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 946: 131

Answer

(a) $5cm$ (b) $-2.5cm$ (c) increase

Work Step by Step

(a) We know that $m=-\frac{d_i}{d_{\circ}}$ $0.5=-\frac{d_i}{d_{\circ}}$ $\implies d_i=-\frac{d_{\circ}}{2}$ Now $\frac{1}{f}=\frac{1}{d_{\circ}}+\frac{1}{d_i}$ We plug in the known values to obtain: $-\frac{1}{5}=\frac{1}{d_{\circ}}+(-\frac{2}{d_{\circ}})$ This simplifies to: $d_{\circ}=5cm$ (b) We can find the image distance as $d_i=-\frac{d_{\circ}}{2}$ $d_i=-\frac{5cm}{2}$ $d_i=-2.5cm$ (c) We know that if we move the object closer to the lens, then the magnification of the image increases.
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