Schaum's Outline of College Physics, Twelfth Edition (Schaum's Outlines)
12th Edition
ISBN: 9781259587399
Author: Eugene Hecht
Publisher: McGraw-Hill Education
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Textbook Question
Chapter 38, Problem 32SP
Determine the two locations of an object such that its image will be enlarged 8.0 times by a thin lens of focal length +4.0 cm.
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Schaum's Outline of College Physics, Twelfth Edition (Schaum's Outlines)
Ch. 38 - 38.15 [I] Draw diagrams to indicate qualitatively...Ch. 38 - 38.16 [I] A thin lens has a focal length of +20.0...Ch. 38 - 38.17 [I] An object is very far from the front of...Ch. 38 - 38.18 [I] You are designing a copy machine using a...Ch. 38 - 38.19 [I] Show that for a thin positive lens
Ch. 38 - 38.20 [I] Where must an object be located with...Ch. 38 - 38.21 [I] A bug on the central axis is 300 cm from...Ch. 38 - 38.22 [I] Considering the bug in the previous...Ch. 38 - 38.23 [I] Where should an object be placed in...Ch. 38 - 38.24 [I] Where should an object be placed in...
Ch. 38 - 38.25 [I] A 1.0-cm-tall object is placed in front...Ch. 38 - 38.26 [I] A 1.0-cm-tall object is placed in front...Ch. 38 - 38.27 [I] Where should a 1.0-cm-tall object be...Ch. 38 - 38.28 [I] What is the separation between the...Ch. 38 - 38.29 [I] An object on the central axis is 200 cm...Ch. 38 - 38.30 [I] We have a thin negative lens with a...Ch. 38 - 38.31 [I] Determine the nature, position, and...Ch. 38 - 38.32 [II] Determine the two locations of an...Ch. 38 - 38.33 [II] What are the nature and focal length of...Ch. 38 - 38.34 [II] Describe fully the image of an object...Ch. 38 - 38.35 [II] Compute the focal length of a lens that...Ch. 38 - 38.36 [II] A luminous object and a screen are 12.5...Ch. 38 - 38.37 [II] A plano-concave lens has a spherical...Ch. 38 - 38.38 [II] A convex-concave lens has faces of...Ch. 38 - 38.39 [II] A double convex glass lens has faces...Ch. 38 - 38.40 [II] Two thin lenses, of focal lengths +12...Ch. 38 - 38.41 [II] What must be the focal length of a...
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Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.Similar questions
- A converging lens made of crown glass has a focal length of 15.0 cm when used in air. If the lens is immersed in water, what is its focal length? (a) negative (b) less than 15.0 cm (c) equal to 15.0 cm (d) greater than 15.0 cm (e) none of those answersarrow_forwardA lamp of height S cm is placed 40 cm in front of a converging lens of focal length 20 cm. There is a plane mirror 15 cm behind the lens. Where would you find the image when you look in the mirror?arrow_forwardYou view an object by holding a 2.5 cm-focal length magnifying glass 10 cm away from it. How far from your eye should you hold the magnifying glass to obtain a magnification of 10 ?arrow_forward
- How far should you hold a 2.1 cm-focal length magnifying glass from an object to obtain a magnification of 10 x ? Assume you place your eye 5.0 cm from the magnifying glass.arrow_forwardWhat is the magnification of a magnifying lens with a focal length of 10 cm if it is held 3.0 cm from the eye and the object is 12 cm from the eye?arrow_forwardWhat is the focal length of a magnifying glass that produces a magnification of 3.00 when held 5.00 cm from an object, such as a rare coin?arrow_forward
- A converging lens has a focal length of 10.0 cm. Locate the object if a real image is located at a distance from the lens of (a) 20.0 cm and (b) 50.0 cm. What If? Redo the calculations if the images are virtual and located at a distance from the lens of (c) 20.0 cm and (d) 50.0 cm.arrow_forwardA converging lens has a focal length of 20.0 cm. Locate the image for object distances of (a) 40.0 cm, (b) 20.0 cm, and (c) 10.0 cm. For each case, state whether the image is real or virtual and upright or inverted. Find the magnification in each case.arrow_forwardUnder what circumstances will an image be located at the focal point of a spherical lens or mirror?arrow_forward
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Convex and Concave Lenses; Author: Manocha Academy;https://www.youtube.com/watch?v=CJ6aB5ULqa0;License: Standard YouTube License, CC-BY