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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 Fermat1 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.2

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. (No diagram? Try Reflection Law without SVG.)

 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.3 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

 $r_{\small A} = \sqrt{h_{\small A}^2 + x^2} \quad\text{ and }\quad r_{\small B} = \sqrt{h_{\small B}^2 + (w-x)^2}$ (2)

where the length designations are as shown in Fig. 1 above. Plugging into Eq. (1) we have

 $s = \sqrt{h_{\small A}^2 + x^2} + \sqrt{h_{\small B}^2 + (w-x)^2}.$ (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 right-hand 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)(w-x)}{\sqrt{h_{\small B}^2v+(w-x)^2}}\\ \text{or}\quad\quad\quad&\\ \frac{\mathrm{d} s}{\mathrm{d} x}&=\frac{x}{\sqrt{h_{\small A}^2 +x^2}} - \frac{w-x}{\sqrt{h_{\small B}^2 + (w-x)^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

 $\frac{x}{\sqrt{h_{\small A}^2 + x^2}} = \frac{w-x}{\sqrt{h_{\small B}^2 + (w-x)^2}}.$ (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 Pierre de Fermat, Encyclopædia Britannica Online. [Return.]

2 Fermat’s principle, Encyclopædia Britannica Online. [Return.]

3 The light ray is traveling in a single medium so its speed is constant (except at reflection). [Return.]

 Get a Mirage &see what reflection can do. 