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E.2 Circular Aperture

Figure E.4 shows the aperture plane with the coordinates $ \tilde{r}$ and $ \tilde{\phi}$.
Figure E.4: Coordinate system in a circular aperture
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The transmission function of the aperture is

$\displaystyle \tilde{\tau}(\tilde{r},\tilde{\phi})= \left\{\begin{array}{r@{\qu...
... & \tilde{r} < \tilde{r_0}  0 & \tilde{r} \ge \tilde{r_0} \end{array} \right.$ (E.30)

and

$\displaystyle \tilde{x}=\tilde{r} \cos{\tilde{\phi}}$    
$\displaystyle \tilde{y}=\tilde{r} \sin{\tilde{\phi}}$ (E.31)

Using polar coordinates for the transformed coordinates $ u$ and $ v$ defined in the previous section

$\displaystyle u=\rho \cos{\phi'}$    
$\displaystyle v=\rho \sin{\phi'}$ (E.32)

gives

$\displaystyle \rho = \sqrt{u^2+v^2}=\frac{1}{\lambda}\sqrt{\left(\frac{x}{R_0}+\frac{x'}{R'_0}\right)^2+\left(\frac{y}{R_0}+\frac{y'}{R'_0}\right)^2}$ (E.33)

We are now reviewing the case with the source being on the optical axis of the aperture ($ x = y = 0$) which yields

$\displaystyle \rho = \frac{1}{\lambda}\frac{r'}{R'_0}=\frac{\sin{\theta'_z}}{\lambda}$ (E.34)

where

$\displaystyle r'=\sqrt{x'^2+y'^2}$ (E.35)

is the radial distance of the projection point from the z-axis in the projection plane. $ \theta'_z$ is the angle of the projection point seen from the aperture (refer to Figure E.2). The FOURIER transform of $ \tilde{\tau}$ can be calculated from (E.27). The differential area element $ d\tilde{\sigma}$ is in polar coordinates

$\displaystyle d\tilde{\sigma} = \tilde{r} d\tilde{\phi} d\tilde{r}$    

as depicted by Figure E.4 The exponential factor in the integral of (E.27) in the above defined coordinate system gives

$\displaystyle \exp[-i2\pi(\tilde{u}x+\tilde{v}y)] = \exp[-i2\pi\rho\tilde{r}(\cos{\phi'}\cos{\tilde{\phi}}+\sin{\phi'}\sin{\tilde{\phi}})]$    
$\displaystyle = \exp[-i2\pi\rho\tilde{r}\cos(\tilde{\phi}-\phi')]$    

Therefore (E.27) is in polar coordinates

$\displaystyle T(\rho,\phi') \cong \int\limits_0^{\tilde{r_0}} \tilde{r} \int\li...
...i} \exp[-i 2 \pi \tilde{r} \cos(\tilde{\phi}-\phi')] d\tilde{\phi} d\tilde{r}$ (E.36)

The inner integrand of this solution is the well known BESSEL function of zeroth order and is defined as

$\displaystyle J_0(w) = \frac{1}{2\pi}\int\limits_0^{2\pi} \exp[-iw\cos(\tilde{\phi}-\phi')] d\tilde{\phi}$ (E.37)

by using this definition with (E.27) the FOURIER transform of a circular aperture gives

$\displaystyle T(\rho) = 2\pi\int\limits_0^{\tilde{r_0}} \tilde{r}J_0(2\pi\rho\tilde{r}) d\tilde{r}$ (E.38)

which is independent of $ \rho'$ as a consequence of the rotational symmetry of the aperture. Applying a coordinate transformation $ w' = 2\pi\rho\tilde{r}$ to the integral of (E.38) yields

$\displaystyle T(\rho)=\frac{1}{2\pi\rho^2}\int\limits_0^{2\pi\rho\tilde{r_0}} w'J_0(w') dw'$ (E.39)

To solve this integral one can use a relation between BESSEL functions of different order

$\displaystyle w\frac{dJ_n}{dw}+nJ_n=wJ_{n-1} $ (E.40)

integrating (E.40) with $ n=1$ gives

$\displaystyle w\frac{J_1}{dw}+J_1 = w J_0$    
$\displaystyle \frac{d(w J_1)}{dw} = w J_0$    
$\displaystyle w J_1(w) = \int\limits_0^w w' J_0(w') dw'$ (E.41)

Using this result in (E.39) gives finally

$\displaystyle T(\rho) = \frac{1}{2\pi\rho^2}2\pi\rho\tilde{r_0}J_1(2\pi\rho\tilde{r_0})$ (E.42)

or

$\displaystyle T(\rho) = 2\pi\tilde{r_0}^2\frac{J_1(2\pi\rho\tilde{r_0})}{2\pi\rho\tilde{r_0}}$ (E.43)

In the center of the diffraction pattern with $ \rho = 0$ the properties of the BESSEL function give

$\displaystyle \lim_{w \to 0} \left(\frac{J_1(w)}{w}\right)=\frac{1}{2}$    

yielding

$\displaystyle T(0) = \pi \tilde{r_0}^2$ (E.44)

Taking (E.29) the electrical field of the diffraction pattern behind a circular aperture is finally

$\displaystyle E'(\rho)=E'(0)\left(\frac{2J_1(2\pi\rho\tilde{r_0})}{2\pi\rho\tilde{r_0}}\right)\left(\frac{D'}{R'_0}\right)e^{i(\phi_0-\phi_{00})}$ (E.45)

The average optical intensity is proportional to the square of the absolute electrical field. Therefore the intensity for diffraction behind a circular aperture is

$\displaystyle I'(\rho) = I'(0) \left(\frac{2J_1(2\pi\rho\tilde{r_0})}{2\pi\rho\tilde{r_0}}\right)^2\left(\frac{D'}{R'_0}\right)^2$ (E.46)

with

$\displaystyle \rho = \frac{\sin \theta'_z}{\lambda} = \frac{r'}{R'_0\lambda}$ (E.47)

By using the (E.14) and (E.35) in (E.46) one obtains

$\displaystyle I'(\rho) = I'(0) \left(\frac{2J_1(2\pi\rho\tilde{r_0})}{2\pi\rho\tilde{r_0}}\right)^2\frac{1}{1+\left(\frac{r'}{D'}\right)^2}$ (E.48)

Substituting $ r'$ from (E.47) into above equation yields

$\displaystyle I'(\rho) = I'(0) \left(\frac{2J_1(2\pi\rho\tilde{r_0})}{2\pi\rho\tilde{r_0}}\right)^2\left(1-\rho^2\lambda^2\right)$ (E.49)

With the substitution

$\displaystyle x=2\pi\rho\tilde{r_0}=\frac{2\pi\tilde{r_0}r'}{R'_0\lambda}$ (E.50)

(which is dimensionless) the intensity is finally

$\displaystyle I'(\rho) = I'(0) \left(\frac{2J_1(x)}{x}\right)^2\left(1-x^2\left[\frac{\lambda}{2\pi\tilde{r_0}}\right]^2\right)$ (E.51)

The (for most cases valid) assumption that the aperture radius is much bigger than the wavelength ( $ \lambda \ll \tilde{r_0}$) the intensity is

$\displaystyle I'(\rho) = I'(0) \left(\frac{2J_1(x)}{x}\right)^2$ (E.52)

Which is the well known intensity distribution for a circular aperture given in many textbooks. The variable $ x$ reduces to

$\displaystyle x=\frac{2\pi\tilde{r_0}r'}{D'\lambda}$ (E.53)

for the assumption mentioned above.
Figure E.5: AIRY-disk with rings
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{\includegraphics{figures/airy_disk_circular.ps}}}


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Next: E.3 Rectangular Aperture Up: E. Diffraction in Far Previous: E.1 Linear Approximation

R. Minixhofer: Integrating Technology Simulation into the Semiconductor Manufacturing Environment