Domanda del libro di testo
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° above the horizon. How deep is the pool?
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Ch 34: Ray Optics
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796
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TOC.tableOfContents
- 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: 9780137344796Non è quello che usi tu?Cambia libro di testo
Capitolo 34, Problema 47a
A light ray in air is incident on a transparent material whose index of refraction is n. Find an expression for the (non-zero) angle of incidence whose angle of refraction is half the angle of incidence.
Guida verificata passo dopo passo1
Start by applying Snell's Law, which relates the angles of incidence and refraction to the indices of refraction of the two media: . Here, is the index of refraction of air (approximately 1), is the index of refraction of the material (), is the angle of incidence, and is the angle of refraction.
Substitute the given condition that the angle of refraction is half the angle of incidence: . This means .
Using the trigonometric identity for the sine of a half-angle, rewrite as .
Substitute this expression for into Snell's Law: .
Square both sides of the equation to eliminate the square root, and solve for . This will give you an expression for the angle of incidence in terms of the index of refraction .
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Concetti chiave
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Snell's Law
Snell's Law describes the relationship between the angles of incidence and refraction when a light ray passes between two different media. It is mathematically expressed as n1 * sin(θ1) = n2 * sin(θ2), where n1 and n2 are the indices of refraction of the two media, and θ1 and θ2 are the angles of incidence and refraction, respectively.
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07:59
Snell's Law
Index of Refraction
The index of refraction (n) is a dimensionless number that describes how much light slows down when it enters a material compared to its speed in a vacuum. It is defined as n = c/v, where c is the speed of light in a vacuum and v is the speed of light in the material. A higher index indicates that light travels slower in that medium.
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03:46Index of Refraction
Angle Relationships
In the context of the problem, the relationship between the angle of incidence (θ1) and the angle of refraction (θ2) is crucial. The question specifies that θ2 is half of θ1, which can be expressed as θ2 = θ1/2. This relationship will be used in conjunction with Snell's Law to derive the expression for the angle of incidence.
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03:43Relationships Between Force, Field, Energy, Potential
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