Skip to content

Fourier Optics

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.

15.2 The Abbe Theory of the Microscope

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.

15.3 Spatial Filtering

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.
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.