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: 21

Answer

a. Concave b. Upright, virtual, magnified c. 1.40 m

Work Step by Step

The object distance is 20.0 cm. The magnification is +1.4. Find the image distance, then the focal length. $$m=-\frac{d_i}{d_o}$$ $$d_i=-md_o$$ Use the mirror equation, 23–2. $$\frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{f}$$ Solve for the focal length. $$f=\frac{d_od_i}{d_o+d_i}=\frac{d_o(-md_o)}{d_o-md_o}$$ $$=\frac{(20.0cm)(-(1.4)20.0cm)}{20.0cm-1.4(20.0cm)}=70cm$$ The focal length is positive, so the mirror is concave, aka a converging mirror. b. Find the image distance. $$d_i=-md_o=-1.4(20.0cm)=-28.0cm$$ The image will be upright (because m is positive), virtual (because the image distance is negative), and magnified (because the magnitude of m is greater than 1). c. The radius is twice the focal length, or r = 1.40m.
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