Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 24 - Optical Instruments - Exercises and Problems - Page 714: 43

Answer

$3.55\;\rm m$

Work Step by Step

We know that the length of the telescope is given by the sum of the two focal lengths of the two lenses. $$L=f_{\rm obj}+f_{\rm eye}\tag 1$$ we are given the focal length of the eye lens, so we have to work on the objective one. We know that the magnification of the telescope is given by $$M=-\dfrac{f_{\rm obj}}{f_{\rm eye}}=\dfrac{\theta_{\rm eye}}{\theta_{\rm obj}}$$ Hence, $$f_{\rm obj}= f_{\rm eye}\dfrac{\theta_{\rm eye}}{\theta_{\rm obj}}\tag 2$$ we ignored the negative sign since we are dealing with lengths. We are given $\theta_{\rm eye}$ and we know that $\theta_{\rm obj}$, for small angle approximation, is given by $$\theta_{\rm obj}=\dfrac{D_{\rm Mars}}{R}$$ where $D$ is the diameter of Mars, and $R$ is the distance between it and our Earth. Plugging into (2), $$f_{\rm obj}=\dfrac{\theta_{\rm eye} f_{\rm eye}R}{D_{\rm Mars}} $$ Plugging into (1), $$L=\dfrac{\theta_{\rm eye} f_{\rm eye}R}{D_{\rm Mars}}+f_{\rm eye} $$ Plugging the known and remember to convert the given angle to rad and other units to SI-system units. $$L=\dfrac{\left[\dfrac{0.5^\circ\cdot \pi}{180^\circ}\right](25\times 10^{-3})( 1.1 \times 10^8\times 10^3)}{(6800\times 10^3)}+(25\times 10^{-3})$$ $$L=\color{red}{\bf 3.55}\;\rm m$$
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