The Prism & Minimum Deviation

A prism turns a beam of light through an angle, bending it toward its base. How much depends on the angle at which the light goes in, and there’s one special angle that gives the least bending of all.

Following a ray through a prism#

A prism’s cross-section is a triangle; two of its faces are the refracting surfaces, meeting at the angle of the prism A. A ray strikes the first face at angle of incidence i and refracts at angle r. Inside, it meets the second face at angle r′ and leaves at angle i′, the angle of emergence, provided r′ is not beyond the critical angle.

Without the prism, the ray would have gone straight on. The angle between that original direction and the final ray is the angle of deviation δ. Each face turns the ray, by (i − r) and (i′ − r′), so

The normals at the two faces meet at some point V. In the four-sided figure formed by the apex, the two points of incidence and V, two angles are right angles, so the angle at V is 180° − A. In the triangle formed by the ray inside and the two normals, r + r′ plus that same angle is 180°. So:

Minimum deviation#

As i changes, so does δ, and for one particular i it is smallest. At that point the ray passes through the prism symmetrically: i = i′ (and r = r′).

Why symmetric?

Light paths are reversible. Suppose the minimum happened with i bigger than i′. Send light backwards along the emerging ray: it retraces the path exactly, with the same deviation, which must also be the minimum. But now the angle of incidence is the old i′ and the emergence the old i, so incidence is smaller than emergence, which contradicts the assumption. The same argument rules out i < i′. If the minimum is unique, it must have i = i′.

Refractive index from minimum deviation#

At minimum deviation δm: r = A/2 and i = (A + δm)/2. Snell’s law μ = sin i / sin r then gives

Thin prisms

If A is small, δm is small too and the sines can be replaced by the angles: