Textbook Question
A 1.0-cm-tall object is 10 cm in front of a converging lens that has a 30 cm focal length. Calculate the image position and height. Compare with your ray-tracing answers in part a.
1
views
Ch 34: Ray Optics
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796
Not the one you use?Change textbookSelect a different textbook
Table of contents
- 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: 9780137344796Not the one you use?Change textbook
Chapter 34, Problem 31
A 1.0-cm-tall candle flame is 60 cm from a lens with a focal length of 20 cm. What are the distance and the height of the flame's image?
Verified step by step guidance1
Identify the given values: The object height \( h_o \) is 1.0 cm, the object distance \( d_o \) is 60 cm, and the focal length \( f \) is 20 cm. We need to find the image distance \( d_i \) and the image height \( h_i \).
Use the lens formula to find the image distance \( d_i \): \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \). Rearrange this equation to solve for \( \frac{1}{d_i} \): \( \frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} \). Substitute the given values for \( f \) and \( d_o \).
After calculating \( \frac{1}{d_i} \), take the reciprocal to find \( d_i \), the image distance. This will tell you where the image is located relative to the lens.
To find the image height \( h_i \), use the magnification formula: \( M = \frac{h_i}{h_o} = -\frac{d_i}{d_o} \). Rearrange this to solve for \( h_i \): \( h_i = M \cdot h_o \). Substitute the values for \( M \), \( h_o \), and \( d_i \).
Interpret the results: The sign of \( d_i \) will indicate whether the image is real or virtual (positive for real, negative for virtual), and the sign of \( h_i \) will indicate whether the image is upright or inverted (positive for upright, negative for inverted).
Verified video answer for a similar problem:
This video solution was recommended by our tutors as helpful for the problem above.
Was this helpful?
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 expressed as 1/f = 1/v - 1/u. This formula is essential for determining the position of the image formed by the lens based on the position of the object.
Recommended video:
05:38
Lens Maker Equation
Magnification
Magnification is the ratio of the height of the image (h') to the height of the object (h), and it can also be expressed as the negative ratio of the image distance (v) to the object distance (u). It is given by the formula M = h'/h = -v/u. Understanding magnification helps in determining how the size of the image compares to the size of the object.
Recommended video:
09:03Mirror Equation
Sign Convention in Optics
In optics, the sign convention dictates how distances are measured in relation to the lens. Typically, distances measured in the direction of the incoming light are considered negative, while those in the direction of outgoing light are positive. This convention is crucial for correctly applying the lens formula and calculating image properties.
Recommended video:
Guided course
08:55Net Torque & Sign of Torque
Related Practice
Textbook Question
Find the focal length of the glass lens in FIGURE EX34.28.
Textbook Question
A 2.0-cm-tall object is 15 cm in front of a plano-convex polystyrene plastic lens that has a 13 cm radius of curvature. What are the (a) position and (b) height of the image?
1
views
Textbook Question
Find the focal length of the plano-concave polystyrene plastic lens in FIGURE EX34.25.
1
views
Textbook Question
A goldfish lives in a 50-cm-diameter spherical fish bowl. The fish sees a cat watching it. If the cat's face is 20 cm from the edge of the bowl, how far from the edge does the fish see it as being? (You can ignore the thin glass wall of the bowl.)
2
views
Textbook Question
A laser beam in air is incident on a liquid at an angle of 53° with respect to the normal. The laser beam's angle in the liquid is 35°. What is the liquid's index of refraction?
1
views