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

(a) $v = 1.9\times 10^8~m/s$ (b) $f = 4.5\times 10^{14}~Hz$
(a) We can find the index of refraction of the solid as: $n = \frac{\lambda_a}{\lambda_s}$ $n = \frac{670~nm}{420~nm}$ $n = 1.6$ We can find the speed of light as it travels through the solid as: $v = \frac{c}{n}$ $v = \frac{3.0\times 10^8~m/s}{1.6}$ $v = 1.9\times 10^8~m/s$ (b) We can find the frequency of the wave as it travels through the solid as: $f = \frac{v}{\lambda}$ $f = \frac{1.9\times 10^8~m/s}{420\times 10^{-9}~m}$ $f = 4.5\times 10^{14}~Hz$