Supplementary Materials: Fourier Optics

Additional examples, review, and visualizations

1. Fourier Transform Review

1.1 Basic Definitions

The Fourier transform decomposes a function into its constituent frequencies. For a function \(f(x)\), its Fourier transform \(F(k)\) is defined as:

\[F(k) = \int_{-\infty}^{\infty} f(x) e^{-i2\pi kx} dx\]

The inverse Fourier transform reconstructs the original function:

\[f(x) = \int_{-\infty}^{\infty} F(k) e^{i2\pi kx} dk\]

In optics, \(x\) typically represents spatial coordinates and \(k\) represents spatial frequencies.

1.2 Important Properties

  1. Linearity: \(\mathcal{F}\{af(x) + bg(x)\} = aF(k) + bG(k)\)

  2. Shift Theorem: \(\mathcal{F}\{f(x-a)\} = e^{-i2\pi ka}F(k)\)

  3. Convolution Theorem: \(\mathcal{F}\{f * g\} = F(k) \cdot G(k)\)

  4. Parseval’s Theorem: \(\int |f(x)|^2 dx = \int |F(k)|^2 dk\)

1.3 Common Fourier Transform Pairs

Function Fourier Transform
\(\delta(x)\) (Delta function) \(1\) (constant)
\(1\) (constant) \(\delta(k)\) (Delta function)
\(\text{rect}(x)\) (Rectangle function) \(\text{sinc}(k)\) (Sinc function)
\(e^{-\pi x^2}\) (Gaussian) \(e^{-\pi k^2}\) (Gaussian)
\(\cos(2\pi ax)\) \(\frac{1}{2}[\delta(k-a) + \delta(k+a)]\)

1.4 Application to Optics

In optics, the Fourier transform relates: - The field distribution in real space to its spatial frequency spectrum - The field at one plane to its distribution in the focal plane of a lens - The aperture function to its diffraction pattern in the far field

2. Worked Examples

2.1 Transmission Examples

Example 1: Thin Lens Transmission Function

Problem: Calculate the transmission function for a thin biconvex lens with focal length \(f = 100\) mm at wavelength \(\lambda = 633\) nm.

Solution:

For a thin lens, the transmission function is purely a phase modulation:

\[t(x,y) = e^{-i\frac{k}{2f}(x^2+y^2)}\]

With \(k = \frac{2\pi}{\lambda} = \frac{2\pi}{633 \times 10^{-9}} \approx 9.92 \times 10^6\) rad/m:

\[t(x,y) = e^{-i\frac{9.92 \times 10^6}{2 \times 0.1}(x^2+y^2)} = e^{-i 4.96 \times 10^7 (x^2+y^2)}\]

The phase shift at position \((x,y) = (1 \text{ mm}, 0) = (10^{-3} \text{ m}, 0)\) is:

\[\phi = -4.96 \times 10^7 \times (10^{-3})^2 = -49.6 \text{ rad} \approx -15.8\pi \text{ rad}\]

This represents approximately 8 complete phase cycles at just 1 mm from the center.

Example 2: Phase Grating

Problem: A phase grating has a transmission function \(t(x) = e^{i\alpha\sin(2\pi x/d)}\) where \(d = 10\) μm and \(\alpha = \pi/2\). Determine the amplitudes of the diffraction orders.

Solution:

This is a sinusoidal phase grating. Using the Jacobi-Anger expansion:

\[e^{i\alpha\sin\theta} = \sum_{n=-\infty}^{\infty} J_n(\alpha)e^{in\theta}\]

Where \(J_n\) is the Bessel function of the first kind.

For our grating:

\[t(x) = e^{i(\pi/2)\sin(2\pi x/d)} = \sum_{n=-\infty}^{\infty} J_n(\pi/2)e^{in(2\pi x/d)}\]

Each term in this expansion corresponds to a diffraction order with amplitude \(J_n(\pi/2)\).

For the first few orders: - 0th order: \(J_0(\pi/2) \approx 0.568\) - ±1st order: \(J_{\pm1}(\pi/2) \approx 0.382\) - ±2nd order: \(J_{\pm2}(\pi/2) \approx 0.129\) - ±3rd order: \(J_{\pm3}(\pi/2) \approx 0.019\)

2.2 Wave Propagation Examples

Example 3: Fresnel Propagation

Problem: A uniformly illuminated circular aperture of diameter \(D = 2\) mm is placed at \(z = 0\). Calculate the intensity at the center of the diffraction pattern at distance \(z = 1\) m for light with wavelength \(\lambda = 500\) nm.

Solution:

For Fresnel diffraction, we need to evaluate:

\[U(x,y,z) = \frac{e^{ikz}}{i\lambda z}e^{i\frac{k}{2z}(x^2+y^2)} \iint U(x',y',0)e^{i\frac{k}{2z}(x'^2+y'^2)}e^{-i\frac{k}{z}(xx'+yy')}dx'dy'\]

For the center of the pattern \((x=0, y=0)\) with uniform illumination \(U(x',y',0) = U_0\) within the aperture:

\[U(0,0,z) = \frac{e^{ikz}}{i\lambda z} \iint_{aperture} U_0 e^{i\frac{k}{2z}(x'^2+y'^2)}dx'dy'\]

For a circular aperture of radius \(a = D/2 = 1\) mm, using polar coordinates:

\[U(0,0,z) = \frac{U_0 e^{ikz}}{i\lambda z} \int_0^{2\pi}\int_0^a e^{i\frac{k}{2z}r'^2}r'dr'd\theta'\]

\[U(0,0,z) = \frac{2\pi U_0 e^{ikz}}{i\lambda z} \int_0^a e^{i\frac{k}{2z}r'^2}r'dr'\]

Evaluating this integral:

\[U(0,0,z) = \frac{2\pi U_0 e^{ikz}}{i\lambda z} \cdot \frac{z}{ik}(1-e^{i\frac{k}{2z}a^2})\]

\[U(0,0,z) = \frac{2\pi U_0 e^{ikz}}{k} \cdot (1-e^{i\frac{k}{2z}a^2})\]

For our values: \(k = 2\pi/\lambda = 2\pi/(500 \times 10^{-9}) = 1.26 \times 10^7\), \(z = 1\) m, \(a = 10^{-3}\) m:

\[\frac{k}{2z}a^2 = \frac{1.26 \times 10^7}{2 \times 1}(10^{-3})^2 = 6.28\]

Therefore: \[U(0,0,z) = \frac{2\pi U_0 e^{ikz}}{k} \cdot (1-e^{i\cdot 6.28})\]

Since \(e^{i\cdot 6.28} \approx e^{i\cdot 2\pi} = 1\), we get: \[U(0,0,z) \approx 0\]

This indicates a dark spot at the center, as expected from Fresnel diffraction for these parameters.

2.3 Spatial Frequency Examples

Example 4: Resolution Limit

Problem: A microscope objective has a numerical aperture NA = 0.65. Calculate the smallest resolvable feature size according to the Abbe diffraction limit for light with wavelength \(\lambda = 488\) nm.

Solution:

The Abbe diffraction limit states that the smallest resolvable feature has a spatial frequency of:

\[\nu_{max} = \frac{2\text{NA}}{\lambda}\]

For NA = 0.65 and \(\lambda = 488\) nm:

\[\nu_{max} = \frac{2 \times 0.65}{488 \times 10^{-9}} = 2.66 \times 10^6 \text{ cycles/m}\]

The corresponding minimum resolvable feature size is:

\[d_{min} = \frac{1}{2\nu_{max}} = \frac{\lambda}{2\text{NA}} = \frac{488 \times 10^{-9}}{2 \times 0.65} = 375 \text{ nm}\]

Example 5: Angular Spectrum Filtering

Problem: A spatial filter blocks all spatial frequencies above \(\nu_{cutoff} = 100\) cycles/mm. What is the minimum spot size achievable after this filter for light with wavelength \(\lambda = 600\) nm?

Solution:

The minimum spot size is related to the highest spatial frequency that can pass through the filter:

\[w_{min} = \frac{1}{2\nu_{cutoff}} = \frac{1}{2 \times 100 \times 10^3} = \frac{1}{2 \times 10^5} = 5 \times 10^{-6} \text{ m} = 5 \text{ μm}\]

The corresponding numerical aperture that this filter effectively creates is:

\[\text{NA} = \lambda \times \nu_{cutoff} = 600 \times 10^{-9} \times 100 \times 10^3 = 0.06\]

3. Angular Spectrum Visualizations

3.1 Conceptual Visualization

The angular spectrum representation can be visualized through several complementary perspectives:

  1. Plane Wave Decomposition: Each spatial frequency component corresponds to a plane wave traveling at a specific angle.

  2. Fourier Space Representation: The 2D Fourier transform of a field shows its angular spectrum.

  3. Evanescent Waves: Spatial frequencies beyond \(1/\lambda\) correspond to evanescent waves that decay exponentially away from the source.

3.2 Interactive Visualization

The widget below shows a Gaussian beam (\(w_0 = 0.5\,\text{mm}\), \(\lambda = 633\,\text{nm}\), Rayleigh range \(z_R \approx 124\,\text{cm}\)) propagated via the exact angular spectrum method. Drag the slider to change the propagation distance and observe the beam broadening due to diffraction.

z = 10 cm    0 – 400 cm

3.3 Physical Interpretation

When light passes through an aperture, we can interpret the resulting diffraction in two equivalent ways:

  1. Spatial Domain: The aperture creates a new light source that diffracts according to the Huygens-Fresnel principle.

  2. Frequency Domain: The aperture acts as a spatial filter, selectively transmitting certain angular components (spatial frequencies).

The angular spectrum approach provides a rigorous mathematical framework that connects these two viewpoints. It shows that:

  • Each spatial frequency corresponds to a plane wave traveling at a specific angle
  • These plane waves propagate independently in free space
  • The complete field at any plane is the coherent superposition of all these plane waves
  • The maximum angle of propagation determines the system’s resolution limit

This perspective elegantly explains why: - Smaller apertures produce wider diffraction patterns (they filter out high spatial frequencies) - Optical systems have finite resolution (they cannot capture infinitely high spatial frequencies) - Evanescent waves decay exponentially (they correspond to “trapped” light that doesn’t propagate)