Physics Technology Update (4th Edition)

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

Chapter 27 - Optical Instruments - Problems and Conceptual Exercises - Page 970: 28

Answer

(a) 12.0 cm to the right of the second lens. (b) $-6.0$

Work Step by Step

(a) We can find the required location of the image as follows: $d_{i1}=\frac{(d_{\circ 1})(f_1)}{d_{\circ 1}-f_1}$ We plug in the known values to obtain: $d_{i1}=\frac{(12.0cm)(8.0cm)}{12.0cm-8.0cm}$ $\implies d_{i1}=24cm$ The object distance is $d_{\circ 2}=20.0cm-24.0cm=-4.0cm$ Now the image formed by the second lens is located at $d_{i2}=\frac{(d_{\circ 2})(f_2)}{d_{\circ 2}-f_2}$ We plug in the known values to obtain: $d_{i2}=\frac{(-4.0cm)(-6.0cm)}{-4.0cm+6.00cm}$ $\implies d_{i2}=12.0cm$ to the right of the second lens. (b) The required magnification can be determined as follows: $m=(-\frac{d_{i1}}{d_{\circ 1}})-(\frac{d_{i2}}{d_{\circ 2}})$ $\implies m=(\frac{d_{i1} d_{i2}}{d_{\circ}1 d_{\circ 2}})$ We plug in the known values to obtain: $m=\frac{(24.0cm)(12.0cm)}{(12.0cm)(-4.0cm)}=-6.0$
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