Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 26 - Geometrical Optics - Problems and Conceptual Exercises - Page 942: 58

Answer

$22.3ns$

Work Step by Step

We know that $d=vt$ $\implies d=\frac{c}{n}t$ This simplifies to: $t=\frac{nd}{c}$.......eq(1) For the first medium, the above equation can be written as $t_1=\frac{n_1 d_1}{c}$ We plug in the known values to obtain: $t_1=\frac{1.51(1.51)}{3.00\times 10^8}=7.60ns$ Now for the second medium, eq(1) can be written as $t_2=\frac{n_2d_2}{c}$ We plug in the known values to obtain: $t_2=\frac{1.33(3.31)}{3.00\times 10^8}$ $t_2=14.67ns$ Now the total time is $t_{total}=t_1+t_2=14.67+7.60=22.3ns$
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