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Fourier Optics

15.1 Fraunhofer Diffraction as a Fourier Transform

Section titled “15.1 Fraunhofer Diffraction as a Fourier Transform”

In the Fraunhofer (far-field) limit, the diffraction pattern of an aperture with transmission function t(x,y)t(x, y) is the Fourier transform:

U(x",y")=eikziλzeik(x2+y2)/(2z)t(x,y)eik(xx+yy)/zdxdyU(x", y") = \frac{e^{ikz}}{i\lambda z}\,e^{ik(x'^2 + y'^2)/(2z)}\iint t(x, y)\,e^{-ik(xx' + yy')/z}\,dx\,dy

Where (x,y)(x', y') are coordinates in the observation plane at distance zz from the aperture.

Defining spatial frequencies fx=x/(λz)f_x = x'/(\lambda z), fy=y/(λz)f_y = y'/(\lambda z):

U(fx,fy)F{t(x,y)}(fx,fy)U(f_x, f_y) \propto \mathcal{F}\{t(x,y)\}(f_x, f_y)

This correspondence between diffraction and Fourier transforms is the foundation of Fourier optics and has profound implications for image processing and optical information processing.

Ernst Abbe (1873) showed that a microscope forms an image by taking two Fourier transforms: the objective lens performs the first Fourier transform (creating the diffraction pattern at its back focal plane), and the eyepiece (or tube lens) performs the inverse transform.

Resolution limit: The finest spatial frequency that can pass through the objective is:

f_{\max} = \frac{\text{NA}{\lambda}}

Where NA=nsinθ\text{NA} = n\sin\theta is the numerical aperture. The minimum resolvable distance (Abbe limit):

dmin=λ2NAd_{\min} = \frac{\lambda}{2\,\text{NA}}

For green light (λ=550\lambda = 550 nm) and NA = 1.4 (oil immersion): dmin196d_{\min} \approx 196 nm.

Since the back focal plane of a lens contains the spatial frequency spectrum of the input, placing a mask (spatial filter) in this plane modifies the image:

  • Low-pass filter: Blocks high spatial frequencies \to smooths the image, removes fine detail
  • High-pass filter: Blocks low frequencies \to enhances edges, removes uniform background
  • Phase contrast microscopy: (Zernike, 1942) Adds a π/2\pi/2 phase shift to the undiffracted (DC) component, converting phase variations into intensity variations. This makes transparent biological specimens visible without staining.
Optical conceptFourier correspondenceApplication
Aperture functionInput signal t(x,y)t(x,y)Diffraction pattern
Fraunhofer patternFourier transform F{t}\mathcal{F}\{t\}Far-field imaging
Lens back focal planeFourier planeSpatial filtering
Image planeInverse Fourier transform4f imaging system
Optical transfer funcNormalised FT of PSFResolution characterisation
  • Confusing Fraunhofer and Fresnel diffraction. Fraunhofer diffraction applies in the far field (za2/λz \gg a^2/\lambda where aa is the aperture size). Fresnel diffraction (near field) uses a quadratic phase factor and is not simply a Fourier transform.
  • Forgetting the quadratic phase factor in the Fresnel regime. The Fraunhofer integral approximation drops the quadratic phase term, but it is present in the full Fresnel diffraction integral.
  • Assuming the Abbe limit is the only resolution limit. The Abbe limit assumes coherent illumination; the Rayleigh criterion (for resolved point sources) gives d=0.61λ/NAd = 0.61\lambda/\text{NA}, slightly different.
  • Thinking spatial filtering only removes noise. Spatial filtering can also introduce artifacts (ringing from hard-edged low-pass filters, edge enhancement exaggeration from high-pass filters).

Problem 1. A 4f imaging system has a lens with focal length f=50f = 50 cm and aperture diameter D=2D = 2 cm. The input is illuminated with λ=633\lambda = 633 nm. What is the cutoff spatial frequency?

Solution. The cutoff frequency is determined by the lens aperture acting as a low-pass filter. The maximum spatial frequency that passes through the system is fmax=D/(2λf)f_{\max} = D/(2\lambda f): fmax=0.02/(2×633×109×0.5)=31,600f_{\max} = 0.02/(2 \times 633\times 10^{-9} \times 0.5) = 31,600 cycles/m. Finer details in the input are blocked, giving a minimum feature size of 1/fmax31.6 μ1/f_{\max} \approx 31.6\ \mum. \blacksquare

Problem 2. Derive the point spread function (PSF) of a circular aperture of diameter DD.

Solution. The amplitude PSF is the Fourier transform of the aperture function (a circle of diameter DD). This gives the Airy pattern: I(r)=I0[2J1(kDr/(2f))/(kDr/(2f))]2I(r) = I_0[2J_1(kDr/(2f))/(kDr/(2f))]^2, where J1J_1 is the Bessel function of the first kind. The first zero occurs at r=1.22λf/Dr = 1.22\lambda f/D, which is the Rayleigh criterion. \blacksquare

  • Microscopy: Structured illumination microscopy (SIM) uses patterned illumination to encode high-frequency information, doubling resolution beyond the Abbe limit. Stimulated emission depletion (STED) microscopy breaks the diffraction barrier entirely.
  • Holography: Digital holography records the full complex field (amplitude and phase) using Fourier optics principles, enabling numerical refocusing and 3D imaging.
  • Astronomy: Adaptive optics corrects wavefront distortions in real time using Fourier optics concepts. Aperture synthesis in radio astronomy reconstructs images from sparse Fourier samples.
  • Optical computing: 4f correlators perform convolution operations optically at the speed of light, used in pattern recognition and optical neural networks.
Worked Example 15.1: Diffraction from a Grating

A diffraction grating with NN slits of width aa and spacing dd has transmission function:

t(x)=n=0N1rect ⁣(xnda)t(x) = \sum_{n=0}^{N-1}\text{rect}\!\left(\frac{x - nd}{a}\right)

The Fraunhofer pattern is:

I(θ)=I0(sinαα)2(sinNβsinβ)2I(\theta) = I_0\left(\frac{\sin\alpha}{\alpha}\right)^2\left(\frac{\sin N\beta}{\sin\beta}\right)^2

Where α=πasinθ/λ\alpha = \pi a\sin\theta/\lambda (single-slit envelope) and β=πdsinθ/λ\beta = \pi d\sin\theta/\lambda (multi-slit interference).

For N=5N = 5, d=3ad = 3a:

  • Principal maxima at β=mπ\beta = m\pi: sinθ=mλ/d\sin\theta = m\lambda/d
  • Between principal maxima: N2=3N - 2 = 3 secondary maxima
  • Width of principal maximum: Δθ=λ/(Ndcosθ)\Delta\theta = \lambda/(Nd\cos\theta)
  • Missing orders: when mm is a multiple of d/a=3d/a = 3 (i.e., 3rd, 6th, … Orders are suppressed by the single-slit zero)

The resolving power: R=mN=m×5R = mN = m \times 5.