Textbook Question
A beam of light has a wavelength of 650 nm in vacuum. What is the wavelength of these waves in the liquid?
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Ch 33: The Nature and Propagation of Light
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610
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Table of contents
- Ch 01: Units, Physical Quantities & Vectors41
- Ch 02: Motion Along a Straight Line67
- Ch 03: Motion in Two or Three Dimensions47
- Ch 04: Newton's Laws of Motion23
- Ch 05: Applying Newton's Laws59
- Ch 06: Work & Kinetic Energy14
- Ch 07: Potential Energy & Conservation8
- Ch 08: Momentum, Impulse, and Collisions18
- Ch 09: Rotation of Rigid Bodies42
- Ch 10: Dynamics of Rotational Motion48
- Ch 11: Equilibrium & Elasticity27
- Ch 12: Fluid Mechanics31
- Ch 13: Gravitation35
- Ch 14: Periodic Motion32
- Ch 15: Mechanical Waves25
- Ch 16: Sound & Hearing32
- Ch 17: Temperature and Heat36
- Ch 18: Thermal Properties of Matter38
- Ch 19: The First Law of Thermodynamics33
- Ch 20: The Second Law of Thermodynamics22
- Ch 21: Electric Charge and Electric Field36
- Ch 22: Gauss' Law24
- Ch 23: Electric Potential31
- Ch 24: Capacitance and Dielectrics18
- Ch 25: Current, Resistance, and EMF26
- Ch 26: Direct-Current Circuits30
- Ch 27: Magnetic Field and Magnetic Forces18
- Ch 28: Sources of Magnetic Field10
- Ch 29: Electromagnetic Induction28
- Ch 30: Inductance17
- Ch 31: Alternating Current21
- Ch 32: Electromagnetic Waves1
- Ch 33: The Nature and Propagation of Light20
- Ch 34: Geometric Optics34
- Ch 35: Interference19
- Ch 36: Diffraction25
- Ch 37: Special Relativity20
- Ch 38: Photons: Light Waves Behaving as Particles15
- Ch 39: Particles Behaving as Waves26
- Ch 40: Quantum Mechanics I: Wave Functions21
- Ch 41: Quantum Mechanics II: Atomic Structure25
- Ch 42: Molecules and Condensed Matter18
- Ch 43: Nuclear Physics16
- Ch 44: Particle Physics and Cosmology23
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
Chapter 33, Problem 9
Light traveling in air is incident on the surface of a block of plastic at an angle of 62.7° to the normal and is bent so that it makes a 48.1° angle with the normal in the plastic. Find the speed of light in the plastic.
Verified step by step guidance1
First, identify the known values: the angle of incidence (θ₁) is 62.7° and the angle of refraction (θ₂) is 48.1°. The speed of light in air (v₁) is approximately 3.00 x 10⁸ m/s.
Use Snell's Law to relate the angles and the indices of refraction: n₁ sin(θ₁) = n₂ sin(θ₂), where n₁ is the refractive index of air (approximately 1) and n₂ is the refractive index of the plastic.
Rearrange Snell's Law to solve for the refractive index of the plastic (n₂): n₂ = n₁ sin(θ₁) / sin(θ₂). Substitute the known values into this equation.
Once you have n₂, use the relationship between the speed of light in a medium and its refractive index: v₂ = v₁ / n₂, where v₂ is the speed of light in the plastic.
Substitute the values for v₁ and n₂ into the equation to find v₂, the speed of light in the plastic.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Snell's Law
Snell's Law describes how light bends when it passes from one medium to another. It is expressed as n1 * sin(θ1) = n2 * sin(θ2), where n1 and n2 are the refractive indices of the respective media, and θ1 and θ2 are the angles of incidence and refraction. This law is crucial for determining the refractive index of the plastic in this problem.
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Snell's Law
Refractive Index
The refractive index of a medium is a measure of how much it reduces the speed of light compared to the speed of light in a vacuum. It is calculated as n = c/v, where c is the speed of light in a vacuum and v is the speed of light in the medium. Knowing the refractive index allows us to find the speed of light in the plastic.
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Speed of Light in a Medium
The speed of light in a medium is determined by the medium's refractive index. It is given by the formula v = c/n, where c is the speed of light in a vacuum (approximately 3.00 x 10^8 m/s) and n is the refractive index of the medium. This concept is essential for calculating the speed of light in the plastic once the refractive index is known.
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Related Practice
Textbook Question
As shown in Fig. E33.11, a layer of water covers a slab of material X in a beaker. A ray of light traveling upward follows the path indicated. Using the information on the figure, find the angle the light makes with the normal in the air.
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Textbook Question
A horizontal, parallel-sided plate of glass having a refractive index of 1.52 is in contact with the surface of water in a tank. A ray coming from above in air makes an angle of incidence of 35.0° with the normal to the top surface of the glass. What angle does the ray refracted into the water make with the normal to the surface?
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Textbook Question
Light of a certain frequency has a wavelength of 526 nm in water. What is the wavelength of this light in benzene?
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Textbook Question
A light beam travels at 1.94 × 108 m/s in quartz. The wavelength of the light in quartz is 355 nm. If this same light travels through air, what is its wavelength there?
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Textbook Question
(a) A tank containing methanol has walls 2.50 cm thick made of glass of refractive index 1.550. Light from the outside air strikes the glass at a 41.3° angle with the normal to the glass. Find the angle the light makes with the normal in the methanol. (b) The tank is emptied and refilled with an unknown liquid. If light incident at the same angle as in part (a) enters the liquid in the tank at an angle of 20.2° from the normal, what is the refractive index of the unknown liquid?
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