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 - Problems - Page 675: 47

Answer

a. The image is 37 cm behind the lens. b. +2.3

Work Step by Step

Use the lens equation, 23–8. The focal length and the object distance are positive. $$\frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{f}$$ a. Solve for the image distance. $$d_i=\frac{d_of}{d_o-f}=\frac{(0.16m)(0.28m)}{0.16m-0.28m}$$ $$=-0.37m$$ The negative sign means that the image is 37 cm behind the lens, i.e., it is a virtual image. b. Find the magnification from Eq. 23–9. $$m=-\frac{d_i}{d_o}=-\frac{-0.3733m}{0.16m}=+2.3$$ The magnification is positive. The image is upright.
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