%\begin{center}
\section{L-9: Wavelength Measurement and Holograms}\label{l09_lambdaHolo}
%\end{center}

\noindent {OBJECTIVES}:  \begin{enumerate}
\item \latex{\vspace*{-.07in}}  
To measure the wavelength of laser light by using a steel 
rule as a diffraction grating; to observe and understand holograms.
\end{enumerate}

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\hspace*{.3in} \framebox[5.2in]{\parbox[b]{5in}{\smallskip {\em CAUTION:} 
The laser is a very bright source.  Do not 
allow the laser beam to 
enter the eye and do not point the beam at anyone!}}
\smallskip
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<IMG SRC="../../lab/gif/caution.gif" ALIGN=center> <br><br> 
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\noindent {APPARATUS}:  \begin{enumerate}
\item[] \latex{\vspace*{-.1in}} 
He-Ne laser; steel rule; tripod stand \& angle clamp; 
holograms; mercury arc; diverging lenses; 15 meter tape.\latex{\vspace*{-.07in}}  
\end{enumerate}

\begin{center}
{\bf Part~I~-~Wavelength~Measurement}\latex{\vspace*{-.07in}}
\end{center}

\noindent {INTRODUCTION}:  \begin{enumerate}
\item[] \latex{\vspace*{-.1in}} 
Specular reflection of the laser beam from a steel 
ruler gives a zero order image.  In Fig.~1 the distance $h_{0}$ and $D$ 
determine $\theta_r$, the angle of reflection: $\theta_r=\tan^{-1}(D/h_{0}$) 
and hence also $\theta_i$, the angle of 
incidence.  

\noindent The markings on the steel ruler form a coarse reflection 
grating which also yields higher order diffracted images.  The angle 
of diffraction for the $n^{th}$ 
order is
\[\delta _{n} = \tan^{-1}(D/h_{n})~~ .\]

\begin{figure}[h]  \centering \latex{\vspace*{-.2in}}
\latex{\includegraphics[height=1.6in]{figs/l9-01.eps}}
\html{\includegraphics[height=1.75in]{figs/l9-01.eps}}
\latex{\vspace*{-.1in}}
\caption {Schematic of diffraction from ruler markings.}
\end{figure} \latex{\vspace*{-.1in}}
\noindent Let $d$ be the separation of 
lines on the steel ruler.  For the $n^{th}$ 
order the path difference, $\Delta $, between waves from successive ruler lines 
is $n \lambda $.  From Fig.~2, $\Delta =\mbox{AC}-\mbox{BE}$.  But 
$\mbox{AC} = d \cos
(90^\circ-\theta_i)=d \sin \theta_i$ and 
$\mbox{BE}= d \cos
(90^\circ-\delta_n)=d \sin \delta_n$.  \\
Hence
\[\Delta  =  d~(\sin \theta_i  -  \sin \delta _{n}) = n \lambda~~ .\]\latex{\vspace*{-.1in}}
\begin{figure}[!h] \latex{\vspace*{-.2in}}\centering 
\latex{\includegraphics[height=1.55in]{figs/l9-02.eps}}
\html{\includegraphics[height=2.25in]{figs/l9-02.eps}}
\latex{\vspace*{-.1in}}\caption {Geometry required for constructive interference.}
\end{figure} 
% \latex{\vspace*{-.1in}}

\item[] \begin{description}
\item{\em QUESTION}: To minimize uncertainty in $\lambda $ should you measure a high 
or low order diffracted image?
\end{description}
\end{enumerate}

\noindent {\em EXPERIMENT}: \begin{enumerate}
\item[] \latex{\vspace*{-.07in}}  
Locate the zero order spectrum (specular reflection) by moving 
ruler sideways so beam misses the small (diffracting) divisions.  Return 
ruler so one has the high order diffraction images.  Make 
all the relevant measurements. Calculate $\lambda $ and compare
with the accepted $\lambda =633$~nm.  
For the  error 
estimate first find the uncertainty in i and $\delta _{n}$ (by estimating 
reasonable ranges of $D$, $h_0$ and $h_n$).  Then find the corresponding 
uncertainties in sin i and in $\sin \delta_n$ (and hence in $\lambda $).
\end{enumerate}

\begin{center}
{\bf Part~II~-~HOLOGRAMS}
\end{center}

\begin{enumerate}
\item {\em To~observe~a~hologram:} For this use a high intensity mercury arc with a 
green color filter.  Have the room reasonably dark and stand about 3 meters 
in front of the mercury arc source.  Look through the hologram but not 
directly at the source.  The images are virtual and 
three-dimensional.  

\noindent Alternatively, diverge a laser beam to illuminate the 
hologram and view it  as in Fig.~3.  \\ 
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<IMG SRC="../../lab/gif/caution.gif" ALIGN=center> <br><br> 
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\fbox{DO~NOT ALLOW
the UNDIFFUSED LASER BEAM TO ENTER THE EYE!}
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<IMG SRC="../../lab/gif/caution.gif" ALIGN=center> <br><br> 
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\begin{figure}[!h] \latex{\vspace*{-.1in}}\centering 
\includegraphics[height=1.75in]{figs/l9-03.eps}
\latex{\vspace*{-.1in}}\caption {}
\end{figure}

\noindent {OPTIONAL}:  Simultaneously illuminate the hologram with a diverged red 
laser light and a green light from the Hg arc.  Note the larger 
red image.  
\item {\em An~explanation~of~the~HOLOGRAM:}
To understand a hologram image, consider in 
Fig.~4 the formation for the simplest possible object, 
a small particle at O$_1$. 
The glass plate B partially reflects the plane waves from the laser and 
partially transmits them to illuminate the object.

\begin{figure}[!h] \latex{\vspace*{-.1in}}\centering 
\includegraphics[height=2.45in]{figs/l9-04.eps}
\latex{\vspace*{-.1in}}\caption {}
\end{figure}

\noindent The reflected plane waves plus the spherical waves scattered from O$_1$ 
meet in the photograhic plate P and produce an interference pattern.  The 
spacing d between regions of constructive interference is
$d = \lambda /\sin \theta $ (see in Fig.~4 the enlargement of the region around I).  This 
interference results in lines with spacing d in the photographic plate and 
constitute a diffraction grating of spacing d.  Note that $\theta $ and $d$ vary 
with position on the plate.  This diffraction grating of variable spacing 
constitutes the hologram for our small particle at O$_1$. 

\noindent If we illuminate this grating with plane waves traveling in the direction 
BD, the same direction as the reflected 
plane waves (from the beam splitter B) used to 
to form the hologram, we observe constructive interference 
(a first order spectrum) in the direction given by the regular grating 
equation $\sin \theta = \lambda /d$:  namely the direction of the original 
wave from O$_1$ which formed the hologram.  Because d varies 
from place to place on the plate, no matter where we look through the plate the 
light in the first order image comes from the direction O$_1$.  
Thus the spherical wave fronts originally from O$_1$ are 
reconstructed, and an observer at E sees a virtual image of O$_1$ at O$_1$.

\noindent A similar treatment holds for a second scattering point O$_2$, and the 
brightness of the image point will depend on the contrast in the 
interference pattern which in turn depends on the brightness of the object 
point.  The image of an extended object thus forms point by point,
and the image is three dimensional because the virtual image 
points coincide with 
the corresponding object points in space.  
If one views the hologram with longer $\lambda $ light, 
all angles increase and a magnified image results.

\noindent A first order diffraction beam also appears on the other side of 
the normal to the hologram, and the diffracted light makes the same angle 
with the normal to the hologram as the original waves from O$_1$.  
Thus a real image of O$_1$ forms at O$'_1$ (Fig.~5) and 
similarly for all other object points.  This real image can be caught on a 
screen, one plane in focus at a time.

\begin{figure}[!h] \latex{\vspace*{-.1in}}\centering 
\includegraphics[height=1.75in]{figs/l9-05.eps}
\latex{\vspace*{-.1in}}\caption {}
\end{figure}
\end{enumerate}
