Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 23 - Light: Geometric Optics - Problems - Page 676: 61

Answer

$d_{id}=10.0cm$ $m=-1$

Work Step by Step

$f_c=20.0cm$ $f_d=-10.0cm$ $d=25cm$ $d_{oc}=60.0cm$ $d_{ic}=\frac{1}{\frac{1}{f}-\frac{1}{d_{oc}}}=\frac{1}{\frac{1}{20.0cm}-\frac{1}{60.0cm}}=30.0cm$ $d_{od}=-5cm$ $d_{id}=\frac{1}{\frac{1}{f}-\frac{1}{d_{od}}}=\frac{1}{\frac{1}{-10.0cm}-\frac{1}{-5cm}}=10.0cm$ $m=m_1\times m_2=\frac{d_{ic}}{d_{oc}}\times\frac{d_{id}}{d_{od}}=\frac{30.0cm}{60.0cm}\times\frac{10.0cm}{-5.0cm}=-1$
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