Angle of Incidence = Angle of Relection
You probably know that when light reflects off of a surface
the angle of incidence is equal to the angle of reflection. It
makes sense since the light seems to bounce like a ball off a
hard surface. But, why isn't the incidence angle larger or
smaller than the reflection angle? What is presented below is
one way to think about it.
In the 1650s the French mathematician Pierre de
Fermat^{2}
made a wonderfully simple statement that came to be known as
Fermat's Principle. In terms of optics this is
[T]he path taken by a ray of
light in traveling between two points requires either a minimum
or a maximum
time.^{3}
Using this principle, we shall show that when light reflects
off a surface the incident angle does indeed equal the
reflection angle.
Consider light reflecting off some surface as depicted in
the Fig. 1 diagram.

Fig.
1. Light ray begins at $\small A$,
reflects off the surface at $\small P$, and
goes to $\small B$. The dotted vertical lines
are perpendicular to the reflecting surface. 
In order to minimize the time from $\small A$ to $\small B$
via $\small P$ we need only compute the minimum distance
traversed.^{4}
That distance
\[
\large s = r_{\small A} + r_{\small B} \] .

(1) 
Please note that the start and ends points
$\small A$ and
$\small B$, and the position of the reflecting material are
fixed throughout.
From the Pythagorean theorem we known that
\[r_{\small
A} = \sqrt{h_{\small A}^2 + x^2} \quad\text{ and }\quad
r_{\small B} = \sqrt{h_{\small B}^2 + (wx)^2}\]

(2)

where the length designations are as shown in Fig. 1 above.
Plugging into Eq. (1) we have
\[s
= \sqrt{h_{\small A}^2 + x^2} + \sqrt{h_{\small B}^2 +
(wx)^2}.\] 
(3)

A rough plot of this distance is shown in Fig. 2.

Fig.
2. Path length $s$ versus $x$. The precise
minimum is found using a derivative.

We are interested in the minimum travel time for the ray,
i.e., the lowest point on the Fig. 2 graph. The most convenient
way to find the minimum distance $s$ is to use elementary
calculus. The sole variable in our problem is the horizontal
position of the reflection point $\small P$, which means that
only $x$ is changing on the righthand side of Eq. (3).
Differentiating this expression with respect to $x$
\[
\begin{aligned} \frac{\mathrm{d} s}{\mathrm{d}
x}&=\frac{(\frac{1}{2})(2x)}{\sqrt{h_{\small
A}^2+x^2}} +
\frac{(\frac{1}{2})(2)(wx)}{\sqrt{h_{\small
B}^2v+(wx)^2}}\\ \text{or}\quad\quad\quad&\\
\frac{\mathrm{d} s}{\mathrm{d}
x}&=\frac{x}{\sqrt{h_{\small A}^2 +x^2}} 
\frac{wx}{\sqrt{h_{\small B}^2 + (wx)^2}}.
\end{aligned} \]

(4)

At an extremum the derivative is zero, i.e.
$\frac{\mathrm{d} s}{\mathrm{d} x } = 0$. Setting Eq. (4b) to
zero gives us
\[
\frac{x}{\sqrt{h_{\small A}^2 + x^2}} =
\frac{wx}{\sqrt{h_{\small B}^2 + (wx)^2}}.
\]

(5)

Comparing Eq. (5) with Fig. (1) diagram, we recognize the
two ratios as sine functions and write
\[
\sin \theta_{\small A} = \sin \theta_{\small B}
\].

(6) 
Therefore, we conclude that
\[\quad\theta_{\small
A} = \theta_{\small B} \].

(7) 
Or, in more familiar terms, the angle of incidence
equals the angle of reflection.
^{1} Scalable Vector Graphics are employed
in the main version of this page at
Reflection Law.
[Return.]
^{2} Pierre de Fermat, Encyclopædia Britannica
Online. [Return.]
^{3} Fermat’s principle, Encyclopædia
Britannica Online. [Return.]
^{4} The light ray is traveling in a
single medium so its speed is constant (except at reflection).
[Return.]
