## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson

# Chapter 16 - Traveling Waves - Exercises and Problems: 23

#### Answer

(a) $f = 8.6\times 10^8~Hz$ (b) $\lambda = 23~cm$

#### Work Step by Step

(a) We can find the frequency of the waves as they travel through the air. $f = \frac{v}{\lambda}$ $f = \frac{3.0\times 10^8~m/s}{0.35~m}$ $f = 8.6\times 10^8~Hz$ We know that the frequency of a wave does not change when moving from one medium to another medium. Therefore, the frequency as the signal travels through the glass is $8.6\times 10^8~Hz$. (b) We can find the speed of the signal as it travels through glass. The index of refraction of glass is $n = 1.5$. Therefore; $v = \frac{c}{n}$ $v = \frac{3.0\times 10^8~m/s}{1.5}$ $v = 2.0\times 10^8~m/s$ We can find the wavelength as: $\lambda = \frac{v}{f}$ $\lambda = \frac{2.0\times 10^8~m/s}{8.6\times 10^8~Hz}$ $\lambda = 0.23~m = 23~cm$

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