1993 AP test. problem #4.
a) determine the speed of the blue light in the glass
n = c / v
v = c / n
  = (3.00 * 10^8 m/s) / 1.6
  = 1.875 * 10^8 m/s

b) determine the wavelength of the red light in the glass
lambda_medium = lambda_vacuum / n
(this follows from v = c / n = f * lambda_medium => lambda_medium = c / (f*n) = (c/f) / n = lambda_vac / n)
 = 700 nm / 1.5
 = 466.67 nm

c) determine the frequency of the red light in the glass
frequency is the same no matter the medium, so f = c / lambda = (3.00*10^8 m/s) / (700 * 10^-9 m) = 4.286 * 101^4 Hz

d) beam is straight when going into left edge. Going out from left edge, both beams are bent up; by Snell's Law, when going from glass (higher n) to vacuum (lower n), the beam is bent away from the normal. the beam that travels faster in the medium (red) will not be bent away as much as the beam that's slower (blue). (sin th1 / sin th2 = v1 / v2, where v1 is in glass, v2 is vacuum; both have equal v2 in vacuum, and sin th1 in medium are equal; so beam with higher v1, in medium, means smaller th2. Or just think about what would happen if one were traveling at vacuum speed inside the medium - there would be no bending when it comes out)

e) beam is straight when going into left edge. They are both bent toward normal when coming out of right edge. The beam that is faster in medium (red) will be bent more toward normal than the slower beam (blue).