Light incident in air along AB at an angle a on a prism of index of refraction n and vertex angle q refracts at B and C, exiting at angle r along CD. The question is what path minimizes the angular deviation d of the light ray. I have seen the result, namely that the path ABCD is symmetric with <a = <r, discussed using a symetry argument (for example, see Physics, volume 2, by Roller and Blum, Holden-Day, 1982, p. 1529) but never actually derived or shown.

In the figure below all non-boldfaced angles are expressed in terms of q, a (the first angle of incidence, and b (the first angle of refraction) by using the facts that the angles of a triangle add to 180°, those of a quadrilateral add to 360°, or that vertical or opposite angles are equal.

To express the deviation angle d, the fact that an exterior angle in a triangle, r in DCHF, equals the sum of the remote interior angles, <CHF and <HFC, yields

r = d + (q - a), or

d = r + a - q. (1)

sina = nsinb (2)

nsin(q - b) = sinr (3)

d = sin-1[nsin(q - b)] + sin-1[nsinb] - q (4)

To show that d is minimized when the path is symmetric, or that b = q - b, b = q/2, requires dealing with three variables, n, q, and b, on the right side of the equation. Mathematically this is entering the realm of partial derivatives and/or Lagrangians, which is "beyond the scope of this course."

Instead, let’s examine Eq. (1) for various values of q and n and use a graphing calculator to find the value(s) of b (x in the graphs below) that minimize d. In the first of the three screen shots below q = 60° and n = 1.3 for the lower curve, n = 1.5 for the higher. In the second q = 45° and n = 1.3 for the lower curve, n = 1.5 for the higher, while in the third q = 30° and n = 1.3 for the lower curve, n = 1.5 for the higher. In all three the horizontal window is from 0 to 90° and the vertical from 0 to 45°. In each case the path that minimizes d is symmetric (b = q/2).

q = 60°, b = q - b = 30° q = 45°, b = q - b = 22.5° q = 30°, b = q - b = 15°