The exhibit “Mirage” shows an interesting optical illusion: A crystal lies in front of an observer as if on a presentation plate. But if you reach for it, you simply — reach into the void. How is this possible? In the following text, we will go into a little more detail about the mathematical background behind this illusion.

The exhibit uses two parabolic reflector to create the strange effect. Such mirrors are also used, for example, in reflecting telescopes and in satellite communications.

Figure 1: The “Mirage” exhibit

And now … the mathematics of it:

What is a parabolic mirror and why is it so useful? To understand this, we first need to take a closer look at the parabola: The standard parabola, familiar from school lessons, is described by the equation y=y(x)=x^2. We now want to examine the properties of this curve in more detail. To do this, trap a beam of light \mathbf{s}_x (x\in\mathbb R) from above; this can be described by the equation

    \[\mathbf{s}_x(t)=\begin{pmatrix} x\ -t\end{pmatrix}=\begin{pmatrix} x\ 0\end{pmatrix}-t\mathbf{e}_2.\]

Here t\in\mathbb R is a real parameter. In the point A=(x,x^2) this ray \mathbf{s}_x is now reflected. At this point the tangent has the normalized direction vector \mathbf{t}=\frac{1}{\sqrt{1+4x^2}}(1,2x). Thus the normal vector is equal to \mathbf{n}=\frac{1}{\sqrt{1+4x^2}}(-2x,1). Thus, we obtain for the direction vector of the outgoing ray \mathbf{s}_x':

    \[v=-\mathbf{e}_2+2\langle\mathbf{e}_2,\mathbf{n}\rangle\mathbf{n}=\frac{1}{1+4x^2}\begin{pmatrix} -4x\ 1-4x^2\end{pmatrix}.\]

Here \langle\mathbf{u},\mathbf{v}\rangle is the dot product of the vectors \mathbf{u} and \mathbf{v}. Thus the precipitating ray \mathbf{s}_x' is given by the equation

    \[\mathbf{s}_x'(t)=\begin{pmatrix} x\ x^2\end{pmatrix}+\frac{t}{1+4x^2}\begin{pmatrix} -4x\ 1-4x^2\end{pmatrix}.\]

If we now set the x coordinate to zero in this equation, we get t_0=\frac{1+4x^2}{4}. This gives

    \[F=\mathbf{s}_x'(t_0)=\begin{pmatrix} 0\ x^2+\frac{1}{1+4x^2}\cdot\frac{1+4x^2}{4}(1-4x^2)\end{pmatrix}=\begin{pmatrix} 0\ 1/4\end{pmatrix}.\]

Thus, all incident rays \mathbf{s}_x (x\in\mathbb R) from above are reflected such that the corresponding outgoing ray \mathbf{s}_x' passes through the focus F. Moreover, we can still calculate the length of the path \overline{AF}:

    \[\lvert\overline{AF}\rvert=\left\lvert\begin{pmatrix} x\ x^2\end{pmatrix}-\begin{pmatrix} 0\ 1/4\end{pmatrix}\right\rvert=\sqrt{x^2+(x^2-1/4)^2}=\sqrt{x^4+1/2x^2+1/16}=x^2+1/4.\]

So this distance has exactly the same length as the distance from the point A to the horizontal y=-1/4 (the so-called directrix of the given parabola) is.

Functionality of the exhibit

The “Mirage” exhibit makes use of the above two properties of the parabola. To do this, one first rotates the standard parabola around the z axis and thus obtains the equation


of a so-called paraboloid of rotation P. This is also used for reflecting telescopes. So what is the idea behind the exhibit? We will describe this in the following. The principle behind it can also be used to transmit other waves (e.g. sound waves):

We take two parabolic mirrors P_1 and P_2. Let us say that the mirror P_1 is identical with the standard rotational paraboloid P. The mirror P_2 now faces the mirror P_1 at a certain distance d\gt 0. It is thus described by the equation z=z(x,y)=d-(x^2+y^2). If one would continue the two parabolic mirrors P_1 and P_2 arbitrarily far (i.e. for all x,y\in\mathbb R) they would intersect in the circle K given by the parametrization

    \[\mathbf{x}(\varphi)=\begin{pmatrix} R_0\cos(\varphi)\ R_0\sin(\varphi)\ d/2\end{pmatrix}\]

for an angle \varphi\in[0,2\pi) with R_0=\sqrt{d/2}. For real purposes, however, we need to continue the two mirrors only up to some radius R\leq R_0 (i.e., for x^2+y^2\leq R^2).

Let F_1 and F_2 be the respective foci of P_1 and P_2. Then, according to the above consideration,

    \[F_1=\begin{pmatrix} 0\ 0\ 1/4\end{pmatrix}\]


    \[F_2=\begin{pmatrix} 0\ 0\ d-1/4\end{pmatrix}.\]

We now assume a ray starting at the focal point F_1. Let us say this is given by the equation

    \[\mathbf{s}(t)=\begin{pmatrix} 0\ 0\ 1/4\end{pmatrix}+t\begin{pmatrix} \cos(\varphi)\ \sin(\varphi)\ a\end{pmatrix}.\]

Here the parameter a measures the rise of the beam \mathbf{s}. For this to hit the mirror P_1, we need a relatively small slope of a\leq\frac{R^2-1/4}{R}. If the ray \mathbf{s} is now reflected at P_1, the outgoing ray runs afterwards — according to the above considerations — parallel to the z-axis at a distance \leq R. This ray reflected at P_1 is then again reflected by P_2, so that the outgoing ray then passes (again as above) exactly through the focal point F_2. By the same reasoning, a ray \mathbf{s}' starting in F_2 — if its slope a' is large enough (namely a'\geq-\frac{R^2-1/4}{R}) — will also arrive in F_1 after two refexions.

For applications it is still important to know how long the distance covered by the above ray \mathbf{s} is on its way from F_1 to F_2. Let us assume that the ray is reflected at the points Q_1 and Q_2 (the straight line Q_1Q_2 is parallel to the z axis). Further let r be the distance from Q_1 (and thus Q_2) to the z-axis. Then, as explained above,


So if we calculate the total distance that the ray \mathbf{s} starting in F_1 travels from F_1 to F_2, we get:


So this distance is independent of r.

Now what are the applications of this construction? For example, sound waves (or other waves) can be transmitted well over a long distance d in this way: to do this, place the sound source at the point F_1. The sound waves emitted by this source are now reflected at the parabolic mirrors P_1 and P_2 in the way we just deduced. Thus they concentrate again in the focal point F_2 after a certain time, which the sound needs to cover the distance from F_1 over the reflection points Q_1 and Q_2 to F_2. But this time is proportional to the distance \lvert\overline{F_1Q_1}\rvert+\lvert\overline{Q_1Q_2}\rvert+\lvert\overline{F_2Q_2}, which has — as derived above — the constant length d+1/2 (independent of the outgoing beam). Thus also no distortion takes place: The transmitted audio signal arrives at point F_2 from all directions at the same time. This mode of operation is exploited, for example, in satellite dishes.

The “Mirage” exhibit also uses this trick — but differently from radio wave transmission. Whereas in the latter the distance d between the two mirrors is particularly large, in our exhibit it is particularly small: Let’s set d=1/4 (i.e. exactly to the focal length). This time we put our “source” not in the focal point F_1, but in F_2=(0,0,0). Similarly, we let the mirrors touch each other, i.e. R=R_0=\frac{1}{2\sqrt{2}}. Now, if a beam goes

    \[\mathbf{s}(t)=\begin{pmatrix} 0\ 0\end{pmatrix}+t\begin{pmatrix} \cos(\varphi)\ \sin(\varphi)\ a\end{pmatrix}\]

from F_2 and if the slope a\geq-\frac{R^2-1/4}{R}=\frac{1}{2\sqrt{2}} is large enough, \mathbf{s} is first reflected at P_2 and then at P_1 and finally “ends” in F_1=(0,0,1/4). Regardless of the rise and direction, the same distance of 3/4 is always covered. If one cuts out a small circular hole around F_1 in P_2, then an object (in the exhibit: a crystal), which is actually in the origin F_2=(0,0,0), suddenly appears to the observer as if it were in the point F_1=(0,0,1/4) above. This simple trick is used in the exhibit.

At this point it should be noted that if the crystal is mirrored twice, the orientation is also reversed twice, so that the crystal does not appear mirror-inverted either.

If you want to see a nice explanatory video about the above topics, have a look at the video [1] of the YouTuber Mathologer.

Figure 2: Actual image and reflection of the crystal


[1] https://www.youtube.com/watch?v=0UapiTAxMXE

[2] https://de.wikipedia.org/wiki/Parabel_(Mathematik)

[3] https://de.wikipedia.org/wiki/Paraboloid

[4] https://de.wikipedia.org/wiki/Parabolantenne