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%TCIDATA{Created=Fri Oct 11 17:28:47 1996}
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%TCIDATA{Language=American English}

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\markright{{\sf Vectorial Diffraction Theory}}
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\begin{document}


\section{9.10\quad Vectorial Diffraction Theory.}

\subsection{9.10.1}

\quad \thinspace \textbf{a. }In principle, all one has to do is apply the
vector Kirchhoff integral of previous section, making an appropriate choice
of the surface $S$ and of the field components, so that plausible
approximations can be introduced on physical grounds. A suitable choice of $%
G $ to meet the particular boundary conditions (Dirichlet or Neumann, for
instance) will also help in obtaining formulas that are both accurate and
compact. The main features of the results, however, are due to the
integration of the factor $\exp (ik\left| \mathbf{x-x}^{\prime }\right| )$
over the effective source region and are independent of the more refined
handling of other factors.

\smallskip\ 

\textbf{b}. In practice, there is an endless variety of tricks that are
applicable to particular situations. We can only give a couple of examples
of the application of the theory to metallic screens.

\smallskip\ 

\textbf{c.} As a first example, consider a flat, thin metallic screen of
finite dimensions at $z=0.$ For fields to the right of the screen, the
surface $S$ is chosen to consist of the plane $S_1$ at $z=0^{+}$ and the
closing half-sphere at infinity; for the fields to the left, $S$ consists
similarly of the plane $z=0^{-},$ and the closing half-sphere at the left of
the screen. The sources of the incident $\mathbf{E}^{(0)}$ are on the left.

\smallskip\ 

\textbf{d. }It is perhaps best to work with the scattered fields, $\mathbf{E}%
^{\prime },\mathbf{B}^{\prime },$ rather than with the total fields $\mathbf{%
E}^{(0)}+\mathbf{E}^{\prime },$ $\mathbf{B}^{(0)}+\mathbf{B}^{\prime },$ for
the simple reason that $\mathbf{E}^{\prime },\mathbf{B}^{\prime }$ have
sources only on the finite screen.

\newpage

\subsection{9.10.2}

\quad \quad \textbf{a. }From the symmetry of the problem, it is seen that $%
B_x^{\prime }$ and $B_y^{\prime }$ vanish at $z=0,$ except that they take
opposite values at $z=0^{+}$ and $z=0^{-}$ where the metallic screen is
present (see figure). If the scattered fields can be expressed in terms of
an integral involving $\mathbf{n\times B}^{\prime }$, this integral will
have to be extended only to (one face of) the screen. For $S_1$ at $z=0^{+},$
$\mathbf{n}$ points along $z$ and $\partial /\partial n\rightarrow \partial
/\partial z.$

\smallskip\ 

\textbf{b. }In this case it is best to work directly with the Kirchhoff
integral for the components of the vector potential $\mathbf{A,}$ using a
Neumann $G$ ($\partial G/\partial z$ vanishes on the plane $z=0).$ We have 
\begin{equation}
A_x^{\prime }(\mathbf{x}^{\prime })=\int_{S_1}G_N\frac{\partial A_x^{\prime }%
}{\partial z}dxdy\quad \quad \quad A_y^{\prime }(\mathbf{x}^{\prime
})=\int_{S_1}G_N\frac{\partial A_y^{\prime }}{\partial z}dxdy  \label{1}
\end{equation}
where $G_N$ reduces simply to 
\[
2\frac{e^{ikR}}{4\pi R},\text{\quad \quad with }R=\sqrt{(x^{\prime
}-x)^2+(y^{\prime }-y)^2+z^{\prime 2}.} 
\]
\textbf{\ }

\smallskip\ 

\textbf{c. }In vector notation, eqs.(\ref{1}) are written (note that $A_z=0)$%
\begin{equation}
\mathbf{A}^{\prime }(\mathbf{x}^{\prime })=\frac 1{2\pi }\int_{screen}\left( 
\mathbf{n\times B}^{\prime }(\mathbf{x)}\right) \frac{e^{ikR}}Rdxdy.
\label{1'}
\end{equation}

\smallskip\ 

\textbf{d. }It should be noted that (\ref{1}) holds (as written) only to the
right of the screen, but (\ref{1'}) holds on both sides (since $\mathbf{%
n\rightarrow -n}$ and $\mathbf{B}^{\prime }\mathbf{\rightarrow -B}^{\prime }$
on the left).

\smallskip\ 

\textbf{e.} $\mathbf{B}^{\prime }$ is found from $\mathbf{\nabla \times A}%
^{\prime }$ and $\mathbf{E}^{\prime }$ from $ik\mathbf{E}^{\prime }=\mathbf{%
\nabla \times B}^{\prime }.$

\smallskip\ 

\textbf{f.} The Kirchhoff \textsl{approximation} is now physically
transparent: for points to the right of the screen we can take $\mathbf{B}%
^{\prime }(\mathbf{x)\approx -B}^{(0)}(\mathbf{x)}$ in (\ref{1'}), so that $%
\mathbf{B}^{(0)}+\mathbf{B}^{\prime }=0;\mathbf{\ }$for the left-side fields
one must then take $\mathbf{B}^{\prime }=\mathbf{B}^{(0)}$ to satisfy the
symmetry.

\newpage 

\subsection{9.10.3}

\quad \thinspace \textbf{a.} As a second example, consider the
``complementary'' case of a flat, thin metallic screen extending to
infinity, with a hole at finite size in it. One can use for this eq.(\ref{1}%
) as it stands, but a formula where the integration is \textsl{over the
aperture only} is clearly better.

\smallskip\ 

\textbf{b. }The quickest way to proceed is to note, as usual, that Maxwell's
equations in empty space are symmetric in $\mathbf{E}$ and $\mathbf{B.}$
Thus, if $\mathbf{B}^{\prime }$ is given by 2.c and 2.e, it must be true
that 
\begin{equation}
\mathbf{E}^{\prime }(\mathbf{x}^{\prime })=\frac{\mathbf{\nabla \times }}{%
2\pi }\int_{S_1}\left( \mathbf{n\times E}^{\prime }(\mathbf{x)}\right) \frac{%
e^{ikR}}Rdxdy.  \label{2}
\end{equation}
(one introduces an electric vector potential $\mathbf{Z}$ analogous to $%
\mathbf{A).}$

\smallskip\ 

\textbf{c. }The difference now is that $\mathbf{E}^{\prime },$ unlike $%
\mathbf{B}^{\prime },$ does not vanish by symmetry at $z=0;$ however, the
total field $\mathbf{E}_{/\!/}=\mathbf{E}_{/\!/}^{(0)}+\mathbf{E}%
_{/\!/}^{\prime }$ vanishes \textsl{on the screen.} Now it is easy to see
that, to the right of the screen 
\begin{equation}
\mathbf{E}^{(0)}(\mathbf{x}^{\prime })=\frac{\mathbf{\nabla \times }}{2\pi }%
\int_{S_1}\left( \mathbf{n\times E}^{(0)}(\mathbf{x)}\right) \frac{e^{ikR}}%
Rdxdy.  \label{3}
\end{equation}
because $S_1$ encloses the sources (at infinity) of $\mathbf{E}^{(0)}.$

\smallskip\ 

\textbf{d.} Adding (\ref{2}) and (\ref{3}) side by side we have, for $z>0$%
\begin{equation}
\mathbf{E}(\mathbf{x}^{\prime })=\frac{\mathbf{\nabla \times }}{2\pi }\int
\left( \mathbf{n\times E}(\mathbf{x)}\right) \frac{e^{ikR}}Rdxdy.  \label{4}
\end{equation}
where the integral is now extended only over the \textsl{holes} in the
screen.

\smallskip\ 

\textbf{e. }Kirchhoff's approximation is now to replace $\mathbf{E}$ by $%
\mathbf{E}^{(0)}$ in (\ref{4}).

\newpage

\subsection{9.10.4}

\quad \thinspace \textbf{a.} The fields on the left of the screen can be
found as follows. $\mathbf{E}^{\prime }$ is still given by (\ref{2}), where
now $S_1$ is just to the left of the screen and the normal must point to the
left. If, following Jackson, we still take $\mathbf{n}$ to point to the
right, we have 
\begin{equation}
\mathbf{E}^{\prime }(\mathbf{x}^{\prime })=-\frac{\mathbf{\nabla \times }}{%
2\pi }\int_{S_1}\left( \mathbf{n\times E}^{\prime }(\mathbf{x)}\right) \frac{%
e^{ikR}}Rdxdy.  \label{2'}
\end{equation}

\smallskip\ 

\textbf{b. }If we consider an infinite metallic screen with \textsl{no}
holes, the reflected fields on the left is given by (note the analogy with (%
\ref{2'})) 
\begin{equation}
\mathbf{E}^{(1)}(\mathbf{x}^{\prime })=-\frac{\mathbf{\nabla \times }}{2\pi }%
\int_{S_1}\left( \mathbf{n\times E}^{(1)}(\mathbf{x)}\right) \frac{e^{ikR}}%
Rdxdy.  \label{3'}
\end{equation}

\smallskip\ 

\textbf{c. }Subtracting (\ref{3'}) from (\ref{2'}) side by side and noting
that $\mathbf{E}^{(0)}+\mathbf{E}^{(1)}=0$ on the plane $S_1,$ we find 
\begin{equation}
\mathbf{E}^{\prime }(\mathbf{x}^{\prime })-\mathbf{E}^{(1)}(\mathbf{x}%
^{\prime })=-\frac{\mathbf{\nabla \times }}{2\pi }\int \left( \mathbf{%
n\times E}(\mathbf{x)}\right) \frac{e^{ikR}}Rdxdy.  \label{4'}
\end{equation}
where the integral is now over the aperture only.

\smallskip\ 

\textbf{d. }To display the results, it is convenient to introduce $\mathbf{E}%
^{\prime \prime },$ the \textsl{diffracted field due to the holes}, given by
definition by the r.h.s. of (\ref{4}) everywhere. Then, on the right of the
screen $\mathbf{E=E}^{\prime \prime },$ while on the left, according to (\ref
{4'}) 
\begin{equation}
\mathbf{E=E}^{(0)}+\mathbf{E}^{\prime }=\mathbf{E}^{(0)}+\mathbf{E}^{(1)}-%
\mathbf{E}^{\prime \prime }.  \label{5}
\end{equation}

\smallskip\ 

\textbf{e.} Kirchhoff's approximation is to replace $\mathbf{E}$ by $\mathbf{%
E}^{(0)}$ in the expression (\ref{4}) for $\mathbf{E}^{\prime \prime }.$

\end{document}
