This problem can be solved using a combination of snell's law and the geometry of the triangle.
We label the angles of incidence and refraction at the surface of entry with
θ1 and θ2, respectively, and the internal angle and external angle to the exit surface with θ3 and θ4, respectively, as shown in the diagram below. The apex angle Φ = 60° is the angle between the surfaces of entry and exit. The ray in the glass forms a triangle with these surfaces in which the interior angles must add to up 180°.
(90° − θ2) + Φ + (90° − θ3) = 180°,
θ3 = Φ − θ2.
Using snell's law we can solve for θ2 first, and then θ3 band θ4.
To solve for θ2, we use the following formula given by snell's law:
sin θ1 = n sin θ2 (*since index of refraction of air is 1, we are left with only sin θ1 on the left side)
solving for θ2, we must solve it twice, once for the value of n (violet light) and then n (red light).
θ2 = inverse sin ( sin θ1 / n)
n (violet) = 1.66
n (red) = 1.62
θ2 (violet) = inverse sin ( sin θ1 / 1.66)
θ2 (red) = inverse sin ( sin θ1 / 1.62)
Now we can solve for θ3
θ3 = Φ − θ2
Φ = 60
θ3 (violet) = 60 - θ2 (violet)
θ3 (red) = 60 - θ2 (red)
To solve for θ4 ,
Use snell's law once again:
n sin θ3 = sin θ4
θ4 = inverse sin( n sin θ3 )
taking into account the two different values of n, we get:
θ4 (violet) = inverse sin( 1.66 * sin θ3 (violet) )
θ4 (red) = inverse sin( 1.62 * sin θ3 (red) )
And so the angular spread is:
θ4 = θ4 (violet) - θ4 (red)