Answer
(a) $v = 1.9\times 10^8~m/s$
(b) $f = 4.5\times 10^{14}~Hz$
Work Step by Step
(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:
$f = \frac{v}{\lambda}$
$f = \frac{1.9\times 10^8~m/s}{420\times 10^{-9}~m}$
$f = 4.5\times 10^{14}~Hz$
