Physics Wave & Modern Optics 100% Free Open Access
Chapter 7 • Theory & Derivations

Fourier Optics & Modern Diffraction Theory

2D Fourier transforms, convolution theorem, thin lens as an optical Fourier transformer, 4f spatial frequency filtering, and the Wiener-Khinchin theorem.

§7.1 Two-Dimensional Spatial Fourier Transforms and the Convolution Theorem

Fourier optics provides a unifying mathematical framework that treats optical imaging and diffraction through linear systems theory, where optical apertures and field distributions are decomposed into continua of spatial frequencies.

1. 2D Spatial Fourier Transform Pairs

Let $u(x, y)$ be the complex optical amplitude distribution in an aperture plane. Its two-dimensional Fourier transform $U(f_x, f_y)$ represents the spatial frequency spectrum of plane wave components: $$U(f_x, f_y) = \mathcal{F}\{u(x, y)\} = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} u(x, y) e^{-i 2\pi (f_x x + f_y y)} dx dy$$ where $f_x = \frac{k_x}{2\pi} = \frac{\sin \theta_x}{\lambda}$ and $f_y = \frac{k_y}{2\pi} = \frac{\sin \theta_y}{\lambda}$ are the spatial frequencies in cycles per unit length (e.g., $\text{lines/mm}$). The inverse Fourier transform reconstructs the original spatial distribution: $$u(x, y) = \mathcal{F}^{-1}\{U(f_x, f_y)\} = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} U(f_x, f_y) e^{i 2\pi (f_x x + f_y y)} df_x df_y$$

2. Key Fourier Transform Theorems in Optics

  • Linearity: $\mathcal{F}\{a u_1 + b u_2\} = a U_1 + b U_2$
  • Spatial Scaling (Similarity): $\mathcal{F}\{u(ax, by)\} = \frac{1}{|ab|} U\left(\frac{f_x}{a}, \frac{f_y}{b}\right)$. Narrowing an aperture spatially spreads its diffracted spectrum in frequency space.
  • Spatial Shift (Shift Theorem): $\mathcal{F}\{u(x - x_0, y - y_0)\} = U(f_x, f_y) e^{-i 2\pi (f_x x_0 + f_y y_0)}$
  • Parseval’s Energy Theorem: $$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} |u(x, y)|^2 dx dy = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} |U(f_x, f_y)|^2 df_x df_y$$ Total optical radiant energy is conserved identically between the physical aperture plane and the spatial frequency Fourier plane.

3. The Convolution Theorem in Optical Imaging

The two-dimensional spatial convolution of an input object $g(x,y)$ with an optical system impulse response (point spread function) $h(x,y)$ is: $$g_{\text{out}}(x, y) = g_{\text{in}}(x, y) * h(x, y) = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} g_{\text{in}}(\xi, \eta) h(x - \xi, y - \eta) d\xi d\eta$$ The **Convolution Theorem** proves that convolution in space is mathematically equivalent to simple algebraic multiplication in frequency space: $$G_{\text{out}}(f_x, f_y) = G_{\text{in}}(f_x, f_y) \cdot H(f_x, f_y)$$ where $H(f_x, f_y) = \mathcal{F}\{h(x, y)\}$ is the **Optical Transfer Function (OTF)** of the imaging system.

§7.2 The Thin Convex Lens as an Optical 2D Fourier Transformer

One of the most remarkable discoveries in physical optics is that an ordinary thin spherical convex glass lens performs an exact, instantaneous two-dimensional mathematical Fourier transform of any coherent light field.

1. Phase Transformation of a Thin Lens

A thin spherical lens of focal length $f$ and refractive index $n$ possesses a variable glass thickness profile: $$\Delta(x, y) = \Delta_0 - \frac{x^2 + y^2}{2} \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = \Delta_0 - \frac{x^2 + y^2}{2(n - 1)f}$$ When a plane wave traverses the lens, the spatially varying phase delay is: $$t_L(x, y) = e^{i k n \Delta(x,y)} e^{i k [\Delta_0 - \Delta(x,y)]} = e^{i k \Delta_0} \exp\left[ -i \frac{k}{2f} (x^2 + y^2) \right]$$ Dropping the constant phase factor $e^{i k \Delta_0}$, the **lens transmission function** is: $$t_L(x, y) = \exp\left[ -i \frac{k}{2f} (x^2 + y^2) \right]$$

2. Field in the Back Focal Plane

Place an input transparency with optical field distribution $u_0(x, y)$ in the front focal plane of the lens ($z = -f$). Applying the Fresnel diffraction integral from the input plane to the lens, multiplying by the lens transmission function $t_L(x, y)$, and propagating to the back focal plane ($z = +f$): The quadratic phase factors cancel completely, leaving: $$u_f(u, v) = \frac{1}{i \lambda f} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} u_0(x, y) \exp\left[ -i \frac{2\pi}{\lambda f} (u x + v y) \right] dx dy$$ This is the exact mathematical Fourier transform of the input object $u_0(x, y)$ evaluated at spatial frequencies: $$f_x = \frac{u}{\lambda f}, \quad f_y = \frac{v}{\lambda f}$$ Every spatial coordinate $(u, v)$ in the back focal plane corresponds to a specific spatial frequency of the input image.

§7.3 Spatial Frequency Filtering in the 4f Correlator and the Wiener-Khinchin Theorem

The Fourier transforming property of lenses enables optical computing and real-time spatial image processing using the classic **4f Optical Correlator**.

1. Architecture of the 4f System

The 4f system consists of two identical lenses $L_1$ and $L_2$ of focal length $f$ separated by distance $2f$:
  1. Input Plane $P_1$ ($z = 0$): Object transparency $u_{\text{in}}(x, y)$.
  2. Fourier Plane $P_2$ ($z = 2f$): Lens $L_1$ forms the optical Fourier transform $U(f_x, f_y)$. A physical spatial filter (mask) $H(f_x, f_y)$ is inserted here.
  3. Output Plane $P_3$ ($z = 4f$): Lens $L_2$ takes the inverse Fourier transform, synthesizing the filtered output image: $$u_{\text{out}}(x, y) = \mathcal{F}^{-1} \{ U(f_x, f_y) \cdot H(f_x, f_y) \}$$

2. Types of Optical Spatial Filters

  • Low-Pass Filter (Circular Pinhole Mask): Blocks high spatial frequencies ($f_r > f_c$). Removes high-frequency noise, speckle, and grain, producing a smooth image.
  • High-Pass Filter (Opaque Central Dot): Blocks the zero-order DC frequency ($f_x = f_y = 0$). Highlights sharp edges, boundaries, and high-frequency textural details while suppressing uniform background illumination.
  • Directional Filter (Slit Mask): Eliminates periodic grating lines oriented along specific angles (e.g., removing raster scanning lines from video images).

3. The Wiener-Khinchin Theorem in Optical Coherence

The mutual coherence function (autocorrelation) of an optical field $\Gamma(\tau)$ is: $$\Gamma(\tau) = \langle E^*(t) E(t + \tau) \rangle = \lim_{T \to \infty} \frac{1}{2T} \int_{-T}^T E^*(t) E(t + \tau) dt$$ The **Wiener-Khinchin Theorem** establishes that the power spectral density $S(\nu)$ of an optical field is the Fourier transform of its temporal autocorrelation function $\Gamma(\tau)$: $$S(\nu) = \int_{-\infty}^{\infty} \Gamma(\tau) e^{-i 2\pi \nu \tau} d\tau$$ $$\Gamma(\tau) = \int_{-\infty}^{\infty} S(\nu) e^{i 2\pi \nu \tau} d\nu$$ This theorem is the physical foundation of Fourier Transform Spectroscopy (FTS), where the emission spectrum of a star or chemical compound is recovered by Fourier transforming the interferogram measured by a Michelson interferometer.

📝 Chapter Worked Examples & Exercises

Complete derivations & analytical proofs
Hard Example 7.1: 4f Optical Low-Pass Spatial Filter Pinhole Diameter Calculation

A 4f optical spatial filtering system uses a laser of wavelength $\lambda = 632.8 ext{ nm}$ and two lenses of focal length $f = 250 ext{ mm}$. The input object is a photograph corrupted by high-frequency periodic grid noise with spatial frequency $f_{ ext{noise}} = 50.0 ext{ lines/mm}$. (a) Calculate the radial position $r_{ ext{noise}}$ of the noise spectral peaks in the Fourier transform plane. (b) What maximum pinhole aperture diameter $D_{ ext{filter}}$ should be placed in the Fourier plane to completely eliminate the grid noise while retaining image content up to $40.0 ext{ lines/mm}$?

Step 1: Calculate the radial position of noise peaks in the Fourier plane
$$r = f_{\text{spatial}} \lambda f$$ $$r_{\text{noise}} = (50.0 \times 10^3 \text{ m}^{-1}) \times (632.8 \times 10^{-9} \text{ m}) \times (0.250 \text{ m}) = 50.0 \times 10^3 \times 1.582 \times 10^{-7} = 7.91 \times 10^{-3} \text{ m} = 7.91 \text{ mm}$$

The noise components appear as bright diffraction spots at radius $7.91 ext{ mm}$ from the optical axis.

Step 2: Determine cutoff radius for 40 lines/mm signal
$$r_{\text{signal, max}} = (40.0 \times 10^3 \text{ m}^{-1}) \times (632.8 \times 10^{-9} \text{ m}) \times (0.250 \text{ m}) = 6.328 \times 10^{-3} \text{ m} = 6.328 \text{ mm}$$

The desired image frequencies extend out to radius $6.33 ext{ mm}$.

Step 3: Choose pinhole filter diameter
$$r_{\text{filter}} \in [6.33 \text{ mm}, 7.91 \text{ mm}] \implies \text{Choose } r_{\text{filter}} = 7.00 \text{ mm}$$ $$D_{\text{filter}} = 2 r_{\text{filter}} = 14.0 \text{ mm}$$

A pinhole diameter of $14.0 ext{ mm}$ transmits all signal up to $40 ext{ lines/mm}$ while completely blocking the $50 ext{ lines/mm}$ noise.