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

The transmission function of the aperture is

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

and

$$\tilde{x}=\tilde{r} \cos{\tilde{\phi}}$$

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

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

$$u=\rho \cos{\phi'}$$

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

gives

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

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

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

where

$$r'=\sqrt{x'^2+y'^2}$$ (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

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

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

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

$$= \exp[-i2\pi\rho\tilde{r}\cos(\tilde{\phi}-\phi')]$$

Therefore (E.27) is in polar coordinates

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

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

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

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

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

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

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

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

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

$$w\frac{J_1}{dw}+J_1 = w J_0$$

$$\frac{d(w J_1)}{dw} = w J_0$$

$$w J_1(w) = \int\limits_0^w w' J_0(w') dw'$$ (E.41)

Using this result in (E.39) gives finally

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

or

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

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

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

yielding

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

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

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

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

with

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

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

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

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

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

With the substitution

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

(which is dimensionless) the intensity is finally

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

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

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

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

for the assumption mentioned above.

Figure E.5: AIRY-disk with rings