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: 85

Answer

(a) farther away (b) $34cm$

Work Step by Step

(a) As $d_i=\frac{1}{\frac{1}{f}-\frac{1}{d_{\circ}}}$ We know that $m=-\frac{d_i}{d_{\circ}}$ $\implies m=-\frac{1}{d_{\circ}(\frac{1}{f}-\frac{1}{d_{\circ}})}$ $\implies m=\frac{1}{1+\frac{d_{\circ}}{f}}$ Thus, if we want to decrease the magnification, the value of $1+\frac{d_{\circ}}{f}$ should be increased -- that is, the distance of the object should increase. (b) We know that $d_{\circ \space initial }=f(\frac{1}{m_{initial}}-1)$ $\implies d_{\circ\space initial}=(-34cm)(\frac{1}{\frac{1}{3}}-1)=-68cm$ Similarly $d_{\circ \space fianl }=f(\frac{1}{m_{final}}-1)$ $\implies d_{\circ\space final}=(-34cm)(\frac{1}{\frac{1}{4}}-1)=-102cm$ Now $|d_{\circ\space final}-d_{\circ\space initial}|=|-102-(-68)|=34cm$
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