# Reflecting on Reflections

**I.** When looking at a shiny ball bearing in my outstretched hand, I see the reflection of my head. This tiny image seems to lie somewhere inside the sphere. Where exactly? More precisely, what is the limiting position of the point \(I\) (*Image*) in Figure 1, as \(MA \parallel NB\) approaches \(NB\)? The answer turns out to be the midpoint of the radius \(OB\).

**Figure 1.**The reflected ray

*AP*seems to emanate from the point

*I*. And

*I*approaches the midpoint of the radius

*OB*as the ray

*MA*approaches the ray

*NB*.

\[\angle 1\stackrel{A}{=} \angle2\stackrel{B}{=} \angle3\stackrel{C}{=} \angle 4,\]

where \(A\) holds because the incoming rays are parallel, \(B\) is the law of reflection, and \(C\) holds because the two angles are vertical. Summarizing, \(\angle1=\angle4 \stackrel{def}{=} \theta\), making the triangle \(OAI\) equilateral and implying that

\[OI= \frac{r}{2 \cos \theta } \rightarrow\frac{r}{2} \ \ \hbox{for} \ \ \theta \rightarrow 0,\]

as claimed. V.I. Arnold’s engaging book [1] contains a derivation of this fact, although it takes slightly more than a page of calculation.

**Figure 2.**Illustration of the mirror formula. If

*d*

_{1}=∞, then

*d*

_{2}=

*f*.

**II. **An even shorter derivation of the image location results from the mirror formula, which states that the source-to-mirror distance \(d_1\) and the image-to-mirror distance \(d_2\) satisfy

\[\frac{1}{d_1}+\frac{1}{d_2} =\frac{1}{f}, \tag1\]

**Figure 3.**If

*d*

_{1}=

*r*(the radius of curvature), then

*d*

_{2}=

*r*.

\[\frac{1}{r} + \frac{1}{r} = \frac{1}{f}\]

of \(f =r/2\), as claimed.

**III.** Figure 4 shows how the mirror formula comes out of the reflection law \(\theta _1= \theta _2.\) We have \(\theta _1= \angle3- \angle 1\), where \(\angle 3= ks+o(s)\) ((\k\) being the curvature at \(B\)), and \(\angle 1 = s/d_1 +o(s)\). With the similar expression for \(\theta_2\), the reflection law amounts to

\[ks - \frac{s}{d_1} = \frac{s}{d_2}-ks+o(s).\]

**Figure 4.**Proof of the mirror formula.

^{1}\[\frac{1}{d_1} + \frac{1}{d_2} = 2k.\]

Substituting \(d_1= \infty\) and \(d_2=f\), we conclude that \(2k= 1/f\). This proves the mirror formula \((1)\) and reproduces the fact that the focal distance is half the radius of curvature, \(f = 1/2k = r/2\). Incidentally, we proved this for any smooth curve, not necessarily a circle.

*The figures in this article were provided by the author.*

**Retaining the names \(d_1\), \(d_2\) for the limiting values of the distances, in a mild abuse of notation.**

^{1}**References**

[1] Arnold, V.I. (2014). *Mathematical Understanding of Nature: Essays on Amazing Physical Phenomena and Their Understanding by Mathematicians*. Providence, RI: American Mathematical Society.