The total deviation ε of a light ray refracted through a prism can be calculated, given the incidence angle α. The same calculation allows for determination of the emergence angle δ.

Deviation Diagram for a Prism

We have:
ε=ζ+η
A=β+γ
=>A+ε=(ζ+β)+(γ+η)
=>A+ε=α+δ
=>ε=α+δ-A (1)
sin(δ)/sin(β)=n=>δ=arcsin{n*sin(β)}
β=A-γ
=>δ=arcsin{n*sin(A-γ)} (2)
β=arcsin{sin(α)/n}, because sin(α)/sin(β)=n again,
so (2)=>δ=arcsin{n*sin[A-arcsin(sin(α)/n)]} (3)
Now sin[A-arcsin(sin(α)/n)]=sin(A)*cos(arcsin(sin(α)/n))-cos(A)*sin(α)/n
So δ=arcsin{n*sin(A)*cos(arcsin(sin(α)/n))-cos(A)*sin(α)}
=arcsin{n*sin(A)*Ö[1-(sin(α)/n)²]-cos(A)*sin(α)}
=>δ=arcsin{sin(A)*Ö[n²-sin²(α)]-cos(A)*sin(α)} (4)

And finally:
(1)(4)=>

(5)

Here's a graph of the above equation, with A=60°, n_D=1.72803 (SF10 Crystal) and α ranging from 45° to 80°. Angles are measured in degrees on the graph. Note that the deviation has a minimum, close to 60°. This fact can also be experimentally demonstrated using the angle readers of a spectroscope or numerically simulated (as in reference [2]).

Deviation Angle ε as a Function of Incidence Angle α

Minimum Deviation Diagram for a Prism

For the position of minimum deviation, the ray travels perpendicular to the bisect of angle A, and incidence and emergence angles are equal. Then:
α=δ and β=γ (6)
Also: A=β+γ, ε_min=(α-β)+(δ-γ) (7)
(7)(6)=>A=2*β, ε_min=2*α-2*β or
β=A/2 and ε_min/2=α-A/2 => α=ε_min/2+A/2 (8)
Now we know that n=sin(α)/sin(β) so from the last equation and (8) we get:

(9)

For n_D=1.72803 and A=60°, expression (9) gives ε_min=59.54084145°, which is exactly the minimum displayed on the graph above.

The actual proof that the angle ε_min given by (9) is in fact the minimum of (5) is very complicated (it can be found in [3], page 81). It can be proved using calculus, that if α is a root of the equation: dε/dα=0, with ε given by (5), then ε(A,n,α)=2*arcin(n*sin(A/2))-A=ε_min. A proof using Maple 9 commands, is provided courtesy of Robert Israel. After you execute the commands, try to understand why this is a perfectly valid proof.

> epsilon:=(A,n,alpha)->alpha+arcsin(sin(A)*sqrt(n^2-sin(alpha)^2)-cos(A)* sin(alpha))-A;
> S:= solve(diff(epsilon(A,n,alpha),alpha),alpha);
> S5:= simplify(S[5]) assuming n > 1, A > 0, A < Pi/2;
> Q:=sqrt((1-cos(epsilon(A,n,S5)+A))/2) - n*sqrt((1-cos(A))/2);
> Q:= simplify(expand(Q)) assuming n>1,A > 0,A < Pi/2;
> simplify(combine(Q)) assuming A > 0,A < Pi/2,n > 1;

Because the light dispersion angle is very small, (on the order of 7°) practically all the rays suffer minimum deviation. When a prism spectroscope is designed, its prism is placed in that position for usually the yellow part of the spectrum. (Sodium D lines). The PHASMATRON's prisms are placed in minimum deviation position for the green line of λ=5615.97363281Angstroms, its resonating wavelength.

References:

1. K.D. Alexopoulos, General Physics: Optics, 1st Edition, Athens 1966 (in Greek).
2. M.A. Peterson's Simulation Displaying the Minimum Deviation for a Prism
3. C.J. Smith, A Degree Physics Part III: Optics, Indian Edition 1985, Radha Publishing House.