Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 23 - Light: Geometric Optics - Search and Learn - Page 678: 2

Answer

They chose different signs for the magnification.

Work Step by Step

Whether the object was to be upright or inverted was not specified, and one student chose m = 3 while the other chose m = -3, that is, one assumed the image was upright, and the other student assumed the image was inverted. Relate the object and image distances to the magnification. $$m=-\frac{d_i}{d_o}$$ If m is positive $$d_i=-3d_o$$ Use the mirror equation, 23–2. The focal length of the lens is f = 12 cm. $$\frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{f}$$ Solve for the object distance. $$\frac{1}{d_o}+\frac{1}{-3d_o}=\frac{1}{12cm}$$ $$d_o=8.0cm$$ If m is negative $$d_i=3d_o$$ Use the mirror equation, 23–2. The focal length of the lens is f = 12 cm. $$\frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{f}$$ Solve for the object distance. $$\frac{1}{d_o}+\frac{1}{3d_o}=\frac{1}{12cm}$$ $$d_o=16cm$$ The images produced are virtual and real, respectively.
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