2 cm from the point directly above the source. If total internal reflection occurs, what can you say about the index of refraction of the liquid?

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