Part III: Optics | Chapter 9

Diffraction

Light bending around obstacles — the ultimate proof of its wave nature

9.1 The Huygens-Fresnel Principle

Every point on a wavefront acts as a source of secondary spherical wavelets. The field at any subsequent point is the superposition of all these wavelets, accounting for their amplitudes and phases:

Huygens-Fresnel Integral

$$U(P) = \frac{1}{i\lambda}\iint_{\text{aperture}} U(Q)\frac{e^{ikr}}{r}K(\chi)\,dS$$

where \(U(Q)\) is the field at point \(Q\) on the aperture,\(r\) is the distance from \(Q\) to the observation point \(P\), and \(K(\chi)\) is the obliquity factor (Kirchhoff's contribution).

Two regimes of diffraction are distinguished by the Fresnel number\(N_F = a^2/(\lambda L)\), where \(a\) is the aperture size and \(L\) is the observation distance:

Fresnel Diffraction (\(N_F \gtrsim 1\))

Near-field regime. The curvature of the wavefronts matters. Analysis uses Fresnel zones and is more complex.

Fraunhofer Diffraction (\(N_F \ll 1\))

Far-field regime. Wavefronts are effectively plane. The diffraction pattern is the Fourier transform of the aperture function — mathematically elegant and experimentally common.

9.2 Fraunhofer Single-Slit Diffraction

Derivation 1: Single-Slit Pattern

Consider a slit of width \(a\) illuminated by a plane wave. In the Fraunhofer limit, the field at angle \(\theta\) is the Fourier transform of the slit function:

plane waveaHuygens waveletsscreen+λ/a-λ/acentral maxθ
Figure 1. Single-slit Fraunhofer diffraction. An incoming plane wave passes through a slit of width a. Huygens wavelets from each point in the slit interfere to produce the characteristic sinc-squared intensity pattern on a distant screen, with a central maximum and first minima at ±λ/a.
$$U(\theta) \propto \int_{-a/2}^{a/2} e^{ikx\sin\theta}\,dx = a\,\text{sinc}\!\left(\frac{\pi a \sin\theta}{\lambda}\right)$$

Single-Slit Intensity

$$I(\theta) = I_0 \left[\frac{\sin(\beta)}{\beta}\right]^2 \quad \text{where } \beta = \frac{\pi a \sin\theta}{\lambda}$$

Minima at \(a\sin\theta = m\lambda\) (\(m = \pm 1, \pm 2, \ldots\)). The central maximum has angular half-width \(\Delta\theta \approx \lambda/a\)and contains about 84% of the total intensity.

9.3 Double-Slit Diffraction

Derivation 2: Combined Interference and Diffraction

Two slits of width \(a\) separated by \(d\) produce a pattern that combines the double-slit interference with the single-slit diffraction envelope:

Double-Slit with Finite Width

$$I(\theta) = I_0 \left[\frac{\sin\beta}{\beta}\right]^2 \cos^2\!\left(\frac{\pi d \sin\theta}{\lambda}\right)$$

The sinc-squared envelope modulates the \(\cos^2\) interference fringes. Some interference maxima may be "missing" where they coincide with diffraction minima: this occurs when \(d/a\) is an integer.

For \(N\) slits (a diffraction grating), the interference factor becomes:

$$I_N(\theta) = I_0 \left[\frac{\sin\beta}{\beta}\right]^2 \left[\frac{\sin(N\pi d\sin\theta/\lambda)}{N\sin(\pi d\sin\theta/\lambda)}\right]^2$$

9.4 Circular Aperture: The Airy Pattern

Derivation 3: Diffraction from a Circular Aperture

For a circular aperture of diameter \(D\), the Fraunhofer diffraction involves a 2D Fourier transform with circular symmetry. The result involves the first-order Bessel function \(J_1\):

Airy Pattern

$$I(\theta) = I_0 \left[\frac{2J_1(x)}{x}\right]^2 \quad \text{where } x = \frac{\pi D \sin\theta}{\lambda}$$

The first dark ring (first zero of \(J_1\)) occurs at\(\sin\theta = 1.22\lambda/D\). The central bright disk is called the Airy disk, and it contains 83.8% of the total light.

Derivation 4: The Rayleigh Criterion

Two point sources are considered just resolved when the central maximum of one falls on the first minimum of the other:

Rayleigh Criterion

$$\theta_{\min} = 1.22\frac{\lambda}{D}$$

This sets a fundamental limit on the angular resolution of any optical instrument. For a telescope, larger diameter means better resolution. The Hubble Space Telescope (\(D = 2.4\) m) resolves \(\sim 0.05\) arcseconds in visible light.

9.5 Diffraction Gratings

Derivation 5: Grating Equation and Resolving Power

A diffraction grating has \(N\) slits with spacing \(d\). The principal maxima occur at the grating equation:

$$d\sin\theta_m = m\lambda \quad (m = 0, \pm 1, \pm 2, \ldots)$$

The angular width of each principal maximum is \(\delta\theta \approx \lambda/(Nd\cos\theta)\), which narrows as \(N\) increases. The resolving power is:

Grating Resolving Power

$$R = \frac{\lambda}{\delta\lambda} = mN$$

where \(m\) is the diffraction order and \(N\) is the total number of slits. A grating with 10,000 lines at second order has \(R = 20{,}000\), meaning it can resolve wavelengths differing by 0.025 nm at 500 nm.

Blazed Gratings

In a blazed grating, the groove profile is tilted at an angle (blaze angle) so that the diffraction envelope is directed toward a specific order, concentrating most of the light into that order rather than the useless zeroth order. The blaze condition is:

$$2d\sin\theta_{\text{blaze}} = m\lambda_{\text{blaze}}$$

9.5b Fresnel Diffraction and the Cornu Spiral

In the near-field (Fresnel) regime, the curvature of the wavefront cannot be neglected. Fresnel diffraction involves the Fresnel integrals:

$$C(u) = \int_0^u \cos\!\left(\frac{\pi t^2}{2}\right)dt, \quad S(u) = \int_0^u \sin\!\left(\frac{\pi t^2}{2}\right)dt$$

Plotting \(S(u)\) vs \(C(u)\) gives the Cornu spiral(also called the clothoid), which spirals inward to the points \((\frac{1}{2}, \frac{1}{2})\)and \((-\frac{1}{2}, -\frac{1}{2})\). The diffraction pattern is determined by the length and orientation of the chord connecting two points on the spiral.

The Poisson/Arago Bright Spot

One of the most dramatic predictions of Fresnel diffraction: a circular obstacle illuminated by a plane wave produces a bright spot at the center of its shadow. This follows from Fresnel zone analysis — the contribution from the first exposed Fresnel zone creates constructive interference at the center.

Historical drama: Poisson, a member of the prize committee and supporter of the particle theory, derived this "absurd" prediction from Fresnel's theory, intending to disprove it. Arago immediately performed the experiment and found the bright spot exactly as predicted, spectacularly confirming the wave theory.

9.5c Babinet's Principle

Babinet's principle states that the diffraction pattern from an aperture and its complement (an obstacle of the same shape) are related:

Babinet's Principle

$$U_{\text{aperture}} + U_{\text{complement}} = U_{\text{unobstructed}}$$

Away from the forward direction (where \(U_{\text{unobstructed}} \neq 0\)), the diffraction patterns of complementary screens have identical intensities. A thin wire produces the same far-field pattern as a slit of the same width, and a circular disk produces the same pattern as a circular aperture (explaining the Poisson spot).

9.5d Worked Example: Telescope Resolution

Problem: The James Webb Space Telescope has a primary mirror of diameter\(D = 6.5\) m and observes at \(\lambda = 2\) \(\mu\)m (near-infrared). (a) Calculate its angular resolution. (b) What is the smallest feature it can resolve on the surface of Mars at closest approach (55.7 million km)?

Solution

(a) Angular resolution by the Rayleigh criterion:

$$\theta_{\min} = 1.22\frac{\lambda}{D} = 1.22 \times \frac{2 \times 10^{-6}}{6.5} = 3.75 \times 10^{-7} \text{ rad} = 0.077 \text{ arcsec}$$

(b) Smallest resolvable feature on Mars:

$$d_{\min} = \theta_{\min} \times L = 3.75 \times 10^{-7} \times 5.57 \times 10^{10} \approx 21 \text{ km}$$

At visible wavelengths (\(\lambda = 0.5\) \(\mu\)m), the resolution improves by a factor of 4, to about 5 km. Ground-based telescopes with adaptive optics can achieve similar resolution.

9.5e Diffraction as Fourier Transform

The deep connection between Fraunhofer diffraction and Fourier transforms is central to modern optics. The far-field pattern is:

Diffraction as Fourier Transform

$$U(k_x, k_y) = \mathcal{F}\{t(x, y)\} = \iint t(x,y)e^{-i(k_x x + k_y y)}dx\,dy$$

where \(t(x,y)\) is the aperture transmission function and\(k_x = 2\pi\sin\theta_x/\lambda\) is the spatial frequency. This means:

Slit \(\to\) Sinc

FT of a rectangle is a sinc function

Circle \(\to\) Airy

FT of a circular aperture is the Airy pattern (\(J_1\))

Grating \(\to\) Comb

FT of a periodic array is a series of sharp peaks

Gaussian \(\to\) Gaussian

A Gaussian beam diffracts into a Gaussian (self-Fourier-transforming)

A converging lens performs the Fourier transform optically: placing an object at the front focal plane and observing the back focal plane reveals the Fraunhofer diffraction pattern. This is the principle behind spatial filtering, optical processing, and the 4f imaging system used in optical computing.

9.6 Applications

Spectrometers

Diffraction gratings are the dispersive elements in optical spectrometers used throughout astronomy, chemistry, and materials science. Modern echelle spectrographs use gratings at high order to achieve resolving powers of \(10^5\) or more, enabling measurement of stellar radial velocities and chemical compositions.

X-ray Crystallography

Crystal lattices act as 3D diffraction gratings for X-rays. Bragg diffraction (\(2d\sin\theta = n\lambda\)) reveals atomic structure. This technique determined the structures of DNA, proteins, and countless materials — arguably the most impactful application of diffraction.

Adaptive Optics

Atmospheric turbulence limits telescope resolution far below the diffraction limit. Adaptive optics uses deformable mirrors to correct wavefront distortions in real time, allowing ground-based telescopes to approach their diffraction-limited performance.

Electron Diffraction

The de Broglie wavelength of electrons (\(\lambda = h/p\)) enables diffraction experiments with matter waves. Electron diffraction, predicted by de Broglie and confirmed by Davisson and Germer (1927), is the basis of electron microscopy with sub-angstrom resolution.

9.6b Beyond the Diffraction Limit

The Rayleigh criterion sets a fundamental limit for far-field imaging, but several techniques can achieve super-resolution:

Near-Field Scanning Optical Microscopy (NSOM)

A tiny aperture (smaller than \(\lambda\)) scanned near the surface collects evanescent waves that carry sub-wavelength information. Resolution of\(\lambda/20\) is achievable.

STED Microscopy

Stimulated Emission Depletion microscopy uses a donut-shaped depletion beam to turn off fluorescence around the focal point, effectively shrinking the excitation spot below the diffraction limit. Stefan Hell received the 2014 Nobel Prize for this technique.

PALM/STORM

Photo-Activated Localization Microscopy and Stochastic Optical Reconstruction Microscopy image individual fluorescent molecules one at a time, localizing each to nanometer precision. The composite image achieves resolution of 10–20 nm.

Structured Illumination Microscopy (SIM)

Patterned illumination shifts high spatial frequencies into the observable passband through Moire effects, doubling the resolution to about \(\lambda/4\).

9.7 Historical Context

Francesco Maria Grimaldi (1665): First described diffraction, observing that light passing through a small aperture produced a spot larger than geometric optics predicted, with fringes at the edges. He coined the term "diffraction" (from Latin diffringere, to break apart).

Augustin-Jean Fresnel (1818): Won the French Academy's prize with his wave theory of diffraction. Poisson, a judge who favored the particle theory, noted that Fresnel's theory predicted a bright spot at the center of a circular shadow (the "Poisson spot"). Arago confirmed this experimentally, providing dramatic evidence for the wave theory.

Joseph von Fraunhofer (1821): Developed the diffraction grating as a spectroscopic tool and used it to catalog absorption lines in the solar spectrum (Fraunhofer lines). He produced gratings with unprecedented precision, launching the field of astronomical spectroscopy.

Lord Rayleigh (1879): Established the resolution criterion for optical instruments, showing that diffraction imposes a fundamental limit on angular resolution. This criterion remains the standard measure of resolving power.

George Biddell Airy (1835): Computed the diffraction pattern of a circular aperture, which now bears his name. The Airy disk is central to understanding the resolution of telescopes, microscopes, and cameras.

9.7b Diffraction in Everyday Life and Technology

Diffraction effects are all around us, though often unnoticed:

CDs and DVDs

The closely spaced data tracks on optical discs act as a reflection grating with spacing \(d \approx 1.6\) \(\mu\)m (CD) or 0.74 \(\mu\)m (DVD). When white light illuminates the disc, different wavelengths are diffracted at different angles, creating the rainbow patterns seen on the surface. Blu-ray discs have even finer track spacing (0.32 \(\mu\)m), requiring a blue laser to read.

Camera Aperture and Depth of Field

Photographers face a fundamental tradeoff: smaller apertures (larger f-number) give greater depth of field but worse diffraction blur. The diffraction-limited spot size is \(\sim 2.44\lambda f/\#\). For visible light, diffraction typically limits resolution for f-numbers above f/11 on full-frame cameras.

Holographic Optical Elements

Holographic gratings and lenses, recorded using interference of laser beams, can combine multiple optical functions (focusing, filtering, beam-splitting) in a single thin element. They are used in heads-up displays, augmented reality glasses, and spectrometers.

Acoustic Diffraction

Sound waves diffract around obstacles and through openings, with audible wavelengths (17 mm to 17 m) comparable to everyday objects. You can hear around corners because sound diffracts through doorways. Concert hall design must account for diffraction from architectural features to ensure uniform sound distribution.

9.8 Python Simulation

This simulation visualizes single-slit diffraction, double-slit diffraction with envelope, the Airy pattern for a circular aperture, and multi-slit grating patterns.

Diffraction: Single/Double Slit, Airy Pattern, and Grating Resolution

Python
script.py151 lines

Click Run to execute the Python code

Code will be executed with Python 3 on the server

Summary

Core Equations

  • Single slit: \(I = I_0[\sin\beta/\beta]^2\)
  • Airy: \(I = I_0[2J_1(x)/x]^2\)
  • Rayleigh: \(\theta_{\min} = 1.22\lambda/D\)
  • Grating: \(d\sin\theta = m\lambda\)
  • Resolving power: \(R = mN\)

Key Insights

  • Diffraction = Fourier transform of aperture
  • Smaller aperture = wider diffraction pattern
  • Diffraction sets fundamental resolution limit
  • More grating lines = sharper spectral features
  • Airy disk determines telescope resolution

Practice Problems

Problem 1:Light of wavelength 600 nm passes through a single slit of width 0.1 mm. Find the angular position and linear width of the central maximum on a screen 2 m away.

Solution:

1. First minimum of single-slit diffraction: $a\sin\theta = m\lambda$ with $m = \pm 1$.

2. $\sin\theta_1 = \frac{\lambda}{a} = \frac{600\times10^{-9}}{0.1\times10^{-3}} = 6\times10^{-3}$ rad (small angle).

3. Angular half-width of central maximum: $\theta_1 = 6\times10^{-3}$ rad = 0.344°.

4. Full angular width: $2\theta_1 = 0.012$ rad = 0.688°.

5. Linear position of first minimum on screen: $y_1 = L\tan\theta_1 \approx L\theta_1 = 2 \times 6\times10^{-3} = 0.012$ m = 12 mm.

6. Full width of central maximum: $2y_1 = 24$ mm. The central maximum is twice as wide as the secondary maxima.

Problem 2:In a double-slit experiment with slit separation d = 0.5 mm, slit width a = 0.1 mm, and λ = 500 nm, find which interference orders are missing due to the single-slit envelope.

Solution:

1. Double-slit maxima occur at: $d\sin\theta = m\lambda$, i.e., $\sin\theta = m\lambda/d$.

2. Single-slit minima (zeros of envelope) occur at: $a\sin\theta = p\lambda$, i.e., $\sin\theta = p\lambda/a$.

3. Missing orders occur when both conditions are satisfied simultaneously: $m\lambda/d = p\lambda/a$.

4. This gives: $m = p \cdot \frac{d}{a} = p \cdot \frac{0.5}{0.1} = 5p$.

5. Missing orders: $m = 5, 10, 15, \ldots$ (every 5th interference maximum is suppressed).

6. The intensity pattern is $I = I_0\cos^2\!\left(\frac{\pi d\sin\theta}{\lambda}\right)\left(\frac{\sin(\pi a\sin\theta/\lambda)}{\pi a\sin\theta/\lambda}\right)^2$. The sinc$^2$ envelope kills the 5th, 10th, etc. orders.

Problem 3:A diffraction grating has 5000 lines/cm. Find the angular positions of the first three orders for λ = 550 nm, and the maximum order observable.

Solution:

1. Grating spacing: $d = 1/5000$ cm $= 2\times10^{-4}$ cm $= 2\times10^{-6}$ m = 2 μm.

2. Grating equation: $d\sin\theta = m\lambda$, so $\sin\theta_m = \frac{m\lambda}{d} = \frac{m \times 550\times10^{-9}}{2\times10^{-6}} = 0.275m$.

3. $m = 1$: $\sin\theta_1 = 0.275$, $\theta_1 = 15.96°$.

4. $m = 2$: $\sin\theta_2 = 0.550$, $\theta_2 = 33.37°$.

5. $m = 3$: $\sin\theta_3 = 0.825$, $\theta_3 = 55.59°$.

6. Maximum order: $m_{\max} = \lfloor d/\lambda \rfloor = \lfloor 2000/550 \rfloor = 3$. The 4th order ($\sin\theta = 1.1 > 1$) is not observable.

Problem 4:Two stars are separated by 0.5 arcseconds. Can a telescope with aperture D = 25 cm resolve them at λ = 550 nm? What is the minimum aperture needed?

Solution:

1. Rayleigh criterion: $\theta_{\min} = 1.22\frac{\lambda}{D}$.

2. For $D = 0.25$ m: $\theta_{\min} = 1.22 \times \frac{550\times10^{-9}}{0.25} = 2.684\times10^{-6}$ rad.

3. Convert to arcseconds: $\theta_{\min} = 2.684\times10^{-6} \times \frac{180}{\pi} \times 3600 = 0.554''$.

4. The angular separation is $0.5'' < 0.554''$, so the stars are not resolved by the Rayleigh criterion.

5. Minimum aperture: $D_{\min} = \frac{1.22\lambda}{\theta} = \frac{1.22 \times 550\times10^{-9}}{0.5/206265} = \frac{6.71\times10^{-7}}{2.42\times10^{-6}} $... solving: $D_{\min} = 0.277$ m $\approx 28$ cm.

6. In practice, atmospheric seeing (~1$''$) typically limits ground-based resolution more than diffraction for telescopes larger than ~15 cm. Adaptive optics can recover diffraction-limited performance.

Problem 5:A grating with 10,000 total lines is used in 2nd order. Calculate its resolving power and the minimum wavelength difference it can resolve near λ = 500 nm.

Solution:

1. Resolving power of a grating: $R = mN$, where $m$ is the diffraction order and $N$ is the total number of lines.

2. $R = 2 \times 10{,}000 = 20{,}000$.

3. By definition, $R = \frac{\lambda}{\Delta\lambda}$, so $\Delta\lambda = \frac{\lambda}{R} = \frac{500}{20{,}000} = 0.025$ nm.

4. This means the grating can distinguish two spectral lines separated by just 0.025 nm (or 0.25 Angstroms) at 500 nm.

5. In terms of frequency: $\Delta\nu = \frac{c}{\lambda^2}\Delta\lambda = \frac{3\times10^8}{(500\times10^{-9})^2} \times 0.025\times10^{-9} = 3\times10^{10}$ Hz = 30 GHz.

6. To increase resolving power, one can either use more lines (larger grating) or work in a higher diffraction order (but higher orders may have lower intensity).

Rate this chapter: