Textbook Question
A camera lens has a focal length of 180.0 mm and an aperture diameter of 16.36 mm. What is the ƒ-number of the lens?
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Ch 34: Geometric Optics
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 34, Problem 46b
You wish to project the image of a slide on a screen 9.00 m from the lens of a slide projector. If the dimensions of the picture on a 35 mm color slide are 24 mm ✖ 36 mm, what is the minimum size of the projector screen required to accommodate the image?
Verified step by step guidance1
Determine the magnification (M) of the image using the lens formula and the given distances. The magnification is given by \( M = \frac{d_i}{d_o} \), where \( d_i \) is the image distance (9.00 m or 9000 mm) and \( d_o \) is the object distance (distance from the slide to the lens).
Calculate the dimensions of the projected image on the screen using the magnification. The width of the image is \( M \times 36 \ \text{mm} \), and the height of the image is \( M \times 24 \ \text{mm} \).
Convert the dimensions of the projected image from millimeters to meters by dividing by 1000, as the screen size is typically measured in meters.
Determine the minimum size of the projector screen required by considering the larger of the two dimensions (width or height) and ensuring the screen can accommodate both dimensions.
Summarize the result by stating the minimum dimensions of the screen in meters, ensuring it is large enough to display the entire projected image.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Magnification
Magnification is the process of enlarging the appearance of an object through optical devices like lenses. It is defined as the ratio of the image size to the object size. In this context, understanding how the slide's dimensions relate to the projected image size on the screen is crucial for determining the minimum screen size needed.
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Lens Formula
The lens formula relates the object distance (u), image distance (v), and focal length (f) of a lens, expressed as 1/f = 1/v + 1/u. This formula is essential for calculating how the slide's image is formed by the projector lens, which directly influences the size of the projected image on the screen.
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Aspect Ratio
Aspect ratio is the ratio of the width to the height of an image or screen. For the slide dimensions given (24 mm by 36 mm), calculating the aspect ratio helps ensure that the projected image maintains its proportions on the screen. This is important for determining the minimum dimensions of the screen to avoid distortion of the image.
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Related Practice
Textbook Question
Repeat Exercise 34.41 using the same lenses except for the following changes: The second lens is a diverging lens having a focal length of magnitude 60.0 cm.
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Textbook Question
Where is the near point of an eye for which a contact lens with a power of +2.75 diopters is prescribed?
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Textbook Question
A 1.20 cm tall object is 50.0 cm to the left of a converging lens of focal length 40.0 cm. A second converging lens, this one having a focal length of 60.0 cm, is located 300.0 cm to the right of the first lens along the same optic axis. Find the location and height of the image (call it I1) formed by the lens with a focal length of 40.0 cm.
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Textbook Question
BIO The Lens of the Eye. The crystalline lens of the human eye is a double-convex lens made of material having an index of refraction of 1.44 (although this varies). Its focal length in air is about 8.0 mm, which also varies. We shall assume that the radii of curvature of its two surfaces have the same magnitude. Find the radii of curvature of this lens.
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Textbook Question
Zoom Lens. Consider the simple model of the zoom lens shown in Fig. 34.43a. The converging lens has focal length f1 = 12 cm, and the diverging lens has focal length f2 = -12 cm. The lenses are separated by 4 cm as shown in Fig. 34.43a. (a) For a distant object, where is the of the converging lens? (c) Where is the final image? Compare your answer to Fig. 34.43a.
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