Textbook Question
It is 165 cm from your eyes to your toes. You're standing 200 cm in front of a tall mirror. How far is it from your eyes to the image of your toes?
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Ch 34: Ray Optics
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796
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Table of contents
- Ch 01: Concepts of Motion35
- Ch 02: Kinematics in One Dimension75
- Ch 03: Vectors and Coordinate Systems58
- Ch 04: Kinematics in Two Dimensions60
- Ch 05: Force and Motion23
- Ch 06: Dynamics I: Motion Along a Line53
- Ch 07: Newton's Third Law27
- Ch 08: Dynamics II: Motion in a Plane52
- Ch 09: Work and Kinetic Energy56
- Ch 10: Interactions and Potential Energy50
- Ch 11: Impulse and Momentum56
- Ch 12: Rotation of a Rigid Body71
- Ch 13: Newton's Theory of Gravity52
- Ch 14: Fluids and Elasticity44
- Ch 15: Oscillations55
- Ch 16: Traveling Waves37
- Ch 17: Superposition39
- Ch 18: A Macroscopic Description of Matter53
- Ch 19: Work, Heat, and the First Law of Thermodynamics60
- Ch 20: The Micro/Macro Connection53
- Ch 21: Heat Engines and Refrigerators34
- Ch 22: Electric Charges and Forces40
- Ch 23: The Electric Field39
- Ch 24: Gauss' Law32
- Ch 25: The Electric Potential63
- Ch 26: Potential and Field57
- Ch 27: Current and Resistance42
- Ch 28: Fundamentals of Circuits39
- Ch 29: The Magnetic Field47
- Ch 30: Electromagnetic Induction60
- Ch 31: Electromagnetic Fields and Waves32
- Ch 32: AC Circuits63
- Ch 33: Wave Optics32
- Ch 34: Ray Optics57
- Ch 35: Optical Instruments30
- Ch 36: Special Relativity38
- Ch 37: The Foundations of Modern Physics25
- Ch 38: Quantization49
- Ch 39: Wave Functions and Uncertainty31
- Ch 40: One-Dimensional Quantum Mechanics35
- Ch 41: Atomic Physics18
- Ch 42: Nuclear Physics27
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
Chapter 34, Problem 13
The glass core of an optical fiber has an index of refraction 1.60. The index of refraction of the cladding is 1.48. What is the maximum angle a light ray can make with the wall of the core if it is to remain inside the fiber?
Verified step by step guidance1
Identify the concept: This problem involves total internal reflection in an optical fiber. The critical angle is the key to determining the maximum angle a light ray can make with the wall of the core while staying inside the fiber.
Calculate the critical angle: Use the formula for the critical angle \( \theta_c \) at the core-cladding interface: \( \sin(\theta_c) = \frac{n_2}{n_1} \), where \( n_1 = 1.60 \) (core) and \( n_2 = 1.48 \) (cladding). Solve for \( \theta_c \) using \( \theta_c = \arcsin\left(\frac{n_2}{n_1}\right) \).
Relate the critical angle to the maximum angle inside the fiber: The maximum angle a light ray can make with the wall of the core is related to the critical angle by geometry. The angle of incidence inside the core, \( \theta_i \), is complementary to the angle the ray makes with the wall. Thus, \( \theta_i = 90^\circ - \theta_c \).
Determine the maximum angle with the wall: The maximum angle a light ray can make with the wall of the core is \( \theta_{max} = 90^\circ - \theta_i \). Substitute \( \theta_i \) from the previous step to find \( \theta_{max} = \theta_c \).
Summarize the process: To find the maximum angle, calculate the critical angle using the indices of refraction, then use the relationship between the critical angle and the geometry of the fiber to determine the maximum angle with the wall.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Index of Refraction
The index of refraction 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 indicates that light travels slower in that medium, affecting how light bends when entering or exiting different materials.
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03:46
Index of Refraction
Total Internal Reflection
Total internal reflection occurs when a light ray traveling in a denser medium hits the boundary of a less dense medium at an angle greater than the critical angle. This phenomenon allows light to be completely reflected back into the denser medium, which is essential for the functioning of optical fibers, ensuring that light remains confined within the core.
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Critical Angle
The critical angle is the minimum angle of incidence at which total internal reflection occurs. It can be calculated using the indices of refraction of the two media involved. For optical fibers, knowing the critical angle helps determine the maximum angle at which light can enter the fiber while still being guided effectively within the core.
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Related Practice
Textbook Question
A 1.0-cm-thick layer of water stands on a horizontal slab of glass. A light ray in the air is incident on the water 60° from the normal. What is the ray's direction of travel in the glass?
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Textbook Question
An underwater diver sees the sun 50° above horizontal. How high is the sun above the horizon to a fisherman in a boat above the diver?
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Textbook Question
A costume jewelry pendant made of cubic zirconia is submerged in oil. A light ray in the oil strikes one face of the zirconia crystal at an angle of incidence of 25°. Once inside, what is the ray's angle with respect to the face of the crystal?
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Textbook Question
A biologist keeps a specimen of his favorite beetle embedded in a cube of polystyrene plastic. The hapless bug appears to be 2.0 cm within the plastic. What is the beetle's actual distance beneath the surface?
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Textbook Question
To a fish in an aquarium, the 4.00-mm-thick walls appear to be only 3.50 mm thick. What is the index of refraction of the walls?
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