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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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 14
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?
Verified step by step guidance1
Step 1: Understand the concept of refraction. When light passes from one medium to another (e.g., water to air), its speed changes, causing the light to bend. This bending is described by Snell's Law: , where n₁ and n₂ are the refractive indices of the two media, and θ₁ and θ₂ are the angles of incidence and refraction, respectively.
Step 2: Identify the refractive indices for water and air. The refractive index of water is approximately , and the refractive index of air is approximately . These values will be used in Snell's Law.
Step 3: Recognize that the angle given (50°) is the angle of refraction, , as seen by the diver underwater. You need to calculate the angle of incidence, , which is the angle of the sun above the horizon as seen by the fisherman in the boat.
Step 4: Rearrange Snell's Law to solve for : . Substitute the known values: , , and .
Step 5: Perform the calculation using the formula derived in Step 4. This will give you the angle of incidence, , which represents the sun's height above the horizon as seen by the fisherman.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Angle of Elevation
The angle of elevation is the angle formed by the horizontal line and the line of sight to an object above that line. In this scenario, the diver sees the sun at an angle of 50° above the horizontal, which indicates how high the sun appears from the diver's perspective underwater.
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Critical Angle
Refraction of Light
Refraction is the bending of light as it passes from one medium to another, such as from water to air. This phenomenon affects how objects appear to observers in different mediums, meaning the angle at which the fisherman sees the sun may differ from the angle perceived by the diver due to the change in medium.
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Geometry of Sight Lines
The geometry of sight lines involves understanding how angles and distances relate to the positions of observers and objects. In this case, both the diver and the fisherman are at different depths, and the geometry helps determine the actual height of the sun above the horizon as seen by the fisherman based on the diver's angle of elevation.
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Related Practice
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
A giant ocean tank at an aquarium has acrylic plastic walls 18 cm thick. The index of refraction of acrylic plastic is 1.49. A fish is 220 cm from the inside wall. To a viewer on the outside, how far does the fish appear to be from the outside wall? Hint: The of the first refraction is the object for the second refraction.
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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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Textbook Question
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?
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