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 3
A student has built a 15-cm-long pinhole camera for a science fair project. She wants to photograph her 180-cm-tall friend and have the image on the film be 5.0 cm high. How far should the front of the camera be from her friend?
Guida verificata passo dopo passo1
Understand the problem: This is a case of similar triangles formed by the object (the friend) and its image (on the film inside the pinhole camera). The height of the object, the height of the image, and the distances from the pinhole to the object and the image are related proportionally.
Write the relationship for similar triangles: \( \frac{h_{\text{image}}}{h_{\text{object}}} = \frac{d_{\text{image}}}{d_{\text{object}}} \), where \( h_{\text{image}} = 5.0 \ \text{cm} \), \( h_{\text{object}} = 180 \ \text{cm} \), and \( d_{\text{image}} = 15 \ \text{cm} \) (the length of the pinhole camera).
Rearrange the formula to solve for \( d_{\text{object}} \): \( d_{\text{object}} = \frac{d_{\text{image}} \cdot h_{\text{object}}}{h_{\text{image}}} \).
Substitute the known values into the equation: \( d_{\text{object}} = \frac{15 \cdot 180}{5.0} \).
Simplify the expression to find \( d_{\text{object}} \), which represents the distance from the front of the camera to the friend. This will give the required distance.
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Concetti chiave
Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.
Similar Triangles
The concept of similar triangles is fundamental in optics, particularly in pinhole cameras. When light rays pass through the pinhole, they create an inverted image on the film. The relationship between the height of the object, the height of the image, and their respective distances from the pinhole can be analyzed using the properties of similar triangles, which states that corresponding angles are equal and the ratios of corresponding sides are proportional.
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Percorso guidato
01:53
Charges In A Triangle (Rank Force Pairs)
Magnification
Magnification in optics refers to the ratio of the height of the image to the height of the object. It is a crucial concept for understanding how the size of an image changes based on the distance from the object to the camera. In this scenario, the magnification can be calculated using the formula: magnification = height of image / height of object, which helps determine the necessary distance to achieve the desired image size.
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09:03Mirror Equation
Pinhole Camera Formula
The pinhole camera formula relates the distances of the object and image to the size of the image produced. It can be expressed as: (height of image / height of object) = (distance from pinhole to image / distance from pinhole to object). This formula allows the student to calculate how far the camera should be positioned from her friend to achieve the desired image size, integrating the concepts of similar triangles and magnification.
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06:51Find Field due to Two Perpendicular Currents
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