Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 25 - Optical Instruments - Problems - Page 740: 29

Answer

The lens was moved 3.9 cm toward the fine print.

Work Step by Step

The focal length f is +12 cm. Use the lens equation to find the object distance. First, the image distance is infinity. $$\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{d_o}+\frac{1}{\infty}$$ $$d_o=f=12cm$$ Second, the image distance is -25 cm. The virtual image is at his near point, and the eye is very close to the lens. Find the object distance now. $$\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}$$ $$d_o=\frac{fd_i}{d_i-f}=\frac{(12cm) (-25cm) }{-25cm-12cm}=8.1cm$$ The lens was moved closer to the fine print, by a distance of $12cm-8.1cm=3.9cm$.
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