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\begin{document}
% ---------------------------- Chapter begins -----------------------------
\chapter{The isotropy and absorption of gamma rays}
\section{Object}
To verify the principle of isotropy of radiation and to verify
Lambert's law and measure the mass absorption coefficient of lead.
\section{Theory A}
Isotropy of radiation claims that there is no preferential direction
... is expressed as a $1/d^2$ variation of intensity where $d$ is the source to detector
distance. However, this formulation only holds when both the source
and detector are points.
%FIGURE 1
\begin{figure}[htbp]
\centerbmp{5in}{3in}{/DirectoryName/Filename.bmp} %x=5in y=3in scale as you wish
\caption{\bf Detector geometry.}
\end{figure}
Thus we need to calculate the solid angle subtended by the detector
face. The solid angle, $\Omega $, subtended by any surface from a
point (we assume the radiation source is a point) is defined as
\begin{equation}
\Omega = {\int\int}_s\frac{\hat{R}\cdot d\vec{s}}{R^2}=\int\int_s\frac{\hat{R}\cdot\hat{n}}{R^2}ds
\end{equation}
where $\hat{R}$ is a unit vector along $\vec{R}, \vec{R}$ is the
position vector of the element of area $ds$, $\hat{n}$ is a unit normal
to $ds$. Since in our case the surface in question is a sphere (see
figure 1) $\hat{R}$ and $\hat{n}$ will be parallel, with the result
that $\hat{R}\cdot\hat{n} = 1$. Thus
$$\Omega = {\int\int}_s\frac{dS}{R^2}$$
If we take $s$ to be a whole sphere, then
$$\Omega = \frac{1}{R^2}{\int\int}_s dS = \frac{1}{R^2} 4\pi R^2 = 4\pi $$
That is to say, $4\pi$ is the maximum possible solid angle with
units of steradians or $sr$. The solid angle subtended by the
detector, a flat disc of radius $r$, with respect to the source at a
distance $d$ from the detector will be given by
\begin{eqnarray*}
\Omega & = & \frac{1}{R^2} \int\int_{face} dS = \frac{1}{R^2} \int^{\arccos(d/R)}_0 \int^{2\pi}_0 R^2 \sin\theta\ d\phi\, d\theta \\
& = & 4 \int^{\arccos (d/R)}_0 \int_0^{\pi/2} \sin\theta \ d\phi\, d\theta\\
& = & 2 \pi \int^{\arccos (d/R)}_0 \sin\theta\ d\theta = -2\pi \cos\theta |_0^{\arccos(d/R)} = 2\pi (1-\frac{d}{R})
\end{eqnarray*}
But $R = \sqrt{d^2 + r^2}$ so
\begin{equation}
\Omega = 2\pi (1-\frac{d}{\sqrt{d^2+r^2}})
\end{equation}
......
\begin{equation}
\Omega_2 = 2\pi [1-\cos\theta_2]
\end{equation}
We may also write $\Omega $ as a solid angle over a variable $\theta $
\begin{equation}
\Omega = 2\pi [1-\cos\theta ]\ \mbox{and}\ d\Omega = 2\pi\sin\theta\, d\theta
\end{equation}
In order to write $x$ in terms of $\theta $, the integral for
$\epsilon_E$ must be divided into two regions: one between 0 and
$\theta_1$ and one between $\theta_1$ and $\theta_2$. Between 0 and
$\theta_1,\ x = \frac{h}{\cos\theta }$ (see figure~5). Note that normally in a
paragraph the previous equation looks better as $x=h/\cos\theta$.
\end{document}