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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Ch 34: Geometric Optics
Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552
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Table of contents
- Ch 01: Units, Physical Quantities & Vectors37
- Ch 02: Motion Along a Straight Line57
- Ch 03: Motion in Two or Three Dimensions45
- Ch 04: Newton's Laws of Motion23
- Ch 05: Applying Newton's Laws54
- Ch 06: Work & Kinetic Energy11
- Ch 07: Potential Energy & Conservation6
- Ch 08: Momentum, Impulse, and Collisions15
- Ch 09: Rotation of Rigid Bodies37
- Ch 10: Dynamics of Rotational Motion44
- Ch 11: Equilibrium & Elasticity27
- Ch 12: Fluid Mechanics27
- Ch 13: Gravitation34
- Ch 14: Periodic Motion26
- Ch 15: Mechanical Waves22
- Ch 16: Sound & Hearing28
- Ch 17: Temperature and Heat28
- Ch 18: Thermal Properties of Matter35
- Ch 19: The First Law of Thermodynamics29
- Ch 20: The Second Law of Thermodynamics18
- Ch 21: Electric Charge and Electric Field28
- Ch 22: Gauss' Law21
- Ch 23: Electric Potential31
- Ch 24: Capacitance and Dielectrics18
- Ch 25: Current, Resistance, and EMF24
- Ch 26: Direct-Current Circuits29
- Ch 27: Magnetic Field and Magnetic Forces16
- Ch 28: Sources of Magnetic Field10
- Ch 29: Electromagnetic Induction22
- Ch 30: Inductance14
- Ch 31: Alternating Current20
- Ch 33: The Nature and Propagation of Light17
- Ch 34: Geometric Optics29
- Ch 35: Interference13
- Ch 36: Diffraction19
- Ch 37: Special Relativity19
- Ch 38: Photons: Light Waves Behaving as Particles12
- Ch 39: Particles Behaving as Waves25
- Ch 40: Quantum Mechanics I: Wave Functions20
- Ch 41: Quantum Mechanics II: Atomic Structure21
- Ch 42: Molecules and Condensed Matter17
- Ch 43: Nuclear Physics13
- Ch 44: Particle Physics and Cosmology17
Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook
Chapter 33, Problem 43a
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.
Verified step by step guidance1
Step 1: Use the lens formula to find the image distance (\(d_i\)) for the first lens. The lens formula is \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\), where \(f\) is the focal length of the lens, \(d_o\) is the object distance, and \(d_i\) is the image distance. Substitute \(f = 40.0\ \text{cm}\) and \(d_o = 50.0\ \text{cm}\) into the formula.
Step 2: Rearrange the lens formula to solve for \(d_i\): \(\frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o}\). Perform the subtraction of the reciprocals of \(f\) and \(d_o\) to find \(\frac{1}{d_i}\), then take the reciprocal to find \(d_i\).
Step 3: Determine the magnification (\(M\)) of the first lens using the formula \(M = -\frac{d_i}{d_o}\). Substitute the values of \(d_i\) and \(d_o\) to calculate \(M\).
Step 4: Calculate the height of the image (\(h_i\)) formed by the first lens using the magnification formula \(h_i = M \cdot h_o\), where \(h_o = 1.20\ \text{cm}\) is the height of the object. Substitute the values of \(M\) and \(h_o\) to find \(h_i\).
Step 5: Summarize the results for the first image (\(I_1\)): its location (\(d_i\)) relative to the first lens and its height (\(h_i\)). These values will be used as inputs for further calculations involving the second lens in subsequent parts of the problem.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
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 determining the position of the image formed by a lens. In this scenario, it will help calculate the image location (I1) created by the first converging lens.
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Lens Maker Equation
Magnification
Magnification (M) is the ratio of the height of the image (h') to the height of the object (h), given by M = h'/h = -v/u. This concept is crucial for understanding how the size of the image relates to the size of the object. In this problem, it will be used to find the height of the image formed by the first lens.
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Converging Lens
A converging lens, or convex lens, is thicker at the center than at the edges and causes parallel rays of light to converge at a focal point. The focal length is the distance from the lens to this point. Understanding the behavior of converging lenses is vital for analyzing how they form images, especially in multi-lens systems like the one described in the question.
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Related Practice
Textbook Question
A lensmaker wants to make a magnifying glass from glass that has an index of refraction n = 1.55 and a focal length of 20.0 cm. If the two surfaces of the lens are to have equal radii, what should that radius be?
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Textbook Question
An object is 16.0 cm to the left of a lens. The lens forms an image 36.0 cm to the right of the lens. Draw a principal-ray diagram.
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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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Textbook Question
An object is 16.0 cm to the left of a lens. The lens forms an image 36.0 cm to the right of the lens. If the object is 8.00 mm tall, how tall is the image? Is it erect or inverted?
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Textbook Question
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?
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