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# 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.
Figure 1 explains why. We have

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

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 d1=∞, then d2=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 d1=r (the radius of curvature), then d2=r.
where $$f$$ is the focal length of the mirror (see Figure 2). But setting $$d_1=r$$ (the radius of curvature of the mirror at $$B$$) yields $$d_2 =r$$ (see Figure 3), since the rays emanating from the center of curvature collect back at the center (infinitesimally speaking, i.e., replacing the reflector by its osculating circle). In short, $$(1)$$ yields

$\frac{1}{r} + \frac{1}{r} = \frac{1}{f}$

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.
Dividing by $$s$$ and taking the limit for $$s \rightarrow 0$$ gives1

$\frac{1}{d_1} + \frac{1}{d_2} = 2k.$

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.

1 Retaining the names $$d_1$$, $$d_2$$ for the limiting values of the distances, in a mild abuse of notation.