Light Takes the Fastest Path

Light crossing from air into water bends. It isn't taking the shortest path; it's taking the fastest one, since it travels slower in water. Minimizing the travel time with a derivative reproduces Snell's law of refraction.

How it works

Optimization

In one sentence

To optimize $f$ on an interval, find the critical points where $f'(x) = 0$ and compare $f$ there and at the endpoints; here $T'(x) = 0$ gives $\frac{\sin\theta_1}{v_1} = \frac{\sin\theta_2}{v_2}$ (Fermat's principle).

General formula

{\htmlData{sym=0}{T(x)}} = \frac{\sqrt{a^2 + x^2}}{{\htmlData{sym=1}{v_1}}} + \frac{\sqrt{b^2 + (d - x)^2}}{{\htmlData{sym=2}{v_2}}}, \qquad T'(x) = 0 \;\Rightarrow\; \frac{\sin{\htmlData{sym=3}{\theta_1}}}{{\htmlData{sym=1}{v_1}}} = \frac{\sin{\htmlData{sym=4}{\theta_2}}}{{\htmlData{sym=2}{v_2}}}

Moving the crossing point a little trades air path for water path. At the best point, the time saved in one equals the time lost in the other.

What the symbols mean

$T(x)$
Travel time
How long light takes if it crosses the surface at x.
$v_1$
Speed in air
Faster medium.
$v_2$
Speed in water
Slower medium.
$\theta_1$
Angle of incidence
Angle from vertical in air.
$\theta_2$
Angle of refraction
Angle from vertical in water.

Light goes from a point at height $a$ above the surface to one at depth $b$ below, crossing the surface at $x$ .
The total time is distance divided by speed in each medium, a function $T(x)$ .
Set $T'(x) = 0$ : $\frac{x}{v_1\sqrt{a^2 + x^2}} = \frac{d - x}{v_2\sqrt{b^2 + (d - x)^2}}$ , which is $\frac{\sin\theta_1}{v_1} = \frac{\sin\theta_2}{v_2}$ .
That's Snell's law: optimization explains why light bends.

Learn the math

OptimizationCalc I · Lesson 33, Applications of Derivatives More Optimization ProblemsCalc I · Lesson 34, Applications of Derivatives