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: 45

Answer

a. 0.106 m b. 0.109 m c. 0.117m d. 0.513m

Work Step by Step

Use the lens equation, 23–8. $$\frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{f}$$ a. Solve for the image distance. $$d_i=\frac{d_of}{d_o-f}=\frac{(10.0m)(0.105m)}{10.0m-0.105m}$$ $$=0.106m$$ b. Solve for the image distance. $$d_i=\frac{d_of}{d_o-f}=\frac{(3.0m)(0.105m)}{3.0m-0.105m}$$ $$=0.109m$$ c. Solve for the image distance. $$d_i=\frac{d_of}{d_o-f}=\frac{(1.0m)(0.105m)}{1.0m-0.105m}$$ $$=0.117m$$ d. Solve for the object distance. $$\frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{f}$$ $$d_o=\frac{d_if}{d_i-f}$$ The smallest object distance corresponds to the largest image distance. $$d_o=\frac{(0.132m)(0.105m)}{0.132m-0.105m}$$ $$=0.513m$$
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