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\title{Answers to Examples Class 1}
\author{}
\begin{document}
\maketitle
\section*{Bragg Law}
Taking dot product of {\bf g} with the vector equation for the Bragg law proves that 2{\bf k.g}=$-g^2$.
\begin{figure}[!ht]
  \begin{center}
    \begin{minipage}[t]{0.32\linewidth}
\raisebox{-5cm}{\includegraphics[width=0.98\linewidth]{Ewald2.eps}} 
    \end{minipage}\hfill
    \begin{minipage}[t]{0.65\linewidth}
 \caption{From the geometry of the triangle in the figure, it is evident that $\sin\theta = \frac{1}{2}g/k$ so that $\sin\theta = (0.5/d)/(1/\lambda)$ which on rearrangement gives the Bragg law.}
    \end{minipage}
  \end{center} \label{fig:ewald}\end{figure} 
The symmetry axes should give the hint that the zone axis of the diffraction pattern is $<111>$ so by identifying two plane normals to this direction, and taking account of systematic absences, it is easy to index the pattern as follows:
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\includegraphics[width=5.0cm]{7-1.eps}}
\caption{(a) b.c.c. (b) f.c.c.}
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\newpage  \section*{Use of sketch stereograms}
  
  The first four faces are cut in the order $(111)$, $(11\overline 1)$, $(\overline111)$ and $(\overline 1 1 \overline 1)$.
  \begin{itemize}
\item The first cut leaves a single triad along [111] and three $\{110\}$ mirror planes in that zone. The crystal system is trigonal, defined by the single triad.
\item After the second cut, the (001) and $(1\overline 1 0)$ mirror planes remain in the zone [110] (the latter is a diad). The crystal system is orthorhombic, defined by three mutually perpendicular diads (which can include inverse diads, equivalent to mirror planes).
\item Only the $(\overline 1 0 1)$ mirror remains after three cuts, leaving the crystal system monoclinic.
\item The fourth cut increases symmetry, with a tetrad along [010] and four mirror planes (100), (001), $(\overline 101)$ and (101). The crystal system is tetragonal.
\end{itemize}


\begin{figure}[!ht]
\begin{center}
\includegraphics[width=0.9\textwidth]{cuts.eps}
\label{fig:mechanism}
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\end{figure} 

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