University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 34 - Geometric Optics - Problems - Exercises - Page 1152: 34.12

Answer

36.0 cm.

Work Step by Step

Use the mirror equation. $$\frac{1}{s}+\frac{1}{s’}=\frac{1}{f}$$ For a real image, $s’ \gt 0$, so $m=-\frac{s’}{s}$ is negative. The image height is the same as the object height, so s’=s. $$\frac{1}{s}+\frac{1}{s’}=\frac{1}{s}+\frac{1}{s}=\frac{2}{s}$$ $$\frac{1}{s}+\frac{1}{s’}=\frac{1}{f}=\frac{1}{18.0 cm}$$ Comparing the right hand sides, we see that s = 36.0 cm.
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