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 - General Problems - Page 677: 82

Answer

a) Diverging, virtual, $-10.5\;\rm cm$ b) Converging, virtual, $238.4\;\rm cm$

Work Step by Step

a) Since the image is between the lens and the object, it must be a $\bf \color{red}{\bf diverging\; lens}$. And since the image is in front of the lens, it is a $\color{red}{\bf virtual\; image}$. Its focal length is given by $$\dfrac{1}{d_o}+\dfrac{1}{d_i}=\dfrac{1}{f}$$ $$f=\left[\dfrac{1}{d_o}+\dfrac{1}{d_i}\right]^{-1}$$ Plugging the known; $$f=\left[\dfrac{1}{37.5}+\dfrac{1}{-8.20}\right]^{-1}=\color{red}{\bf -10.5}\;\rm cm$$ --- b) Now the object is between the image and the lens and both of them, the image and the object, are on the same side, so it must be a $\color{red}{\bf converging\; lens}$. And since the image is in front of the lens, it is a $\color{red}{\bf virtual\; image}$. Its focal length is given by $$\dfrac{1}{d_o}+\dfrac{1}{d_i}=\dfrac{1}{f}$$ $$f=\left[\dfrac{1}{d_o}+\dfrac{1}{d_i}\right]^{-1}$$ Plugging the known; $$f=\left[\dfrac{1}{37.5}+\dfrac{1}{-44.5}\right]^{-1}=\color{red}{\bf238.4}\;\rm cm$$
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