Chapter 32, Problem 79GP

When light passes through a prism, the angle that the refracted ray makes relative to the incident ray is called the deviation angle δ, Fig. 32-64. Show that this angle is a minimum when the ray passes through the prism symmetrically, perpendicular to the bisector of the apex angle ϕ, and show that the minimum deviation angle, δ_m, is related to the prism’s index of refraction n by

$n=\frac{\sin \frac{1}{2}\left(\varphi +{\delta }_{\text{m}}\right)}{\sin \varphi /2}.$

[Hint: For θ in radians, (d/dθ)(sin⁻¹θ) = $1/\sqrt{1-{\theta }^2}.$]

FIGURE 32-64

Problems 79 and 80.