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: 34.49

Answer

a. 11. b. 2.1 ms.

Work Step by Step

a. The f-number of a lens is the ratio of its focal length to its diameter. It is an integer. $$\frac{180.0mm}{16.36mm}=f/11$$ b. The amount of light passing through the lens during the exposure time should remain the same. Moving from f/11 to f /2.8 is four steps of 2 in intensity, so one needs 1/16 of the exposure time. In other words, the area increases by a factor of 16. To let in the same amount of light, the exposure time decreases by a factor of 16. $$\frac{1}{16}\frac{1}{30}s=\frac{1}{480}s=2.1ms$$
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