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: 5

Answer

110mm, 220 mm.

Work Step by Step

A single converging lens produces images that are inverted. We’re told that the object height equals the image height, so the magnification m is –1. Apply equation 23–9 to relate the image and the object distance. $$m=-1=-\frac{d_i}{d_o}$$ $$d_i=d_o$$ Now use the lens equation to find the object distance. $$\frac{1}{d_o}+\frac{1}{d_i}=\frac{2}{d_o}=\frac{1}{f}$$ $$d_o=2f=2(55mm)=110mm$$ We’re asked to find the distance, L, between the object and the image on the film. The object is in front of the lens and the image is behind the lens, and both object distance and image distance are positive. Find the sum of the object and image distances. $$L=d_o+d_i=d_o+d_o=110mm+110mm=220mm$$
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