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\title{Brief Article}
\author{The Author}
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\begin{flushleft}
D. Rubin\\
April 12, 2018\\
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\begin{center}
Comment on Magnetometer Calibration\\
(see C. Stoughton docdb 11178-v2 and S. Kim docdb 7655)
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\noindent
The angular dependence of intensity is
$$I=I_0\cos^2(\frac{\pi}{4}+\theta)$$
For $\theta\ll 1$
$$I\sim \frac{1}{2}(I_0-2\theta)$$
and 
\begin{eqnarray}
\frac{I-I_0}{I_0} &\sim& -\frac{1}{2}(1+2\theta)\sim  -\frac{1}{2}+\theta \label{intensity}
\end{eqnarray}
Since  $\theta = VBd$, $I_0\ne I$ when $B=0$. 

\vskip 0.12in
\noindent
But SeungCheon's line (below) goes through zero.
Perhaps he is not plotting $\frac{\Delta I}{I_0}$ but rather $\frac{\Delta V}{V_0}$
where $V_0 = V(\theta=0)$.  
\vskip 0.1in
\noindent
Instead of Equation 1 he would write
$$V  = 2V_0\cos^2(\frac{\pi}{4}-\theta)$$ so that $V(\theta=0)=V_0$.
Then
$$V= 2V_0(\frac{1}{2}-\theta)=V_0(1-2\theta)$$
and $$\frac{V-V_0}{V_0} = -2\theta$$
Again, this equation has the virtue that when $B=0$, $V=V_0$, consistent with is plot.

\begin{figure}[!htb]
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   \includegraphics*[width=\textwidth]{SeungCheonFig7.pdf}
   \label{sc}
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\end{figure}

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