Chapter 30

Problem 27
  r2
i
r1
 v1 
 r3
 v3
†
red
v2
violet
Given
nred = 1.582
nviolet = 1.633
θi = 45◦
Solution
Find the angular dispersion of the outgoing beam.
From Snell’s law
n1 sin θ1 = n2 sin θ2
Solving for θ2 gives
n1
−1
sin θ1
θ2 = sin
n2
The angle of refraction for the red light once it enters the prism is
1
◦
−1
sin 45 = 26.55◦
θr1 = sin
1.582
The angle of refraction for the violet light once it enters the prism is
1
◦
−1
sin 45 = 25.66◦
θv1 = sin
1.633
Next the angle of incidence θ2 on the exiting interface must be determined. The line indicating the light
travelling through the prism forms a triangle from the top portion of the prism. This triangle consists of
the complement of θ1 , the complement of θ2 , and a 60◦ angle. The sum of all three angles must be 180◦ .
This gives the equation
θ10 + θ20 + 60◦ = 180◦
The complement of an angle is 90◦ minus the original angle. Therefore,
(90◦ − θ1 ) + (90◦ − θ2 ) + 60◦ = 180◦
Solving for θ2 gives
θ2 = 60◦ − θ1
For the red light
θr2 = 60◦ − θr1 = 60◦ − 26.55◦ = 33.45◦
†
Problem from Essential University Physics, Wolfson
For the violet light
θv2 = 60◦ − θv1 = 60◦ − 25.66◦ = 34.34◦
The refractive angle on the exiting interface for red light is then
1.582
−1
◦
θr3 = sin
sin 33.45 = 60.69◦
1
The refractive angle on the exiting interface for violet light is
1.633
−1
◦
θv3 = sin
sin 34.34 = 67.10◦
1
The difference between these two angles is the angular dispersion of the outgoing beam.
γ = θv3 − θr3 = 67.10◦ − 60.69◦ = 6.41◦