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 674: 10

Answer

a. 12 cm b. 18 cm c. Inverted

Work Step by Step

a. The focal length is half the radius of curvature. $$f=\frac{1}{2}r=\frac{1}{2}(24cm)=12cm$$ b. Use the mirror equation, 23–2. $$\frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{f}$$ Solve for the image distance. $$d_i=\frac{d_of}{d_o-f}=\frac{(38cm)(12cm)}{38cm-12cm}$$ $$=17.54cm\approx 18cm$$ The image is located 18 cm in front of the mirror. c. Calculate the magnification. $$m=-\frac{d_i}{d_o}=-\frac{17.54cm}{38cm}=-0.46$$ The magnification is negative. The image is inverted.
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