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 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 40b
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?
Verified step by step guidance1
Identify the given values: The object distance \( d_o \) is \( 16.0 \; \text{cm} \), the image distance \( d_i \) is \( 36.0 \; \text{cm} \), and the object height \( h_o \) is \( 8.00 \; \text{mm} \) (convert to cm: \( h_o = 0.800 \; \text{cm} \)).
Recall the magnification formula: \( M = -\frac{d_i}{d_o} \), where \( M \) is the magnification, \( d_i \) is the image distance, and \( d_o \) is the object distance. Substitute the given values to calculate \( M \).
Use the relationship between magnification and height: \( M = \frac{h_i}{h_o} \), where \( h_i \) is the image height and \( h_o \) is the object height. Rearrange the formula to solve for \( h_i \): \( h_i = M \cdot h_o \). Substitute the calculated \( M \) and the given \( h_o \) to find \( h_i \).
Determine whether the image is erect or inverted: The sign of \( M \) indicates the orientation of the image. If \( M \) is negative, the image is inverted. If \( M \) is positive, the image is erect.
Summarize the results: Report the calculated image height \( h_i \) and state whether the image is erect or inverted based on the sign of \( M \).
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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. It is given by the equation 1/f = 1/v - 1/u. Understanding this formula is crucial for determining the characteristics of the image formed by the lens, including its position and size.
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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), expressed as M = h'/h. It also relates to the object and image distances by the formula M = -v/u. This concept helps determine whether the image is erect or inverted based on the sign of the magnification value.
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09:03Mirror Equation
Sign Convention for Lenses
The sign convention for lenses dictates how distances are measured in optics. For a converging lens, distances measured in the direction of the incoming light (toward the lens) are negative, while distances measured in the direction of the outgoing light (away from the lens) are positive. This convention is essential for correctly applying the lens formula and determining the nature of the image.
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06:37Ray Diagrams for Converging Lenses
Related Practice
Textbook Question
For each thin lens shown in Fig. E34.37, calculate the location of the image of an object that is 18.0 cm to the left of the lens. The lens material has a refractive index of 1.50, and the radii of curvature shown are only the magnitudes.
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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
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
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
A converging lens with a focal length of 9.00 cm forms an image of a 4.00 mm tall real object that is to the left of the lens. The image is 1.30 cm tall and erect. Where are the object and image located? Is the image real or virtual?
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