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: 43

Answer

a. Converging; 3.08 D b. Diverging; -0.148 m

Work Step by Step

a. The power of the lens is defined by equation 23–7. $$P=\frac{1}{f}=\frac{1}{0.325m}=3.08D$$ The lens power is positive; this is a converging lens. b. Use equation 23–7. $$P=\frac{1}{f}$$ $$f=\frac{1}{D}=\frac{1}{-6.75D}=-0.148m$$ The focal length is negative; this is a diverging lens.
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