Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 34 - Images - Problems: 39a

Answer

$2.00$

Work Step by Step

First we consider equation (34-8): $\frac{n_1}{p}+\frac{n_2}{i}=\frac{n_2-n_1}{r}$ We note that the object distance is infinite ($p=\infty$), and obtain: $\frac{n_1}{\infty}+\frac{n_2}{i}=\frac{n_2-n_1}{r}$ $\frac{n_2}{i}=\frac{n_2-n_1}{r}$ Solving for the index of refraction of the sphere, $n_2$, we obtain: $n_2=n_1(1-\frac{r}{i})^{-1}$ We know the first index of refraction is that of air, so $n_1=1.00$ and the image distance is twice the radius of the sphere ($i=2r$) Using this information with the equation, we obtain: $n_2=1.00(1-\frac{r}{2r})^{-1}$ $n_2=1.00(1-\frac{1}{2})^{-1}$ $n_2=2.00$
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