{"id":4277,"date":"2022-09-09T11:09:53","date_gmt":"2022-09-09T09:09:53","guid":{"rendered":"https:\/\/erlebnisland-mathematik.de\/?page_id=4277"},"modified":"2023-01-26T14:42:59","modified_gmt":"2023-01-26T13:42:59","slug":"advanced-text-mirage","status":"publish","type":"page","link":"https:\/\/erlebnisland-mathematik.de\/en\/advanced-text-mirage\/","title":{"rendered":"Advanced text Mirage"},"content":{"rendered":"

[vc_row drowwidth=”sidebar-biest-default sidebar-biest”][vc_column][vc_column_text]<\/p>\n

Mirage<\/h1>\n

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

The exhibit uses two parabolic<\/em><\/a> mirrors to create the strange effect. Such mirrors are also used, for example, in reflecting telescopes<\/a><\/em> and in satellite communication.[\/vc_column_text][vc_single_image image=”1088″ img_size=”large” alignment=”center”]Figure 1: The “Mirage” exhibit[\/vc_single_image][vc_column_text]<\/p>\n

And now … the mathematics:<\/h3>\n

[\/vc_column_text][vc_column_text]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<\/a><\/em>: The standard parabola<\/em>, familiar from school lessons, is described by the equation \"y(x) = x^2\". Let us now examine the properties of this curve in more detail. Trap for this purpose a light beam \"\mathbf{s}_x\" (\"x\in\mathbb R\") from above; this can be described by the equation <\/p>\n

  <\/span>   <\/span>\"\[\mathbf{s}_x(t)=\begin{pmatrix} x\\ -t\end{pmatrix}=\begin{pmatrix} x\\ 0\end{pmatrix}-t\mathbf{e}_2.\]\"<\/p>\n

Here \"t\in\mathbb R\" is a real parameter. At the point \"A=(x,x^2)\" this ray \"\mathbf{s}_x\" is now reflected. At this point the tangent has the normalised direction vector<\/em><\/a> \"\mathbf{t}=\frac{1}{\sqrt{1+4x^2}}(1,2x)\". Thus the normal vector<\/em><\/a> is equal to \"\mathbf{n}=\frac{1}{\sqrt{1+4x^2}}(-2x,1)\". We thus obtain for the direction vector of the outgoing beam \"\mathbf{s}_x'\" <\/p>\n

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

Here \"\langle\mathbf{u},\mathbf{v}\rangle\" is the scalar product<\/a><\/em> of the vectors \"\mathbf{u}\" and \"\mathbf{v}\". This makes the failing beam \"\mathbf{s}_x'\" given by the equation <\/p>\n

  <\/span>   <\/span>\"\[\mathbf{s}_x'(t)=\begin{pmatrix} x\\ x^2\end{pmatrix}+\frac{t}{1+4x^2}\begin{pmatrix} -4x\\ 1-4x^2\end{pmatrix}.\]\"<\/p>\n

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

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

All rays \"\mathbf{s}_x\" (\"x\in\mathbb R\") incident from above are thus reflected in such a way that the corresponding outgoing ray \"\mathbf{s}_x'\" passes through the focal point<\/em><\/a> \"F\". In addition, we can still calculate the length of the route\u00a0\"\overline{AF}\": <\/p>\n

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

This distance therefore has exactly the same length as the distance from the point \"A\" to the horizontal \"y=-1/4\" (the so-called directrix<\/em> of the given parabola).[\/vc_column_text][vc_column_text]<\/p>\n

Functionality of the exhibit<\/h3>\n

[\/vc_column_text][vc_column_text]The “Mirage<\/a>” 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 <\/p>\n

  <\/span>   <\/span>\"\[z=z(x,y)=x^2+y^2\]\"<\/p>\n

of a so-called paraboloid of rotation<\/a><\/em> \"P\". This is also used for reflecting telescopes. So what is the idea behind the exhibit? We will describe this below. The principle behind this can also be used to transmit other waves (e.g. sound waves):<\/p>\n

We take two parabolic mirrors \"P_1\" and \"P_2\". Let us say that the mirror \"P_1\" is identical to the standard rotational paraboloid \"P\". The mirror \"P_2\" is now opposite 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 were to continue the two parabolic mirrors \"P_1\" and \"P_2\" arbitrarily far (i.e. for all \"x,y\in\mathbb R\") they intersect in the circle \"K\" given by the parametrisation <\/p>\n

  <\/span>   <\/span>\"\[\mathbf{x}(\varphi)=\begin{pmatrix} R_0\cos(\varphi)\\ R_0\sin(\varphi)\\ d/2\end{pmatrix}\]\"<\/p>\n

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

Let \"F_1\" and \"F_2\" be the respective foci of \"P_1\" and \"P_2\". Then according to the above consideration <\/p>\n

  <\/span>   <\/span>\"\[F_1=\begin{pmatrix} 0\\ 0\\ 1/4\end{pmatrix}\]\"<\/p>\n

and<\/p>\n

  <\/span>   <\/span>\"\[F_2=\begin{pmatrix} 0\\ 0\\ d-1/4\end{pmatrix}\]\"<\/p>\n

holds.\u00a0We now assume a beam that starts at the focal point \"F_1\". Let us say that this is given by the equation<\/p>\n

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

Here the parameter \"a\" measures the rise of the beam \"\mathbf{s}\". In order for this to hit the mirror \"P_1\", we need a relatively small<\/em> slope of \"a\leq\frac{R^2-1/4}{R}\". If the ray \"\mathbf{s}\" is now reflected at \"P_1\", the outgoing ray then runs — according to the above considerations — parallel to the \"z\"-axis at a distance \"\leq R\". This ray reflected at \"P_1\" is then reflected again 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}'\" that starts in \"F_2\" — if its slope \"a'\" is large<\/em> enough (namely \"a'\geq-\frac{R^2-1/4}{R}\") — will also arrive in \"F_1\" after two reflections.<\/p>\n

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 this is reflected at points \"Q_1\" and \"Q_2\" (the straight line \"Q_1Q_2\" is therefore parallel<\/em><\/a> to the \"z\" axis). Let \"r\" be the distance from \"Q_1\" (and thus \"Q_2\") to the \"z\"-axis. Then, as explained above: <\/p>\n

  <\/span>   <\/span>\"\[\lvert\overline{F_1Q_1}\rvert=r^2+1/4=\lvert\overline{F_2Q_2}\rvert.\]\"<\/p>\n

So if we calculate the total distance travelled by the ray \"\mathbf{s}\" starting in \"F_1\" from \"F_1\" to \"F_2\", we get: <\/p>\n

  <\/span>   <\/span>\"\[\lvert\overline{F_1Q_1}\rvert+\lvert\overline{Q_1Q_2}\rvert+\lvert\overline{F_2Q_2}\rvert=(r^2+1/4)+(d-2r^2)+(r^2+1/4)=d+1/2.\]\"<\/p>\n

This distance is therefore independent<\/em> of \"r\".<\/p>\n

So what are the applications of this construction? For example, this is a good way to transmit sound waves (or other waves) over a long distance \"d\": 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 have just derived. Thus, after a certain time, which the sound needs to cover the distance from \"F_1\" via the reflection points \"Q_1\" and \"Q_2\" to \"F_2\", they concentrate again in the focal point \"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}\rvert\", which — as derived above — has the constant length \"d+1/2\" (independent of the outgoing beam). This means that there is also no distortion: 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<\/em>.<\/p>\n

The “Mirage<\/a>” exhibit also uses this trick — but differently from radio wave transmission. While in this case the distance \"d\" between the two mirrors is particularly large, in our exhibit it is particularly small: let us set \"d=1/4\" (i.e. exactly to the focal length<\/em><\/a>). This time we do not place our “source” in the focal point \"F_1\", but in \"F_2=(0,0,0)\". In a similar way we let the mirrors touch \u00a0each other , i.e. \"R=R_0=\frac{1}{2\sqrt{2}}\". If a beam<\/p>\n

  <\/span>   <\/span>\"\[\mathbf{s}(t)=\begin{pmatrix} 0\\ 0\end{pmatrix}+t\begin{pmatrix} \cos(\varphi)\\ \sin(\varphi)\\ a\end{pmatrix}\]\"<\/p>\n

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. Thus, if one cuts out a small circular hole around \"F_1\" in \"P_2\", an object (in the exhibit: a crystal) that is actually located at the origin \"F_2=(0,0,0)\" suddenly appears to the observer as if it were located at the point \"F_1=(0,0,1/4)\" above it. This simple trick is used in the exhibit.<\/p>\n

At this point it should be noted that if the crystal is mirrored twice, the orientation<\/em><\/a> is also reversed twice, so that the crystal does not appear mirror-inverted either.<\/p>\n

If you want to see another nice explanatory video on the above topics, have a look at the video [1]a, <\/a> by YouTuber Mathologer<\/a><\/em>.a, [\/vc_column_text][vc_single_image image=”1092″ img_size=”large” alignment=”center”]Figure 2: Actual image and reflection of the crystal[\/vc_single_image][vc_column_text]<\/p>\n

Literature<\/h3>\n

[\/vc_column_text][vc_column_text][1] https:\/\/www.youtube.com\/watch?v=0UapiTAxMXE<\/a><\/p>\n

[2] https:\/\/de.wikipedia.org\/wiki\/Parabel_(Mathematik)<\/a><\/p>\n

[3] https:\/\/de.wikipedia.org\/wiki\/Paraboloid<\/a><\/p>\n

[4] https:\/\/de.wikipedia.org\/wiki\/Parabolantenne<\/a>[\/vc_column_text][\/vc_column][\/vc_row]<\/p>\n<\/div>","protected":false},"excerpt":{"rendered":"

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