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\begin{document}

\newpage
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\title{\ifnum\TeacherEdition=1 TEACHER VERSION
%\footnote{DELETE THIS FOOTNOTE BEFORE SUBMISSION.  In the \TeX \, source code there is a toggle/option to produce a Teacher Version or a Student Version by ``carving'' out with  $\backslash$teacher\{\} that material which should only appear in the Teacher version. See the source code.}
\else STUDENT VERSION \fi\\Mirror, Mirror}

\teacher{
    \keywords{Optics, first order, ordinary differential equations, numerical methods}

%\tagwords{Here you can enter a wider set of words which serve as tags (as do keywords) in the SIMIODE web site. Keep this list to ten words. The Editor may suggest/change/add/delete some for more refinement as necessary.}

\tagwords{Astronomy, telescopes, mirrors, Foucault Knife Edge Test, interpolation}
}

\markboth{Mirror, Mirror}{Mirror, Mirror}

\author{Kurt Bryan\\
Rose-Hulman Institute of Technology\\
Terre Haute IN 47840 USA\\
kurt.bryan@rose-hulman.edu 
}

\makeStitlePDFLaTex


\begin{abstract}

This project models the ``Foucault Knife Edge Test,'' an optical test commonly used by amateur astronomers who make their own mirrors for reflecting telescopes. The goal of the test is to estimate the shape of the surface of a mirror from optical reflection data.  The model results in a very simple first order nonlinear ordinary differential equation (ODE) that has no closed-form solution. Moreover, one of the components of the ODE is a function that must be estimated from the collected data; some real data from a mirror the author made is included. A computer algebra system can be helpful for performing computations, and sample Maple and Mathematica worksheets are provided.


\end{abstract}


\section*{SCENARIO DESCRIPTION -- TELESCOPES AND MIRROR MAKING}

It is a time-honored tradition among amateur astronomers to build a telescope from scratch, right down to grinding the optical
elements.
Building a telescope is an interesting exercise in ``multiple precision'' construction. The telescope's tube and base are often made of wood, which is
typically cut to an accuracy of a few hundredths of an inch. Machined metal parts might be accurate within a few ten-thousandths of an inch. But
in order to obtain good images,
the optical elements themselves must be made to tolerances on the order of a \textit{millionth of an inch}. How can such precision be
attained at home (or anywhere else)?
Remarkably, this can be done by using only bare human hands, aided by a bit of clever optics and geometry.

In this project we'll look at one facet of constructing an accurate parabolic mirror for use in a reflecting telescope. A schematic
for a typical Newtonian reflecting telescope is shown in Figure \ref{fig1}. Incoming parallel light rays enter the telescope and reflect off
of the \textit{primary} mirror at the back of the scope (shown in red). They then reflect off of the \textit{secondary} mirror or \textit{diagonal} mirror
(also in red) and exit the side of the telescope, where they are focused by an eyepiece to form an image for visual inspection or photography. \begin{figure}[ht]
\centerline{\includegraphics[width=1\textwidth]{reflect_teles.png}}
\caption{A typical reflecting telescope.}
\label{fig1}
\end{figure}
(Despite being in
the optical path, the secondary mirror doesn't really appear in or otherwise ruin the image, as long as the secondary is not too big.)
In order to focus these incoming parallel rays to a point and form a good image, the primary mirror must be parabolic in shape, to extremely high precision. The secondary mirror doesn't alter this fact, but merely serves to bring the
light rays to a focus outside of the telescope.

One of the standard hallmarks of a good mirror is that the surface is parabolic to an accuracy of at least 1/8 that of the wavelength
of light in the visual spectrum, though more stringent figures of 1/10, 1/20, or finer are often cited. Green light has a wavelength of
about $5.5\times 10^{-7}$ meters (this is the ``middle'' of the visual spectrum), so $1/8$ wave corresponds to about $6.9\times 10^{-8}$ meters,
comparable to the size of a virus! We obviously need a very accurate method to test the shape of the mirror. It should
also be noted that a statistic like ``$1/10$ wave accuracy'' is only one figure of merit for the quality of a mirror. Moreover, the mirror
is only one component in the optical system that makes up a telescope. Still, the knife edge test for the shape of the primary
mirror is a traditional staple for amateur mirror makers.

How does one make a parabolic mirror?
Very briefly, the traditional process of grinding a mirror consists of starting with two glass disks. One is placed flat-side down on an immobile waist-high stand or
barrel; this disk is called the ``tool.'' The other disk is placed on top of the tool, flat face to flat face, with water and abrasive in between;
this disk will become the mirror. The mirror-maker then grinds the top disk back-and-forth while walking around the barrel for $20$ to $100$ hours (or more!),
periodically refreshing the abrasive and moving to finer and finer abrasives. Consistent grinding combined with the natural randomness in the
process results in the top face of the tool becoming a convex portion of a sphere and the bottom of the top disk becoming the corresponding concave spherical piece, to very high precision, perhaps a millionth of an inch. Unfortunately, a spherical mirror (unless quite small) does not adequately focus light rays. We have to reshape the spherical mirror into a parabola!
This is called ``figuring'' the mirror. For an introduction to building telescopes see \cite{berry} (a good introduction)
or \cite{texereau} (a classic, slightly more technical reference).

Taking the tiny amounts of glass off of a spherical mirror to turn it into a parabola isn't the difficulty, since glass is very hard. With a fine abrasive and gentle pressure the amount of glass removed with each stroke may be only a few atoms thick. The problem is that we need to know where to remove glass, and this requires that we can measure and monitor the shape of the emerging mirror to very high precision as we form it into a paraboloid. This is where the Foucault Knife Edge Test comes into play.

\section*{THE KNIFE EDGE TEST}

To understand the test, we're going to restrict our attention to a two-dimensional setting. The process of mirror-making, if done with any care, results in a mirror that is rotationally symmetric about its central point, even if the mirror is not precisely spherical or parabolic in shape. This means that to determine the shape of the mirror we really only need to consider a cross-section of the mirror through the center, or even just a radial portion.
%(So Figure \ref{fig1} might be considered as a cross-section of a three-dimensional telescope along its central axis.)

Refer to Figure \ref{fig4a}, in which the mirror to be tested is the curve on the left. The Foucault Knife Edge Test consists of placing a light source (the green dot) on the central axis of the mirror. If the light
source is at just the right distance from the mirror, light rays that strike the mirror at certain points will reflect orthogonally and return to the light source; this is illustrated by the solid blue lines. Note that due to the symmetry assumption, rays that hit the
mirror at the same distance above or below the central axis both return to the source. Contrast this to rays that strike the mirror
elsewhere, shown as dashed lines. These rays may not reflect back to the light source (but will if the light source is moved
closer to or further from the mirror).
\begin{figure}[ht]
\centering

\begin{minipage}{0.48\textwidth}

\vspace*{-.5 true cm}
\centering
\includegraphics[width=1.0\textwidth]{fouc5.jpg}\vspace*{-0.77in}\ \\
\caption{Mirror with incident and reflected rays.}
\label{fig4a}
\end{minipage}
\begin{minipage}{0.5\textwidth}
\centering
\includegraphics[width=0.7\textwidth]{foucault.jpeg}
\caption{Geometry of orthogonal reflection.}
\label{fig4b}
\end{minipage}

\end{figure}

To quantify this, see Figure \ref{fig4b} in which we suppose the mirror is described by a curve $x=S(y)$ for some function $S$, where we'll assume $S(0)=0$ for convenience.
We'll also assume the vertical extent of the mirror is given by $-R<y<R$ for some $R$.
The symmetry assumption means that $S(y)=S(-y)$. The goal is to find the function $S$. The knife edge test consists
of limiting our attention to a specific point $(x,y)=(S(y_0),y_0)$ on the mirror, then adjusting the light source
so that rays reflecting off the mirror at this point (or its symmetric partner $(S(y_0),-y_0)$) return to the light source.
We then measure the horizontal distance ``$d$'' from the light source to the center of the mirror (at $x=0$).
This is repeated for each value of $y_0$ in the range $0<y_0\leq R$.
For the reader who is interested in how one can determine the value of $d$ for any fixed $y_0$ using a simple homemade
apparatus (which explains where a ``knife edge'' comes into play) see the Appendix.


In Figure \ref{fig4b} the light source is at the point $(d,0)$. Consider the displacement vector from $(d,0)$ to the point $(S(y_0),y_0)$ on the mirror, given by
$\vec{v} = <S(y_0)-d,y_0>$. This vector is orthogonal to the surface of the mirror. Equivalently, $\vec{v}$ is orthogonal to the vector
$\vec{T} = <S'(y_0),1>$, which is tangent to the mirror.\\\ \\
{\bf Problem 1:} Show that $\vec{T} = <S'(y_0),1>$ is in fact tangent to the mirror at the point $(S(y_0),y_0)$.\\\ \\

Since $\vec{v}$ and $\vec{T}$ are orthogonal we conclude that $\vec{v}\cdot\vec{T}=0$. When written out explicitly this
yields
$(S(y_0)-d)S'(y_0) + y_0 = 0$. In this Knife Edge Test we perform the above measurement of $d$ for ``every'' $y_0$ in the range $0<y_0\leq R$. Since $d$ depends on $y_0$, we'll write $d(y_0)$, so we have
$$
(S(y_0)-d(y_0))S'(y_0) + y_0 = 0
$$
for $0<y_0\leq R$. Let's drop the subscript on $y$
and just write $(S(y)-d(y))S'(y) + y = 0$. A bit of rearrangement gives
\begin{equation}
\label{maineqn}
S'(y) = \frac{y}{d(y)-S(y)}.
\end{equation}
If we know (measure) $d(y)$ for all $y\in (0,R)$ we can solve the differential equation (\ref{maineqn}) with initial condition $S(0)=0$ to find $S(y)$, the shape of the mirror!

There are some additional practical issues that are addressed below (we can't take data for every $y$, for example), but the above is the basic idea. Before going on, let's stop to try some problems:\\\ \\
{\bf Problem 2:} Suppose we measure $d(y) = 1.5$ meters for every $y$ (that is,
when the light source is at a distance of $1.5$ meters from the mirror, all incident rays reflect orthogonally
off of the mirror, at every point on the mirror, and return to the source). Solve (\ref{maineqn}) with $S(0)=0$. What shape is the mirror, and how does the
light source position relate to this shape? (The answer should seem ``obvious'' after the fact.)\\\ \\
{\bf Problem 3:} For a general (unspecified) function $d(y)$, is (\ref{maineqn}) a linear ODE in the unknown $S$? Is it separable? Is it autonomous? Is it exact? Does it fit into any standard category that you know of? (Try feeding it to a computer algebra system with $d(y)$ left undefined.)\\\ \\
{\bf Problem 4:} Look up the Existence-Uniqueness Theorem for ODEs in a DE textbook. If we assume that $d(y)$ is a ``reasonable'' function,
(say $d$ is continuously-differentiable, and that $d(y)\geq M>0$ for some constant $M$), is (\ref{maineqn}) guaranteed to
have a unique solution with $S(0)=0$? That would be good---it means the data $d(y)$ we collect is sufficient for the purpose
of determining $S$.\\\ \\
{\bf Problem 5:} Suppose $d(y)=\frac{1}{2}+y^2$. Solve (\ref{maineqn}) with $S(0)=0$ to find the shape of the mirror. (You may need a computer algebra system).\\\ \\
{\bf Problem 6:} Suppose $d(y)=\frac{1}{4}+y^2$. Solve (\ref{maineqn}) with $S(0)=0$ to find the shape of the mirror.
(You WILL need a computer algebra system). Compare to Problem 5.

\section*{SOME PRACTICAL ISSUES AND EXAMPLES}

The first obvious difficulty with the knife edge test as outlined above is that we cannot measure $d(y)$ for every $y$. In practice one typically measures $d(y)$ at $3$ to $7$ values of $y$
between $y=0$ and $y=R$.
One approach is to then interpolate this data to form a function $d_p(y)$, then solve the ODE (\ref{maineqn}) using $d_p(y)$
in place of the true $d(y)$. This solution would be computed numerically using any standard numerical
ODE solver and provides an estimate of $S$.

\begin{example}
\label{examp1}
As an example, let's suppose the mirror shape is given by $S(y) = 0.205052y^2$ for $-R<y<R$ and $R=0.076$ meters. This corresponds
to a classic six-inch diameter parabolic mirror with a focal length of 48 inches. In this case the ``true'' function $d(y)$ is given by
$d(y) = 2.438405 + 0.205052y^2$ with $y$ and $d(y)$ in meters (to see this just plug the given $S(y)$ into (\ref{maineqn}) and solve for $d(y)$.)
Suppose we measure $d(y)$ at $y$ values $0.1R, 0.3R, 0.5R, 0.7R, 0.9R$. This yields the data in Table \ref{table1}.
These points are shown (black dots) in Figure \ref{fig6}. Note that $d(y)$ is measured to an accuracy of $10^{-4}$ meters,
just $1/10$ of a millimeter, with the light source over $2$ meters from the mirror. This is not typically possible, but not necessary either! We'll address this a bit later in Problem 7.
\begin{table}
\centering
\begin{tabular}{l|l}
$y$ (meters) & $d(y)$ (meters)\\\hline
$7.62\times 10^{-3}$ & $2.4384$\\
$2.29\times 10^{-2}$ & $2.4385$\\
$3.81\times 10^{-2}$ & $2.4387$\\
$5.33\times 10^{-2}$ & $2.4390$\\
$6.68\times 10^{-2}$ & $2.4394$
\end{tabular}
\caption{Knife edge data at $y=0.1R, 0.3R, 0.5R, 0.7R, 0.9R$, with $R=0.076$ meters.}\label{table1}
\end{table}

We can fit a curve, say a cubic spline, to these data points.
The computation is made easy by using a computer algebra system; see the allied notebook {\tt knifeedge.mw} (Maple) or
{\tt knifeedge.m} (Mathematica). The graph of the spline $d_p(y)$ that interpolates this data is shown as the red curve in Figure \ref{fig6}.
\begin{figure}[ht]
\centerline{\includegraphics[width=0.4\textwidth]{fouc6.jpg}}
\caption{Measured data and interpolating curve (in this case, a cubic spline).}
\label{fig6}
\end{figure}
We use the interpolating function $d_p(y)$ in (\ref{maineqn}) (instead of $d(y)$, which we don't really know). In this
example the actual formula for
$d_p(y)$ is a bit messy and not shown. The numerical solution to (\ref{maineqn}) turns out to be quite close to the correct parabola $S(y) = 0.205052y^2$,
but it's more illuminating to plot the difference of this estimate from the true $S(y)$ as in the left panel of Figure \ref{fig7}. Even more informative is this same difference as shown in the right panel, but with the vertical units as wavelengths of green light, $5.5\times 10^{-7}$ meter. The error
introduced by using only $5$ measurements of $d(y)$, interpolating, and then solving (\ref{maineqn}) for $S(y)$ is negligible (anything
under $1/20=0.05$ of a wavelength is generally considered negligible in this setting.)
\begin{figure}[ht]
\centerline{\includegraphics[width=0.4\textwidth]{fouc7.jpg}\ \ \includegraphics[width=0.4\textwidth]{fouc8.jpg}}
\caption{Error in estimated mirror shape (in meters on the left, wavelengths on the right).}
\label{fig7}
\end{figure}
\end{example}

\begin{example}
\label{examp2}
Most mirror makers are not trying to obtain a very specific parabola, e.g., $S(y)=0.205052y^2$ (corresponding to about a $48$ inch
focal length). Any parabola that is ``close'' will do; the mirror maker would be just about as happy if the mirror under construction
had a $48.2$ inch focal length (corresponding to $S(y) = 0.204201y^2$), since this is easily accommodated into the telescope's construction.
However, a 6 inch mirror with shape $S(y) = 0.204201y^2$ deviates from the mirror with shape $S(y) = 0.205052y^2$ by almost 10 wavelengths at the mirror edge. Hence in determining the quality of a parabolic mirror's shape we shouldn't compare it to a pre-chosen hypothetical parabola,
but rather to that parabola that it best fits in some quantifiable sense. The quality of the mirror will be judged by how far it deviates from this profile.

To illustrate, consider the data in Table \ref{table2}.
Here we have used $7$ points
for the knife edge measurements, at locations $y=(k+0.5)R/7$ for $k=0,1,\ldots,6$. We reconstruct an estimate of $S(y)$
as in Example \ref{examp1}, by fitting a cubic spline to the data in Table \ref{table2}, then solving (\ref{maineqn}) numerically with $S(0)=0$. This gives us an estimate $\tilde{S}(y)$ for the shape of the mirror.

To determine its quality we can then fit a parabola $S(y)=Ay^2$ to the resulting estimate $\tilde{S}(y)$ by minimizing the least squares function
$$
Q(a) = \sum_{j=0}^N (\tilde{S}(y_j)-A y_j^2)^2
$$
with respect to $A$, where the $y_j$ span the interval $[0,R]$, say $y_j=Rj/N$ with $N\approx 10$. In the present example this yields
$A\approx 0.201049$. The resulting parabola $S(y)=0.201049$ (corresponding to a perfect parabola with focal length $48.96$ inches) deviates from $\tilde{S}(y)$ by at most $0.28$ wavelengths. That is, the mirror that gives rise to the data in Table
\ref{table2} deviates from even the best fitting parabola by $0.28$ wavelengths (which means this mirror needs additional work). A plot of the difference $\tilde{S}(y)-A y^2$ is shown in Figure \ref{fig8}.
\begin{table}
\centering
\begin{tabular}{l|l}
$y$ (meters) & $d(y)$ (meters)\\\hline
$5.44\times 10^{-3}$ & $2.4857$\\
$1.63\times 10^{-2}$ & $2.4859$\\
$2.72\times 10^{-2}$ & $2.4863$\\
$3.81\times 10^{-2}$ & $2.4869$\\
$4.90\times 10^{-2}$ & $2.4877$\\
$5.99\times 10^{-2}$ & $2.4884$\\
$7.08\times 10^{-2}$ & $2.4892$
\end{tabular}
\caption{Knife edge data at points $y=(k+0.5)R/7$ for $k=0,1,\ldots,6$.}\label{table2}
\end{table}
\begin{figure}[ht]
\centerline{\includegraphics[width=0.4\textwidth]{fouc9.jpg}}
\caption{Error in estimated mirror shape from best fit parabola.}
\label{fig8}
\end{figure}
\end{example}
{\bf Problem 7:} How can the data for $d(y)$, as in Table \ref{table1} or Table \ref{table2}, be measured to such accuracy ($0.1$ mm) over a $2$ or $3$ meter span? We don't need to!
It's not the absolute value for the various $d(y)$ that's critical, it's their relative values. To illustrate, redo Example \ref{examp1}
but first add an ``error'' of $5$ mm ($0.005$ meter) to each value of $d(y)$, then reconstruct an estimate $\tilde{S}(y)$ of the mirror shape, and compare the error to the ``best fit'' parabola. How is this error manifest in the estimated $S(y)$? Try adding $1$ cm or even $5$ cm to each entry in $d(y)$ and repeat. What conclusion do you draw?\\\ \\
{\bf Problem 8:} In Table \ref{table3} there is some data from a 4.25" mirror the author made (so $R=0.054$ meters, roughly).
\begin{table}
\centering
\begin{tabular}{l|l}
$y$ (meters) & $d(y)$ (meters)\\\hline
$1.44\times 10^{-2}$ & $0.9208$\\
$2.50\times 10^{-2}$ & $0.9211$\\
$3.23\times 10^{-2}$ & $0.9214$\\
$3.82\times 10^{-2}$ & $0.9215$\\
$4.33\times 10^{-2}$ & $0.9216$\\
$4.78\times 10^{-2}$ & $0.9217$\\
$5.20\times 10^{-2}$ & $0.9221$
\end{tabular}
\caption{Knife edge data for $4.25$ inch mirror.}\label{table3}
\vspace*{-.6 true cm}
\end{table}


The relative accuracy between the $d(y)$ offsets is probably around $2\times 10^{-4}$ meters.
Estimate the mirror's profile $S(y)$, and in particular, how far it deviates from a best-fit parabola. What is the maximum
error, in wavelengths of green light?

\section*{CONCLUSION}

The Knife Edge Test was developed by the French physicist Foucault in the mid 19th century. It is one of a large number
of optical tests that can be made on a mirror. It's advantage is that it is relatively simple to build a tester and analyze the
data, and the test is quite sensitive. Most treatments of the test do not formulate the ODE (\ref{maineqn}), since this would be beyond what many amateur
telescope builders are familiar with. The references \cite{berry} and \cite{texereau} give alternative quantitative recipes
for analyzing knife edge data, but for those familiar with differential equations, the ODE (\ref{maineqn}) provides a
simple conceptual framework for determining the shape of the mirror.

\teacher{
\section*{NOTES FOR TEACHERS}

I've used this project in an introductory ODE course several times, and occasionally in our multivariable calculus class
at the point when I talk about using vectors to compute optical reflections, since the students have already seen ODE's in
Calculus 2. Equation (\ref{maineqn}) is an example of a DE that arises in a very simple and practical setting and yet
has no closed form solution---something that's often hard to come by.

I usually take the students through the derivation of (\ref{maineqn}) in class, which doesn't talk long, then turn them loose
with Maple to try solving the DE with a variety of choices of $d(y)$ (with $d(y)$ given explicitly, not as data). One natural
break point for this project is after Problem 6, if you don't want to get involved with interpolation or numerical solutions.
I give Problem 7 as homework, and sometimes I give an additional data set using data collected from mirrors I've made.

The material in the Appendix is entirely optional, but I always cover it in class so they understand how one can make the
required measurements (and it's very easy---no equations!)

}

\section*{APPENDIX: PERFORMING THE KNIFE EDGE TEST}

\subsection*{Spherical Mirrors}

How can we determine if light rays that reflect off the surface of a mirror do so orthogonally? Let's
first consider the case of a ``spherical mirror'' (in the two-dimensional setting, a mirror that is a portion of a circle, as illustrated in Figure \ref{fig2} in the left panel). Let $C$ on the $x$ axis denote the center of the circle of which the mirror is a part, and imagine a ``point'' source of light located precisely
at $C$ (shown in green). Intuitively, all light rays (shown in blue) that emanate from the source and strike the mirror do so at a right angle, and so are reflected back to $C$. The rays will continue past $C$; if an observer is properly positioned behind $C$ these rays will enter the observers ``eye'' (or eyepiece) and the mirror will appear to be brightly and uniformly illuminated.

\begin{figure}[ht]
\centerline{\includegraphics[width=0.4\textwidth]{fouc1.jpg}\ \ \ \ \includegraphics[width=0.4\textwidth]{fouc2.jpg}}
\vspace*{-.8 true cm}
\begin{center}
\parbox{11 true cm}{\caption{Light source (green dot), incident and reflected rays (blue) and spherical mirror (left panel), and with knife edge (right panel).}}
\end{center}
\label{fig2}
\end{figure}


Now imagine a ``knife'' positioned as illustrated in the right panel of Figure \ref{fig2}, with the knife tip slightly below $C$ so that it
doesn't block any of the reflected rays. The situation is essentially the same as in the left panel, and the observer sees a brightly and uniformly
illuminated mirror. However, if the knife edge is advanced vertically, it will eventually block the reflected rays and the observer will see the
mirror go dark. This will happen ``instantaneously'' when the knife tip passes the central (horizontal) axis of the mirror at $C$. (It's not really
instantaneous due to the small imperfections in the apparatus and the wave nature of light.)

Consider now what happens if we do the same experiment, but with the light source and knife edge positioned horizontally somewhere other than $C$,
say closer to the mirror, as illustrated in the left panel of Figure~\ref{fig3}.
\begin{figure}[ht]
\centerline{\includegraphics[width=0.4\textwidth]{fouc3.jpg}\ \ \ \ \includegraphics[width=0.4\textwidth]{fouc4.jpg}}

\vspace*{-.8 true cm}
\begin{center}
\parbox{11 true cm}{\caption{Light source (green dot), incident and reflected rays (blue) and spherical mirror (left panel), and with knife edge (right panel).}}
\end{center}
\label{fig3}
\end{figure}
In this case the reflected rays do not return to $C$; in fact, they don't converge to any single point, but we don't need to prove this
right now. As the knife edge is advanced upward it cuts off light rays from the bottom of the mirror first, then rays from the middle, finally
rays from the top. The observer will see the mirror go dark from bottom to top (assuming he/she accounts for the reversed image behind $C$).
Contrast this to the case in which the light source/knife edge are moved farther from the mirror, illustrated in the right panel of
Figure \ref{fig3}. In this case as the knife edge advances
it cuts off light from the top of the mirror first; the observed will see the mirror go dark from top to bottom, the opposite of the case
in the left panel. In the Knife Edge Test the light source and knife edge are always moved (horizontally) together.

The moral of the above examples is this: If the mirror is circular/spherical
and the light source/knife is closer to the mirror than the center $C$, then as the knife edge is advanced upward the mirror will appear to go dark in the same direction that the knife edge advances (bottom to top). If the light source/knife is farther from the mirror than $C$, the mirror will appear to go dark from top to bottom (opposite the motion of the knife edge). When the source/knife are exactly at $C$ the mirror appears to go dark ``instantly.''

If one wanted to determine the center of curvature of a spherical mirror, a test such as this could be used. Simply place the source/knife
somewhere on the central axis and advance the knife edge upward while observing from behind. If the mirror goes dark in the direction of the knife
edge travels, you're closer to the mirror than $C$; move farther out. If the mirror goes dark the other way, move closer. An iterative
process will lead you to the point $C$, when the entire mirror will go dark simultaneously. If no such point can be found, the mirror is
not spherical!

\subsection*{Mirrors of General Shape}

Refer to Figure \ref{fig10}.
\begin{figure}[ht]
\centerline{\includegraphics[width=0.8\textwidth]{fouc10b.jpg}}
\caption{Knife edge test on a non-spherical mirror.}
\label{fig10}
\end{figure}
Consider what the observer will see as the knife edge advances upward. The first ray to be cut off is a dashed one, and the observer will see the region of the mirror off of which that ray reflected go dark.
However, the other dashed ray will not be cut off, and that diametrically opposite portion of the mirror remains light. As the knife edge continues
upward it will eventually cut off both solid rays simultaneously, since both reflect off the mirror at a right angle and return to the source.
The observer will see the corresponding portions of the mirror (diametrically opposite each other) go dark simultaneously. Only after
the knife edge has advanced further will the region of the mirror off of which the second dashed line go dark.

The grand conclusion:\\\ \\
\textit{If the light source/knife edge is placed at a given distance from the mirror on the central axis and the knife edge
is advanced, and two diametrically opposite regions on the mirror go dark simultaneously, then the light rays that reflected from
that region did so at a right angle.}\\\ \\
We can use the above observation to measure $d(y_1)$ for any specific $y_1$. A Foucault tester consists of a point
light source and knife edge on an adjustable carriage that is placed in front of the mirror.  One places a non-reflective ``mask'' over the mirror that allows only light from zones near $y=\pm y_1$ to pass, and then adjusts the distance from
the test to the mirror so that as the knife edge advances the two zones near $y=\pm y_1$ dim simultaneously. This
provides a measurement of $d(y_1)$.  See \cite{astrobanana} for some photographs of a Foucault test and what a mirror
looks like through the instrument as measurements are taken.



\teacher{

\section*{SOME ANSWERS}
\begin{enumerate}
\item Consider the two nearby points $(S(y_0),y_0)$ and $(S(y_0+dy),y_0+dy)$ on the curve $x=S(y)$. The displacement
vector $<S(y_0+dy)-S(y_0),dy>$ is tangent to the curve in the limit $dy\rightarrow 0$, as is the rescaled version
$<(S(y_0+dy)-S(y_0))/dy,1>$. The latter limits to $<S'(y_0),1>$ as $dy\rightarrow 0$. Alternatively, parameterize
$x=S(y)$ as $\mathbf{r}(t) = <S(t),t>$ and note that $\mathbf{r}'(t) = <S'(t),1>$ is always tangent to the curve, in particular, at $t=y_0$.

\item In this case (\ref{maineqn}) becomes
$$
S'(y) = \frac{y}{1.5-S(y)}
$$
with $S(0)=0$, a separable ODE. Separation of variables and a bit of algebra shows that
$$
S(y) = \frac{3}{2} - \sqrt{(3/2)^2-y^2}.
$$
This is in fact a portion of the circle $(x-3/2)^2 + y^2 = (3/2)^2$ of radius $3/2$, centered at $x=3/2,y=0$.
This makes perfect sense---the light source is at the center of a perfect circle and all rays return to the source!

\item This ODE is not linear, nor separable, nor autonomous, for a general $d$. Written in the form $(d(y)-S)\,dS = y\,dy$,
we can see the ODE is exact precisely when $\frac{\partial (d(y)-S)}{\partial y} = \frac{\partial y}{\partial S}$, leading to
$d'(y)=0$. The DE will be exact (and so ``easily'' solvable) only when $d(y)$ is constant, as in Problem 2.

\item If we write $S'(y) = f(y,S(y))$ with $f(y,S)=y/(d(y)-S)$ then the Existence-Uniqueness Theorem guarantees a unique solution through $y=0,S=0$, since $f$ and it's partial derivatives are continuous near here.

\item The DE (\ref{maineqn}) is $S'(y) = y/(1/2+y^2-S(y)$. With $S(0)=0$ one can easily verify that $S(y)=y^2$ is the solution. Maple also returns this. It's not clear what general technique will make this easy to do by hand though!

\item The DE (\ref{maineqn}) is $S'(y) = y/(1/4+y^2-S(y)$. With $S(0)=0$ Maple returns
$$
S(y) = y^2-\frac{1}{4}-\frac{1}{2}W(-e^{2y^2-1/2}/2)
$$
where $W$ is the ``Lambert W'' function.

\item In this case adding $0.005$ to the knife edge data, reconstructing by solving (\ref{maineqn}), then
fitting a parabola to the reconstructed $\tilde{S}(y)$ yields best-fit parabola $x\approx 0.204631y^2$. The difference
between $\tilde{S}$ and this parabola is tiny, on the order of $1/100$ wavelength.  The only thing that adding $0.005$
to the knife edge data did is to change the best-fit parabola (the original parabola, $x=0.205052y^2$, corresponds to a
focal length of $48$ inches, while the parabola $x=0.204631y^2$ has focal length $48.1$ inches.) Adding $0.01$ or even $0.05$
to the knife edge data doesn't change anything, except the best-fit parabola. The discrepancy between the mirror and this
best-fit parabola remains (almost) the same.

\item Reconstructing a best-fit parabola yields a surface error (in wavelengths) shown in Figure \ref{fig11}.
\begin{figure}[ht]
\centerline{\includegraphics[width=0.4\textwidth]{small_mirror.jpg}}
\caption{Error with regard to best-fit parabola for mirror with knife edge data in Table \ref{table3}.}
\label{fig11}
\end{figure}
\end{enumerate}
}

%\section*{ACKNOWLEDGEMENTS}



\begin{thebibliography}{99}

\bibitem{astrobanana} Astrobananas. \href{http://astrobananasenglish.blogspot.com/2011/07/foucault-test.html}{http://astrobananasenglish.blogspot.com/2011/07/foucault-test.html}. Accessed 31 July 2019. 
\bibitem{berry} Berry, R.  2001. \textit{Build Your Own Telescope, 3$^{\rm rd}$ Edition}. Richmond  VA: Willmann-Bell.

\bibitem{texereau} Texereau, J.  1984. \textit{How to Make a Telescope, 2$^{\rm nd}$  English Edition}. Richmond  VA: Willmann-Bell.


\end{thebibliography}

\end{document}
