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Fourier Optics | Physics - Wyatt's Notes

7.1 Fraunhofer Diffraction as a Fourier Transform

Section titled “7.1 Fraunhofer Diffraction as a Fourier Transform”

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

E(θx,θy)t(x,y)ei(kxx+kyy)dxdyE(\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)t(x, y) is the transmission function of the aperture, and kx=ksinθxk_x = k\sin\theta_x ky=ksinθyk_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)=t1(x,y)t2(x,y)t(x, y) = t_1(x, y) \cdot t_2(x, y) the diffraction pattern is the convolution of their individual transforms:

F{t1t2}=F{t1}F{t2}\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:

fmax=2NAλf_{\mathrm{max}} = \frac{2\mathrm{NA}}{\lambda}

Where NA=nsinθmax\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(fx,fy)=F{h(x,y)}H(f_x, f_y) = \mathcal{F}\{h(x, y)\}

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

MTF(f)=2π[arccos(ffc)ffc1(ffc)2]\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 ffcf \leq f_c, where fc=2NA/λf_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 ff:

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

Uout(x,y)=F1{F(u,v)F{Uin}}U_{\mathrm{out}}(x, y) = \mathcal{F}^{-1}\{F(u, v) \cdot \mathcal{F}\{U_{\mathrm{in}}\}\}

This is a convolution: Uout=UinF1{F}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 aa at the Fourier plane passes only frequencies f<a/(λf)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 f0f_0. A low-pass filter with cutoff fc<f0f_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 π/2\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 Uin(x)=eiϕ(x)1+iϕ(x)U_{\mathrm{in}}(x) = e^{i\phi(x)} \approx 1 + i\phi(x) (for small phase ϕ\phi), the Fourier plane has a DC term δ(u)\delta(u) and diffracted terms F{ϕ}\mathcal{F}\{\phi\}. The phase ring multiplies the DC term by eiπ/2e^{i\pi/2}, giving:

Uout(x)i+iϕ(x)I(x)1+2ϕ(x)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)=Uobj+Uref2=Uobj2+Uref2+UobjUref+UobjUrefI(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)h(x, y) is known, the original image can be recovered by inverse filtering:

Uin=F1{F{Uout}/H(fx,fy)}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)g(x, y) in an image, use a filter F(u,v)=F{g}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+mcos(2πf0x))/2t(x) = (1 + m\cos(2\pi f_0 x))/2.

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

F{t}(u)=12δ(u)+m4δ(uf0)+m4δ(u+f0)\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=0u = 0 and two side orders at u=±f0u = \pm f_0, with intensity ratio I±/I0=m2/4I_{\pm}/I_0 = m^2/4. \blacksquare

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

Solution. Frequency fx=u/(λf)f_x = u/(\lambda f), so u=fxλf=50×103×500×109×0.2=5u = f_x \lambda f = 50 \times 10^3 \times 500 \times 10^{-9} \times 0.2 = 5 mm. A circular aperture of radius a=5a = 5 mm passes all frequencies fx50f_x \leq 50 cycles/mm. \blacksquare

flowchart TD
A[7_Fourier Optics] --> B[Key Concepts]
A --> C[Core Principles]
A --> D[Practical Applications]
B --> E[Fundamental definitions]
C --> F[Design patterns]
D --> G[Real-world usage]

Fourier optics treats lenses as analog computers that perform Fourier transforms at the speed of light. A lens converts spatial patterns into their frequency components at its focal plane. The convolution theorem explains why complex apertures produce predictable patterns: multiplying apertures convolves their spectra. This is why a grating produces discrete spots while a single slit produces a continuous spread. The optical transfer function describes how well an imaging system preserves different spatial frequencies, like an equaliser for images.

Mistake 1: Confusing the convolution theorem for products and convolutions If the aperture function is a product t=t1t2t = t_1 \cdot t_2, the diffraction pattern is the convolution of their Fourier transforms: F{t1t2}=F{t1}F{t2}\mathcal{F}\{t_1 \cdot t_2\} = \mathcal{F}\{t_1\} * \mathcal{F}\{t_2\}. Students often apply the theorem backwards, assuming that a product in real space gives a product in Fourier space. The correct statement is that convolution in real space corresponds to multiplication in Fourier space.

Mistake 2: Assuming a lens performs an exact Fourier transform without considering the paraxial approximation A thin lens adds a quadratic phase factor exp(ik(x2+y2)/(2f))\exp(-ik(x^2+y^2)/(2f)). The Fourier transform relationship at the back focal plane holds only in the paraxial approximation (small angles). For high-NA lenses or large off-axis distances, aberrations and non-paraxial effects distort the transform. Students sometimes assume the Fourier relationship is exact for all configurations.

  • Diffraction: Establishes the Huygens-Fresnel principle and Fraunhofer diffraction integral that the Fourier transform relationship formalises.
  • Coherence: The spatial and temporal coherence of the source determines whether the Fourier transform relationship between aperture and far-field pattern holds.
  • Fourier Optics: Provides the mathematical framework for diffraction as Fourier decomposition, including the convolution theorem and Airy pattern.

Mistake 3: Forgetting that the 4f system inverts the image The second lens in a 4f system performs another Fourier transform, which introduces a coordinate inversion: Uout(x,y)Uin(x,y)U_{\text{out}}(x,y) \propto U_{\text{in}}(-x,-y) when no filter is present. Students sometimes forget this inversion when designing spatial filtering experiments, leading to unexpectedly mirrored output images.

Fourier optics treats lenses as analog computers that perform Fourier transforms at the speed of light. A lens converts spatial patterns into their frequency components at its focal plane. The convolution theorem explains why complex apertures produce predictable patterns: multiplying apertures convolves their spectra. This is why a grating produces discrete spots while a single slit produces a continuous spread. The optical transfer function describes how well an imaging system preserves different spatial frequencies, like an equalizer for images.