Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 26 - Geometrical Optics - Problems and Conceptual Exercises - Page 943: 84

Answer

a) $d=56.4cm$ b) $m_{total}=-0.12$

Work Step by Step

(a) We know that $d_{i1}=(\frac{1}{f_1}-\frac{1}{d_{\circ1}})^{-1}$ $d_{i1}=(\frac{1}{-7.0cm}-\frac{1}{24cm})^{-1}=-5.42cm$ and $d_{\circ2}=35.0cm+d_{i1}=40.42cm$ Similarly $d_{i2}=(\frac{1}{14.0cm}-\frac{1}{40.42cm})^{-1}=21.4cm$ Now $d=35cm+d_{i2}=35cm+21.4cm=56.4cm$ (b) We know that $m_{total}=(-\frac{d_{i1}}{d_{\circ1}})(-\frac{d_i2}{d_{\circ2}})$ We plug in the known values to obtain: $m_{total}=(-\frac{-5.42cm}{24cm})(\frac{-21.4cm}{40.42cm})=-0.12$
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