E.1 Linear Approximation

For the following discussion, Figure E.1 shows the situation.

Figure E.1: Standard geometry of theory of diffraction

The aperture is positioned in the $\tilde{x}\tilde{y}$ plane, which is normal to the z-axis and at the position of point $\tilde{O}$. The point source is positioned in the $xy$ plane which is parallel to the $\tilde{x}\tilde{y}$ plane in a distance D in the negative z direction. The coordinates of the point source are therefore $(x,y,-D)$. The projection point P' with the coordinates $(x',y',D')$ is in the projection plane $x'y'$, which is parallel to the aperture plane $\tilde{x}\tilde{y}$ also. The positions $\tilde{P}$ in the aperture are defined by the coordinates $(\tilde{x},\tilde{y},0)$. The directions of the incoming and outgoing light waves are given with respect to the center of the aperture plane (see Figure E.2).

Figure E.2: Viewing angles of source point and projection point as seen from the center of the aperture.

The directional vectors of the light waves form the angles $\Theta_x$, $\Theta_y$ and $\Theta_z$ with their respective axes x, y and z. Here we define the functions:

$$\alpha = \cos \Theta_x = \frac{-x}{R_0} = \sin \Theta$$ (E.1)

$$\beta = \cos \Theta_y = \frac{-y}{R_0}$$ (E.2)

$$\gamma = \cos \Theta_z = \sqrt{1-\alpha^2-\beta^2} = \frac{D}{R_0}$$ (E.3)

$$\alpha' = \cos \Theta'_x = \frac{-x'}{R'_0} = \sin \Theta'$$ (E.4)

$$\beta' = \cos \Theta'_y = \frac{-y'}{R'_0}$$ (E.5)

$$\gamma' = \cos \Theta'_z = \sqrt{1-\alpha'^2-\beta'^2} = \frac{D'}{R'_0}$$ (E.6)

$$\Theta = \frac{\pi}{2}-\Theta_x$$ (E.7)

$$\Theta' = \frac{\pi}{2}-\Theta'_x$$ (E.8)

If one defines the transmission function:

$$\tilde{\tau}(\tilde{x},\tilde{y})= \left\{\begin{array}{r@{\quad:... ...\sigma_0} 0 & (\tilde{x},\tilde{y}) \in \tilde{\sigma_c} \end{array} \right.$$ (E.9)

with $\sigma=\sigma_0\vee\sigma_c$ and $\tilde{\sigma_0}$ and $\tilde{\sigma_c}$ being the aperture area which are transmitting and not transmitting light respectively. The resulting electric field on the projection side after diffraction at an aperture given by the transmission function $\tilde{\tau}$ yields

$$E'=C \int\!\!\!\int\limits_{\tilde{\sigma}}\tilde{\tau}\tilde{E}\frac{e^{-ikR'}}{R'} d\tilde{\sigma}$$ (E.10)

taking the solution of a spherical wave for diffraction of a planar wave at an infinite small aperture opening

$$\tilde{E}=\frac{A}{R} e^{i (\omega t - kR)}$$ (E.11)

with A as the Amplitude of the incident planar wave. Equation (E.10) yields

$$E'=C A e ^ {i \omega t} \int\limits_{\tilde{\sigma}}\tau\frac{e^{-ik(R+R')}}{RR'} d\tilde{\sigma}$$ (E.12)

The lengths $R$ and $R'$ are given by

$$R = \sqrt{(\tilde{x}-x)^2+(\tilde{y}-y)^2+D^2}$$

$$R' = \sqrt{(x'-\tilde{x})^2+(y'-\tilde{y})^2+D'^2}$$ (E.13)

For the far field approximation $R$ and $R'$ are linear functions of $\tilde{x}$ and $\tilde{y}$. Therefore we calculate the series expansion of $R$ and $R'$ around $R_0$ and $R'_0$, with $R_0$ and $R'_0$ given by

$$R_0 = \sqrt{x^2+y^2+D^2}$$

$$R'_0 = \sqrt{x'^2+y'^2+D'^2}$$ (E.14)

These distances are no functions of $\tilde{x}$ and $\tilde{y}$. Using (E.14) in (E.13) yields

$$R=\sqrt{R_0^2-2(x\tilde{x}-y\tilde{y})+\tilde{x}^2+\tilde{y}^2}=R... ...1-\frac{2}{R_0^2}(x\tilde{x}+y\tilde{y})+\frac{\tilde{x}^2+\tilde{y}^2}{R_0^2}}$$ (E.15)

assuming that the second and third term is much smaller than the first one, the squareroot in (E.15) can be written as

$$R=R_0\sqrt{1+\varepsilon}\simeq R_0(1+\frac{1}{2}\varepsilon-\frac{1}{8}\varepsilon^2)$$ (E.16)

and

$$\varepsilon=-\frac{2}{R_0^2}(x\tilde{x}+y\tilde{y})+\frac{\tilde{x}^2+\tilde{y}^2}{R_0^2}$$ (E.17)

Finally R is calculated by the series expansion to the second order in $\tilde{x}$ and $\tilde{y}$

$$R \simeq R_0(1-\frac{x\tilde{x}+y\tilde{y}}{R_0^2}+\frac{\tilde{x}^2+\tilde{y}^2}{2 R_0^2}-\frac{(x\tilde{x}+y\tilde{y})^2}{2 R_0^4})$$ (E.18)

This result is only valid if $\tilde{x} \ll R_0$ and $\tilde{y} \ll R_0$. This holds especially for the far field where $R_0$ is big in relation to the aperture size

$$\vert\tilde{x}\vert\ll\sqrt{R_0\lambda}$$

$$\vert\tilde{y}\vert\ll\sqrt{R_0\lambda}$$ (E.19)

Therefore the third term in (E.18) is negligible and this gives

$$R \simeq R_0-(\frac{x}{R_0})\tilde{x}-(\frac{y}{R_0})\tilde{y}$$

$$R' \simeq R'_0-(\frac{x'}{R'_0})\tilde{x}-(\frac{y'}{R'_0})\tilde{y}$$ (E.20)

Using the approximation

$$(RR')^{-1} \simeq (R_0R'_0)^{-1}$$ (E.21)

which is valid for the far field, and the transformation of the coordinates

$$u \equiv -(\frac{x}{R_0}+\frac{x'}{R'_0})\frac{1}{\lambda}=\frac{(\alpha-\alpha')}{\lambda}=\frac{\cos{\theta_x}-\cos{\theta'_x}}{\lambda}$$

$$v \equiv -(\frac{y}{R_0}+\frac{y'}{R'_0})\frac{1}{\lambda}=\frac{(\beta-\beta')}{\lambda}=\frac{\cos{\theta_y}-\cos{\theta'_y}}{\lambda}$$ (E.22)

the total optical path can be written as

$$R+R' = (R_0+R'_0)+(u\tilde{x}+v\tilde{y})\lambda$$ (E.23)

substituting this optical path in the phase factor of the diffraction integral in (E.12) yields

$$e^{i \omega t} e^{-i k (R+R')} = e^{i \phi_0} e^{-i 2 \pi (u\tilde{x}+v\tilde{y})}$$ (E.24)

with

$$\phi_0 = \omega t - k (R_0+R'_0)$$ (E.25)

$\phi_0$ is not a function of $\tilde{x}$ and $\tilde{y}$. The coordinates $(x,y)$ of the source and $(x',y')$ of the projection point are now included in $(u,v)$. Finally the electrical field at the projection point P' emanating from the source point P and diffracted at the aperture with the transmission function $\tilde{\tau}(\tilde{x},\tilde{y})$ can be given as

$$E'(u,v) = \frac{C A e^{i \phi_0}}{R_0 R'_0} \int\!\!\!\int\limits... ...tilde{x},\tilde{y}) e^{i 2 \pi (u\tilde{x}+v\tilde{y})} d\tilde{x} d\tilde{y}$$ (E.26)

The integral of (E.26) is exactly the FOURIER transform of the transmission function $\tilde{\tau}(\tilde{x},\tilde{y})$

$$T(u,v) \equiv \int\!\!\!\int\limits_{-\infty}^{+\infty} \tilde{\t... ...tilde{x},\tilde{y}) e^{i 2 \pi (u\tilde{x}+v\tilde{y})} d\tilde{x} d\tilde{y}$$ (E.27)

This important result implies, that under the given assumptions, the electrical field distribution after a diffracting aperture is proportional to the FOURIER transform of the transmission function of the aperture.
If as shown in Figure E.3 the points $P$,$\tilde{O}$ and $P'$ are on a line, then $(u,v)=(0,0)$ and (E.26) together with (E.27) reduces to

$$E'(0,0)=\frac{C A e^{i \phi_{00}}}{R_{00}R'_{00}} T(0,0)$$ (E.28)

Figure E.3: Direct light propagation through aperture

With (E.28), (E.26) can be normalized to $E'(0,0)$

$$E'(u,v)=E'(0,0)\left[\frac{R_{00}R'_{00}}{R_0R'_0}\right]\frac{T(u,v)}{T(0,0)}e^{i(\phi_0-\phi_{00})}$$ (E.29)