Wave Optics: Interference and Diffraction

The Wave Nature of Light

The wave theory of light, developed by Christiaan Huygens and confirmed by Thomas Young's landmark 1801 experiment, describes light as a transverse electromagnetic wave. Wave optics explains phenomena that the particle (ray) model cannot — interference, diffraction, and polarization.

Coherence and Superposition

Two sources are coherent if they maintain a constant phase difference over time. Only coherent sources produce observable interference patterns.

The Principle of Superposition states:

$\psi_{\text{total}}(\mathbf{r}, t) = \psi_1(\mathbf{r}, t) + \psi_2(\mathbf{r}, t)$

For two waves of equal amplitude $A$ and a phase difference $\delta$:

$I = 4I_0 \cos^2\!\left(\frac{\delta}{2}\right)$

Where $I_0 = A^2$ is the intensity of each individual wave.

Young's Double-Slit Experiment

Setup and Path Difference

Two narrow slits $S_1$ and $S_2$ separated by distance $d$ are illuminated by a monochromatic source. At a point $P$ on a screen at distance $D$:

$\Delta = d\sin\theta \approx \frac{dy}{D} \quad (\text{for small } \theta)$

Where $y$ is the vertical distance from the central maximum.

Conditions for Bright and Dark Fringes

| Condition | Path Difference $\Delta$ | Result | |-----------|--------------------------|--------| | Constructive | $\Delta = m\lambda$ | Bright fringe | | Destructive | $\Delta = \left(m + \frac{1}{2}\right)\lambda$ | Dark fringe |

where $m = 0, \pm 1, \pm 2, \ldots$

Fringe Width

The spacing between consecutive bright (or dark) fringes:

$\beta = \frac{\lambda D}{d}$

Key relationships:

• Increasing $D$ → wider fringes ($\beta \propto D$)
• Increasing $d$ → narrower fringes ($\beta \propto 1/d$)
• Longer wavelength → wider fringes ($\beta \propto \lambda$)

Intensity Distribution

The intensity at angle $\theta$:

$I(\theta) = I_{\max} \cos^2\!\left(\frac{\pi d \sin\theta}{\lambda}\right)$

Single-Slit Diffraction

When light passes through a single slit of width $a$, it spreads into a diffraction pattern.

Minima Condition

$a\sin\theta = m\lambda \quad (m = \pm 1, \pm 2, \ldots)$

The angular half-width of the central maximum:

$\theta_{\text{min}} = \arcsin\!\left(\frac{\lambda}{a}\right) \approx \frac{\lambda}{a}$

Intensity Formula

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

This $\text{sinc}^2$ function produces the characteristic central bright maximum flanked by progressively weaker secondary maxima.

Diffraction Grating

A diffraction grating with $N$ slits separated by $d$ (grating constant) produces sharp, bright principal maxima at angles satisfying:

$d\sin\theta = m\lambda$

Resolving Power

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

where $\Delta\lambda$ is the minimum resolvable wavelength difference.

Example: A grating with 600 lines/mm and $m = 2$ has:

$d = \frac{1}{600} \text{ mm} = 1.667 \times 10^{-6} \text{ m}$

$R = mN = 2 \times 600 \times 10^3 = 1.2 \times 10^6 \text{ per metre}$

Numerical Simulation: Young's Double-Slit

import numpy as np
import math

def youngs_intensity(
    y_values: np.ndarray,
    wavelength: float,  # metres
    slit_sep: float,    # d in metres
    screen_dist: float  # D in metres
) -> np.ndarray:
    """
    Calculate normalised intensity pattern for Young's double-slit.
    y_values: array of screen positions (metres)
    """
    delta = slit_sep * y_values / screen_dist  # path difference
    phase = (2 * np.pi / wavelength) * delta
    intensity = np.cos(phase / 2) ** 2
    return intensity

# Parameters
lam = 550e-9   # 550 nm (green light)
d   = 0.5e-3   # 0.5 mm slit separation
D   = 1.0      # 1 m screen distance

y = np.linspace(-5e-3, 5e-3, 1000)  # ±5 mm
I = youngs_intensity(y, lam, d, D)

fringe_width = lam * D / d
print(f"Fringe width β = {fringe_width * 1e3:.3f} mm")
# β = 1.100 mm ✓

# Find fringe positions (maxima)
threshold = 0.99
maxima_y = y[I > threshold] * 1e3  # convert to mm
print(f"Bright fringe positions (mm): {maxima_y[:5].round(2)}")

Rayleigh's Criterion for Resolution

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

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

Where $D$ is the aperture diameter. The factor 1.22 arises from the Bessel function zero for a circular aperture.

Telescope example: Hubble Space Telescope ($D = 2.4\text{ m}$, $\lambda = 500\text{ nm}$):

$\theta_{\text{min}} = 1.22 \times \frac{500 \times 10^{-9}}{2.4} = 2.54 \times 10^{-7} \text{ rad}$

At the Moon's distance ($d = 3.84 \times 10^8\text{ m}$), the minimum resolvable separation is:

$\Delta x = \theta_{\text{min}} \times d \approx 97.5 \text{ m}$