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\usepackage{SEART}

%TCIDATA{TCIstyle=article/art4.lat,SEART,SEART}

%TCIDATA{Created=Mon Sep 01 15:10:12 1997}
%TCIDATA{LastRevised=Fri Nov 07 11:00:23 1997}

\input{tcilatex}
\begin{document}

\SetTitle{Parabolic mirrors and Fermat's principle}
\SetAuthor{T. Curtright}
\Setdate{4 November 1997}
\TitlePage{}

\section{\protect\bigskip Parabolas}

Algebraically, a parabola symmetrical about the $y$ axis (hence unchanged by 
$x\rightarrow -x$) is described by the formula 
\begin{equation}
y=ax^{2}+b\;.  \label{parabola1}
\end{equation}
We plot one particular case: $a=1/4,\;b=0,\;$so $y=x^{2}/4$ .

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Word";type "MAPLEPLOT";width 6.0001in;height 3.9998in;depth 0pt;display
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"black";linestyle 1;linethickness 1;pointstyle "point";xmin "-5";xmax
"5";xviewmin "-4";xviewmax "4";yviewmin "-2";yviewmax
"4";viewset"XY";rangeset"X";phi 45;theta 45;plottype 4;numpoints
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"T";tempfilename 'C:/SWP25/docs/EJA0FQ05.wmf';tempfile-properties "XP";}}

\eject{}Alternatively, any such parabola may be defined by requiring that
each point on the parabola is equidistant from a point on the $y$ axis (at
the point $(0,S)$) and from a straight line below and parallel to the $x$
axis (defined as all points $(x,y=-D)$ for all $x$) where the distance to
the latter line is measured vertically, along lines parallel to the $y$
axis. With this more geometrical definition of the parabola, let's consider
an arbitrary point $(x,y)$ and reduce these distance requirements to
algebraic conditions which may then be compared to the previous, algebraic
definition of the parabola.

The distance between the points $(0,S)$ and $(x,y)$ is

\[
s(x,y)=\sqrt{x^{2}+(y-S)^{2}}\;, 
\]
while the ``vertical'' distance from $(x,y)$ to the line $y=-D$ is 
\[
d(x,y)=y+D\;. 
\]
Equating these two distances we have our ``equidistance condition'' 
\begin{equation}
s(x,y)=\sqrt{x^{2}+(y-S)^{2}}=y+D=d(x,y)\;.  \label{equidistant}
\end{equation}
When this is squared 
\[
s^{2}(x,y)=x^{2}+(y-S)^{2}=\left( y+D\right) ^{2}=d^{2}(x,y)\;, 
\]
and then expanded out 
\[
x^{2}+y^{2}-2yS+S^{2}=y^{2}+2yD+D^{2}\;, 
\]
we see that the $y^{2}$ terms cancel leaving a relation linear in $y$. Thus
the condition becomes $x^{2}-2yS+S^{2}=2Dy+D^{2}\;$. That is $2\left(
S+D\right) y=x^{2}+S^{2}-D^{2}=x^{2}+(S-D)(S+D)$ or 
\begin{equation}
y=\frac{1}{2\left( S+D\right) }\;x^{2}+\frac{1}{2}\left( S-D\right) \;.
\label{parabola2}
\end{equation}
Hence this geometrical description of a parabola is exactly the same as the
previous algebraic one provided that we identify 
\begin{equation}
a=\frac{1}{2\left( S+D\right) }\;,\;\mathtt{and\;}b=\frac{1}{2}\left(
S-D\right) \;.  \label{agreement}
\end{equation}
For the particular case that we plotted above, $a=1/4,\;b=0,\;y=x^{2}/4,$ we
have $S=D=1$ . We add the line $y=-1$ to our previous plot for comparison
purposes.

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Word";type "MAPLEPLOT";width 6.0001in;height 3.9998in;depth 0pt;display
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"black";linestyle 1;linethickness 1;pointstyle "point";function
\TEXUX{$-1$};linecolor "blue";linestyle 1;linethickness 1;pointstyle
"point";function \TEXUX{$(0,1)$};linecolor "red";linestyle 1;linethickness
1;pointstyle "cross";xmin "-5";xmax "5";xviewmin "-4";xviewmax "4";yviewmin
"-2";yviewmax "4";viewset"XY";rangeset"X";phi 45;theta 45;plottype
4;numpoints 49;axesstyle "normal";xis \TEXUX{x};var1name
\TEXUX{$x$};valid_file "T";tempfilename
'C:/SWP25/docs/EJA0SF07.wmf';tempfile-properties "XP";}}The student should
use a measuring device (ruler, drawing compass, piece of string, etc.) to check
that points on the parabola are indeed equidistant from the fixed point $%
(0,1)$ and the line $y=-1$ .

\section{\protect\bigskip Mirrors}

Now, what has this got to do with light and optics? The answer is provided
by Fermat's Principle of Least Time, which leads to the familiar laws of
reflection and refraction, as we showed in class. 

Consider vertical light rays. Any two such rays will obviously take the same
amount of time to go from the top of the page, say, to the line $y=-D$. But
since the distance from the parabola to the point $(0,S)$ is the same as the
vertical distance from the parabola to the line $y=-D$, then any two such
rays will take the same amount of time to go vertically from the top of the
page to the parabola and then from the parabola to the point $(0,S)$. Hence
a parabolic mirror will reflect all such vertical light rays, by Fermat's
Principle, and ``focus'' them at the point $(0,S)$. This is the basis for
the parabolic mirror.

We add to our previous plot a few sample light rays to illustrate all this.%
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49;axesstyle "normal";xis \TEXUX{x};var1name \TEXUX{$x$};valid_file
"T";tempfilename 'C:/SWP25/docs/EJA1R109.wmf';tempfile-properties "XP";}} 

The student should pencil-in a few sample "rays" in this diagram, going
from the top of the page vertically to the parabola and thence to the
point $(0,a)$.  Then check, say by using a protractor, that the angle
of incidence is equal to the angle of reflection where the ray strikes
the parabola. 

It is important to note that we have satisifed Fermat's least time condition
exactly using a parabolic mirror, without making approximations such as
those used in the text to describe spherical mirrors, but we have done this
only for light rays which were initially vertical. That is, only parallel
light rays such as those emanating from a \emph{very} \emph{distant }object
will be focussed exactly by the parabolic mirror.\footnote{%
If we wanted to arrange things to focus light rays from a point-like object
at a fixed but finite distance, we could use an elliptical mirror. But that
is another story ...} Also, since we are using a mirror, and not a
refracting lens, we have avoided the ``chromatic aberrations'' inherent in a
lens due to the variation of the index of refraction with wavelength.


\eject{}Another (the HARD!!) way to check that the law of reflection
holds
is to do so analytically. The straight line from the point $(0,S)$ to
the
parabola at $y=x^{2}/(4S)$ has a slope given by 
\[
t_{1}(x)=\frac{\frac{1}{4S}\,x^{2}-S}{x}\;=\frac{x^{2}-4S^{2}}{4xS}\equiv
\tan \theta _{1}\;. 
\]
On the other hand, the slope of the parabola itself at that same point
is $%
dy/dx=x/(2S)$, which implies that the slope of the normal to the
parabola at
that same point is 
\[
t_{\perp }=\frac{-1}{dy/dx}=\frac{-2S}{x}\equiv -\tan \theta _{\perp
}\;. 
\]
Now considering the angles introduced above, and the fact that the
incoming
ray is vertical, we find the angles of incidence and reflection to be 
\[
\theta _{i}=\frac{\pi }{2}-\theta _{\perp }\;,\;\;\;\theta _{r}=\theta
_{1}+\theta _{\perp }\;. 
\]
Take the $\tan $ of these equations to get 
\[
\tan \theta _{i}=\tan \left( \frac{\pi }{2}-\theta _{\perp }\right)
\;,\;\;\;\tan \theta _{r}=\tan \left( \theta _{1}+\theta _{\perp
}\right)
\;. 
\]
Now, unfortunately, we must use one of those despicable identities that
everyone hates, namely: $\tan \left( \alpha +\beta \right) =\frac{\tan
\left( \alpha \right) +\tan \left( \beta \right) }{1-\tan \left( \alpha
\right) \tan \left( \beta \right) }\;.$ This allows us to conclude that 
\[
\tan \theta _{i}=\frac{1}{\tan \theta _{\perp }}\;,\;\;\;\tan \theta
_{r}=%
\frac{\tan \left( \theta _{1}\right) +\tan \left( \theta _{\perp
}\right) }{%
1-\tan \left( \theta _{1}\right) \tan \left( \theta _{\perp }\right)
}\;. 
\]
Finally we substitute the previous results to find first 
\[
\tan \theta _{i}=\frac{x}{2S}\;, 
\]
and then, with some algebraic manipulations, 
\[
\tan \theta _{r}=\frac{\frac{x^{2}-4S^{2}}{4xS}+\frac{2S}{x}}{1-\frac{%
x^{2}-4S^{2}}{4xS}\frac{2S}{x}}\;=\frac{\frac{x^{2}-4S^{2}+8S^{2}}{4xS}}{%
\frac{4x^{2}S-2x^{2}S+8S^{3}}{4x^{2}S}}=x\,\frac{x^{2}+4S^{2}}{2x^{2}S+8S^{3}%
}=\frac{x}{2S}\;. 
\]
Therefore with a little algebra and trigonometry, we have shown $\tan
\theta
_{i}=\tan \theta _{r}$ which implies $\theta _{i}=\theta _{r}$ .\
Ta-da!!!
The vertically incoming rays reflected by the parabolic mirror do indeed
obey the law of reflection.

\end{document}