Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 25 - Optical Instruments - Problems - Page 741: 42

Answer

$\approx 7\bf\times$

Work Step by Step

First, we need to find the focal length of the eyepiece by using the mentioned equation of 25-3, as the author told us. $$M= \dfrac{f_o}{f_e}$$ Thus, $$f_e=\dfrac{f_o}{M}$$ Plugging the given; $$f_e=\dfrac{26}{6.5}=\bf 4\;\rm cm$$ Now we need to use the following formula to find the image distance, $$\dfrac{1}{f_o}=\dfrac{1}{d_{i1}}+\dfrac{1}{d_{o1}}$$ Solving for $d_{i1}$; $$d_{i1}=\left[\dfrac{1}{f_o}-\dfrac{1}{d_{o1}}\right]^{-1}$$ Plugging the known; $$d_{i1}=\left[\dfrac{1}{26}-\dfrac{1}{400}\right]^{-1}=\bf 27.8\;\rm cm$$ Now we can calculate the angular magnification which is given by $$M= \dfrac{\theta'}{\theta}$$ And from the figure below; $$M=\dfrac{\theta'}{\theta}=\dfrac{h/f_e}{h/d_{i1}}=\dfrac{d_{i1}}{f_e}$$ Plugging the known; $$M =\dfrac{d_{i1}}{f_e}=\dfrac{27.8}{4}=\color{red}{\bf 6.95\bf \times }$$
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