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 1154: 34.50

Answer

0.710 cm.

Work Step by Step

The image and object distances are related to the indices of refraction and to the radius of curvature. $$\frac{n_a}{s}+\frac{n_b}{sā€™}=\frac{n_b-n_a}{R}$$ $$\frac{1.00}{40.0cm}+\frac{1.40}{2.60cm}=\frac{1.40-1.00}{R}$$ $$R=0.710cm$$
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