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 - Search and Learn - Page 678: 8

Answer

$f=10\;\rm cm$

Work Step by Step

Since the same lens was used in both cases, so we have the same magnification factor $m$. $$m=\dfrac{-d_i}{d_o}$$ In the first case, case (a), [the magnification here is positive] $$m=\dfrac{-d_{i1}}{d_{o1}}=\dfrac{-d_{i1}}{5}$$ Thus, $$d_{i1}=-5m\tag 1$$ In the second case, case (b), [the magnification here is negative] $$-m=\dfrac{-d_{i2}}{d_{o2}}=\dfrac{-d_{i2}}{15}$$ Thus, $$d_{i2}= 15m\tag 2$$ Divide (2) by (1); $$\dfrac{d_{i2}}{d_{i1}}=\dfrac{ 15m}{-5m}=-3$$ Thus, $$d_{i2}=-3d_{i1}\tag 3$$ Now we need to use the formula of thin lenses; $$\dfrac{1}{f}=\dfrac{1}{d_o}+\dfrac{1}{d_i}$$ And since we are using the same lens, $$\dfrac{1}{f}=\dfrac{1}{d_{o1}}+\dfrac{1}{d_{i1}}=\dfrac{1}{d_{o2}}+\dfrac{1}{d_{i2}}$$ Plugging from (3) and plugging the known; $$ \dfrac{1}{5}+\dfrac{1}{d_{i1}}=\dfrac{1}{15}+\dfrac{1}{-3d_{i1}}$$ Solving for $d_{i1}$; $$ \dfrac{1}{d_{i1}}+\dfrac{1}{3d_{i1}}=\dfrac{1}{15}-\dfrac{1}{5}$$ $$ \dfrac{4}{3d_{i1}}= -\dfrac{2}{15}$$ So, $$d_{i1}=\color{red}{\bf -10}\rm\;cm$$ Therefore, from (3); $$d_{i2}=-3d_{i1}=-3\times -10=\color{red}{\bf 30}\rm\;cm$$ Now we can find the focal length of the lens which is given by $$ f =\left[\dfrac{1}{d_o}+\dfrac{1}{d_i}\right]^{-1}$$ Plugging the known for any case, a or b, $$ f =\left[\dfrac{1}{5}+\dfrac{1}{-10}\right]^{-1}=\color{red}{\bf 10}\rm\;cm$$
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