Chapter 43: Problem 40

Question: Prove that the optical path lengths for reflection and refraction in Figs 10 and 17 (Chap 43) are each a minimum when compared with other nearby paths connecting the same two points. (HINT: Examine the quantity ${d^2 L\over dx^2}$.

Solution: For refraction we have that the time required for light to travel from point 1 to point 2 a particular path is

$$\Delta t~=~{L_1\over v_1}~+~{L_1\over v_1}~=~{\sqrt{x_1^2+y^2}\over v_1}~+~{\sqrt{x_2^2+(d-y)^2} \over v_2}$$

The first derivative wrt y is

$${d\Delta t\over dy}~=~{y\over v_1\sqrt{x_1^2+y^2} }~+~{y-d\over v_2\sqrt{x_2^2+(d-y)^2}}$$

This leads to Snells law of refraction (assuming a minimum for $\Delta t$). The second derivative of $\Delta t$ is

$${d^2\Delta t\over dy^2}~=~{1\over v_1\sqrt{x_1^2+y^2} }~-~{y^2\ov... ...+~{1\over v_2\sqrt{x_2^2+(d-y)^2}}~-~{(y-d)^2\over v_2(\sqrt{x_2^2+(d-y)^2})^3}$$

$${d^2\Delta t\over dy^2}~=~{x_1^2\over v_1\sqrt{x_1^2+y^2} }~+~{x_2^2\over v_2\sqrt{x_2^2+(d-y)^2}}~\ge~0$$

Since ${d^2\Delta t\over dy^2}\ge 0$ for all values of y we see that Snells law corresponds to a global minimum for the time required for the light to propagate from 1 to 2. The reflection relation can be found trivially by setting v₁=v₂.