Fourier Optics

In the Fraunhofer limit, the diffraction pattern is the Fourier transform of the aperture function:

$E(\theta_x, \theta_y) \propto \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} t(x,y)\, e^{-i(k_x x + k_y y)}\,dx\,dy$

Where $t(x, y)$ is the transmission function of the aperture, and $k_x = k\sin\theta_x$ $k_y = k\sin\theta_y$.

A lens placed one focal length after the aperture produces the Fraunhofer pattern at its back focal plane, performing an optical Fourier transform.

If the aperture is a product $t(x, y) = t_1(x, y) \cdot t_2(x, y)$ the diffraction pattern is the convolution of their individual transforms:

$\mathcal{F}\{t_1 \cdot t_2\} = \mathcal{F}\{t_1\} * \mathcal{F}\{t_2\}$

Where $*$ denotes convolution. This explains, for example, why the double-slit pattern with finite slit width is the product of a sinc function (single slit) and a cosine-squared (double slit).

Abbe showed that a lens images by collecting the diffracted orders and recombining them. The resolution limit arises because high spatial frequencies (large diffraction angles) are lost if they fall outside the lens aperture.

The minimum resolvable spatial frequency is:

$f_{\mathrm{max}} = \frac{2\mathrm{NA}}{\lambda}$

Where $\mathrm{NA} = n\sin\theta_{\mathrm{max}}$ is the numerical aperture.

The optical transfer function (OTF) characterizes how an imaging system transmits spatial frequencies. It is the Fourier transform of the point spread function (PSF):

$H(f_x, f_y) = \mathcal{F}\{h(x, y)\}$

The modulation transfer function (MTF) is the magnitude $|H(f_x, f_y)|$. For a diffraction-limited system with a circular aperture, the MTF is:

$\mathrm{MTF}(f) = \frac{2}{\pi}\left[\arccos\left(\frac{f}{f_c}\right) - \frac{f}{f_c}\sqrt{1 - \left(\frac{f}{f_c}\right)^2}\right]$

for $f \leq f_c$, where $f_c = 2\mathrm{NA}/\lambda$ is the cutoff frequency.

The 4f system is a standard optical setup for spatial filtering, consisting of two identical lenses separated by twice their focal length $f$:

1. Input plane (front focal plane of first lens)
2. Fourier plane (common focal plane between lenses, where a spatial filter is placed)
3. Output plane (back focal plane of second lens)

If $U_{\mathrm{in}}(x, y)$ is the input field and $F(u, v)$ is the filter at the Fourier plane, the output field is:

$U_{\mathrm{out}}(x, y) = \mathcal{F}^{-1}\{F(u, v) \cdot \mathcal{F}\{U_{\mathrm{in}}\}\}$

This is a convolution: $U_{\mathrm{out}} = U_{\mathrm{in}} * \mathcal{F}^{-1}\{F\}$.

Spatial filtering modifies the Fourier spectrum of an image:

Low-pass filter. Block high spatial frequencies to remove noise or sharp edges. A circular aperture of radius $a$ at the Fourier plane passes only frequencies $f < a/(\lambda f)$.

High-pass filter. Block low spatial frequencies to enhance edges. A small opaque stop at the center of the Fourier plane removes the DC component.

Band-pass filter. Select a specific range of spatial frequencies to reveal periodic structures or specific texture scales.

Example 7.1 (Removing raster lines). A printed image has periodic raster lines at frequency $f_0$. A low-pass filter with cutoff $f_c < f_0$ removes the raster while preserving the image content.

Phase contrast microscopy converts phase variations in a transparent specimen into intensity variations, making otherwise invisible structures visible.

Zernike phase contrast works by shifting the phase of the undiffracted (zero-order) light by $\pi/2$ relative to the diffracted light. This is achieved with a phase ring at the Fourier plane that alters the phase of the central spot.

If the input field is $U_{\mathrm{in}}(x) = e^{i\phi(x)} \approx 1 + i\phi(x)$ (for small phase $\phi$), the Fourier plane has a DC term $\delta(u)$ and diffracted terms $\mathcal{F}\{\phi\}$. The phase ring multiplies the DC term by $e^{i\pi/2}$, giving:

$U_{\mathrm{out}}(x) \approx i + i\phi(x) \quad \Rightarrow \quad I(x) \approx 1 + 2\phi(x)$

The intensity is linearly proportional to the phase, revealing transparent structures.

Holography records both amplitude and phase of a wavefront by interfering it with a reference beam. The hologram is the intensity pattern:

$I(x, y) = |U_{\mathrm{obj}} + U_{\mathrm{ref}}|^2 = |U_{\mathrm{obj}}|^2 + |U_{\mathrm{ref}}|^2 + U_{\mathrm{obj}}^* U_{\mathrm{ref}} + U_{\mathrm{obj}} U_{\mathrm{ref}}^*$

When illuminated by the reference beam, the third term reconstructs the original object wavefront, creating a three-dimensional image. The fourth term produces a conjugate image.

Example 7.2 (Gabor holography). In in-line holography, the object and reference beams share the same axis. This requires a small, sparse object so the twin images separate.

Application 1: Image Deconvolution. If the PSF $h(x, y)$ is known, the original image can be recovered by inverse filtering:

$U_{\mathrm{in}} = \mathcal{F}^{-1}\{\mathcal{F}\{U_{\mathrm{out}}\} / H(f_x, f_y)\}$

Wiener filtering adds a regularization term to handle noise.

Application 2: Matched Filtering. To detect a known pattern $g(x, y)$ in an image, use a filter $F(u, v) = \mathcal{F}\{g\}^*$ at the Fourier plane. The output shows correlation peaks at locations where the pattern appears.

Application 3: Optical Pattern Recognition. VanderLugt filters store complex filter functions holographically for real-time pattern recognition.

Problem 1. Find the Fraunhofer pattern of a sinusoidal amplitude grating: $t(x) = (1 + m\cos(2\pi f_0 x))/2$.

Solution. The Fourier transform of $t(x)$ is:

$\mathcal{F}\{t\}(u) = \frac{1}{2}\delta(u) + \frac{m}{4}\delta(u - f_0) + \frac{m}{4}\delta(u + f_0)$

The diffraction pattern consists of a central order at $u = 0$ and two side orders at $u = \pm f_0$, with intensity ratio $I_{\pm}/I_0 = m^2/4$. $\blacksquare$

Problem 2. In a 4f system with $f = 20$ cm and $\lambda = 500$ nm, what filter radius $a$ is needed to pass spatial frequencies up to 50 cycles/mm?

Solution. Frequency $f_x = u/(\lambda f)$, so $u = f_x \lambda f = 50 \times 10^3 \times 500 \times 10^{-9} \times 0.2 = 5$ mm. A circular aperture of radius $a = 5$ mm passes all frequencies $f_x \leq 50$ cycles/mm. $\blacksquare$