E.3 Rectangular Aperture

Figure E.6 shows the aperture plane with the coordinates $\tilde{x}$ and $\tilde{y}$.

Figure E.6: Coordinate system in a rectangular aperture

The transmission function is

$$\tilde{\tau}(\tilde{x},\tilde{y})= \left\{\begin{array}{r@{\quad:... ...} \wedge \vert\tilde{y}\vert < \tilde{y_0} 0 & otherwise \end{array} \right.$$ (E.54)

The area of the aperture is given by $4\tilde{x_0}\tilde{y_0}$. The FOURIER transform can be calculated from (E.27) and splitted into two integrals

$$T(u,v) = T(u)T(v) = \int\limits_{-\tilde{x_0}}^{\tilde{x_0}} e^{-... ...e{x}\int\limits_{-\tilde{y_0}}^{\tilde{y_0}} e^{-i2\pi v \tilde{y}} d\tilde{y}$$ (E.55)

One stripe in Figure E.6 is thereby given by

$$e^{-i2\pi u \tilde{x}} d\tilde{x}\int\limits_{-\tilde{y_0}}^{\tilde{y_0}} e^{-i2\pi v \tilde{y}} d\tilde{y}$$ (E.56)

Integration of (E.55) is simple and straightforward

$$T(v) = \frac{1}{-i2\pi v}\left(e^{-i2\pi v\tilde{y_0}}-e^{+i2\pi ... ...0})}{\pi v}=2 \tilde{y_0} \frac{\sin (2\pi v \tilde{y_0})}{2 \pi v \tilde{y_0}}$$ (E.57)

The rightmost term can be defined as a new function

$$\sinc (w) \equiv \frac{\sin(w)}{w}$$ (E.58)

With this substitution (E.57) yields

$$T(v) = 2\tilde{y_0}\sinc (2\pi v\tilde{y_0})$$ (E.59)

and

$$T(u) = 2\tilde{x_0}\sinc (2\pi u\tilde{x_0})$$ (E.60)

in an analogous way. Therefore in Point $P'$ the electric field is according to (E.29)

$$E'(u,v) = E'(0,0)\frac{2x_0 \sinc (2\pi u\tilde{x_0})2y_0\sinc (2... ...0}\tilde{y_0}}\left[\frac{R_{00}R'_{00}}{R_0R'_0}\right]e^{i(\phi_0-\phi_{00})}$$

$$= E'(0,0) \sinc (2\pi u\tilde{x_0})\sinc (2\pi \tilde{y_0})\left[\frac{R_{00}R'_{00}}{R_0R'_0}\right]e^{i(\phi_0-\phi_{00})}$$ (E.61)

and the intensity as the square of the electrical field is then

$$I'(u,v) = I'(0,0) \sinc ^2(2\pi u \tilde{x_0}) \sinc ^2(2\pi v\tilde{y_0})\left[\frac{R_{00}R'_{00}}{R_0R'_0}\right]^2$$ (E.62)

For the special case of the source being located on the z-axis, the coordinates $x,y$ are zero and the coordinates $x',y'$ are the following functions of $u,v$

$$x' = -R'_0\lambda u \quad y' = -R'_0\lambda v \quad.$$ (E.63)

For $R_{00}$ and $R'_{00}$ (the direct beam) the coordinates are $u=v=0$ and according to (E.63) $x'=y'=0$. The direct beam is therefore in the z-axis and $R_{00},R'_{00}$ can be written as

$$R_{00} \equiv D$$

$$R'_{00} \equiv D'$$ (E.64)

Together with (E.14) and (E.64) the square of the fraction in (E.62) gives

$$\left(\frac{R_{00}R'_{00}}{R_0R'_0}\right)^2 = \frac{D^2D'^2}{(x^2+y^2+D^2)(x'^2+y'^2+D'^2)}$$

$$= \frac{1}{(\frac{x^2+y^2}{D^2}+1)(\frac{x'^2+y'^2}{D'^2}+1)} \Rightarrow_{(x=y=0)}$$

$$\Rightarrow \left(\frac{R_{00}R'_{00}}{R_0R'_0}\right)^2 = \frac{1}{\frac{x'^2+y'^2}{D'^2}+1}$$ (E.65)

Therefore this fraction can be set to unity if $\vert x'\vert \ll D' \wedge \vert y'\vert \ll D'$. This assumption yields finally for the intensity behind a rectangular aperture

$$I'(x',y') = I'(0,0) \sinc ^2\left(\frac{-2\pi x' \tilde{x_0}}{R'_0\lambda}\right) \sinc ^2\left(\frac{-2\pi y'\tilde{y_0}}{R'_0\lambda}\right)$$

$$\cong I'(0,0) \sinc ^2\left(\frac{2\pi x' \tilde{x_0}}{D'\lambda}\right) \sinc ^2\left(\frac{2\pi y'\tilde{y_0}}{D'\lambda}\right)$$ (E.66)

Figure E.7: Comparison of different intensity distributions after diffraction at a rectangular aperture (a) square aperture (b) detail of square aperture (c) rectangular aperture with $\tilde{x_0}=2\tilde{y_0}$ (d) rectangular aperture with $\tilde{y_0}=2\tilde{x_0}$