Domanda del libro di testo
A point source of light illuminates an aperture 2.0 m away. A 12.0-cm-wide bright patch of light appears on a screen 1.0 m behind the aperture. How wide is the aperture?
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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 5
The mirror in FIGURE EX34.5 deflects a horizontal laser beam by 60°. What is the angle ϕ?
Guida verificata passo dopo passo1
Identify the relationship between the angles of incidence and reflection. According to the law of reflection, the angle of incidence is equal to the angle of reflection.
Recognize that the total deflection angle of the laser beam is given as 60°. This deflection angle is the sum of the angle of incidence and the angle of reflection.
Let the angle of incidence be denoted as θ. Since the angle of incidence equals the angle of reflection, the total deflection angle can be expressed as 2θ = 60°.
Solve for θ by dividing both sides of the equation by 2: θ = 60° / 2.
The angle ϕ is the angle between the mirror's surface and the incident beam. Since the angle of incidence θ is measured relative to the normal to the mirror, ϕ can be calculated as ϕ = 90° - θ.
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Reflection of Light
Reflection of light occurs when a light beam strikes a surface and bounces off. The angle of incidence, which is the angle between the incoming beam and the normal (a perpendicular line to the surface), is equal to the angle of reflection. This principle is crucial for understanding how mirrors work and how they redirect light beams.
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Percorso guidato
06:25
Law of Reflection
Angle of Deflection
The angle of deflection refers to the angle by which a beam of light is redirected after striking a surface, such as a mirror. In this context, if a laser beam is deflected by a certain angle, it is essential to understand how this angle relates to the angles of incidence and reflection to determine the resulting angles in the system.
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Percorso guidato
01:04Critical Angle
Geometry of Angles
Understanding the geometry of angles is vital for solving problems involving reflections. In this scenario, the total angle change can be calculated by considering the initial angle of the beam, the angle of incidence, and the angle of reflection. This geometric approach allows for the determination of unknown angles, such as ϕ, based on the known deflection.
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Percorso guidato
02:01Flux Through Angled Surface
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