EBK PHYSICS FOR SCIENTISTS AND ENGINEER
16th Edition
ISBN: 8220100546716
Author: Katz
Publisher: CENGAGE L
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Textbook Question
Chapter 37, Problem 1PQ
A camera obscura is used to form an image of a distant object. If the object is 10.0 m away from the aperture and the magnification of the image is –0.15, what is the image distance?
Expert Solution & Answer
To determine
The distance of the image.
Answer to Problem 1PQ
The image distance is
Explanation of Solution
Write the expression for the magnification.
Here,
From the above equation we get,
Conclusion:
Substitute
Therefore, the image distance is
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Students have asked these similar questions
When two lenses are used in combination, the first one forms an image that then serves as the object for the second lens. The magnification of the combination is the ratio of the height of the final image to the height of the object. This is equal to the product of the separate magnification values mm of each lens. In equation form
mtotal=(m1)(m2).
A 1.20 cm tall object is 50.0 cm to the left of a lens of focal length of magnitude 40.0 cm . A second lens, this one having a focal length of magnitude 60.0 cm, is located 300 cm to the right of the first lens along the same optic axis.
A)
Find the location and height of the image (call it I2) formed by the lens with a focal length of 40.0 cm if the first lens is diverging and the second lens is a converging.
B)
I2 is now the object for the second lens. Find the location and height of the image produced by the second lens.
A convergent lens of focal length 19 cm gives a real image that is 3.1 times larger than the object. Calculate the distance between the object and the lens, in centimeters.
A concave lens of focal length 9.7 cm has an
object located on its principal axis, 41.8 cm
from the vertical axis of the lens. What is the
Magnification of the image produced? Your
answer must have at least 4 decimal places and
be within 1% of the correct value.
Chapter 37 Solutions
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
Ch. 37.2 - A beam in air strikes a glass ball as shown in...Ch. 37.3 - Prob. 37.2CECh. 37.4 - Prob. 37.3CECh. 37.4 - Prob. 37.4CECh. 37.6 - Prob. 37.5CECh. 37.6 - Prob. 37.6CECh. 37.6 - Prob. 37.7CECh. 37 - A camera obscura is used to form an image of a...Ch. 37 - Because you should never stare directly into the...Ch. 37 - Prob. 3PQ
Ch. 37 - Prob. 4PQCh. 37 - Prob. 5PQCh. 37 - Prob. 6PQCh. 37 - Prob. 7PQCh. 37 - Prob. 8PQCh. 37 - Prob. 9PQCh. 37 - Prob. 10PQCh. 37 - Prob. 11PQCh. 37 - Prob. 12PQCh. 37 - Prob. 13PQCh. 37 - Prob. 14PQCh. 37 - Light rays strike a plane mirror at an angle of...Ch. 37 - Prob. 16PQCh. 37 - Prob. 17PQCh. 37 - Prob. 18PQCh. 37 - Prob. 19PQCh. 37 - Prob. 20PQCh. 37 - Prob. 21PQCh. 37 - Prob. 22PQCh. 37 - Prob. 23PQCh. 37 - Prob. 24PQCh. 37 - Prob. 25PQCh. 37 - Prob. 26PQCh. 37 - Prob. 27PQCh. 37 - Prob. 28PQCh. 37 - A convex mirror with a radius of curvature of 25.0...Ch. 37 - The magnitude of the radius of curvature of a...Ch. 37 - Prob. 31PQCh. 37 - The image formed by a convex spherical mirror with...Ch. 37 - An object is placed 25.0 cm from the surface of a...Ch. 37 - Prob. 34PQCh. 37 - Prob. 35PQCh. 37 - Prob. 36PQCh. 37 - Prob. 37PQCh. 37 - Prob. 38PQCh. 37 - Prob. 39PQCh. 37 - Prob. 40PQCh. 37 - Prob. 41PQCh. 37 - Prob. 42PQCh. 37 - Prob. 43PQCh. 37 - Prob. 44PQCh. 37 - Prob. 45PQCh. 37 - Prob. 46PQCh. 37 - Prob. 47PQCh. 37 - Prob. 48PQCh. 37 - Prob. 49PQCh. 37 - Prob. 50PQCh. 37 - Prob. 51PQCh. 37 - Prob. 52PQCh. 37 - Prob. 53PQCh. 37 - Prob. 54PQCh. 37 - Prob. 55PQCh. 37 - Prob. 56PQCh. 37 - You see the image of a sign through a camera...Ch. 37 - Prob. 58PQCh. 37 - Prob. 59PQCh. 37 - Prob. 60PQCh. 37 - An object is placed midway between two concave...Ch. 37 - Prob. 62PQCh. 37 - Prob. 63PQCh. 37 - Prob. 64PQCh. 37 - Prob. 65PQCh. 37 - Prob. 66PQCh. 37 - Observe your reflection in the back of a spoon....Ch. 37 - Prob. 68PQCh. 37 - A small convex mirror and a large concave mirror...Ch. 37 - Prob. 70PQCh. 37 - Prob. 71PQCh. 37 - Prob. 72PQCh. 37 - Prob. 73PQ
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Convex and Concave Lenses; Author: Manocha Academy;https://www.youtube.com/watch?v=CJ6aB5ULqa0;License: Standard YouTube License, CC-BY