3.1.2 Physical Optics#
Prompts
State Huygens’ principle. How does it derive the laws of reflection and refraction from wavefront geometry?
Why does frequency stay the same across media but wavelength does not? How does this explain the factor of \(n\) in the optical path length?
How does Young’s double-slit experiment decide between the particle and wave theories of light?
How does Fermat’s principle of stationary time emerge from wave optics in the short-wavelength limit?
Lecture Notes#
Overview#
The wave theory of light explains what the particle picture cannot. Huygens’ principle — every point on a wavefront is a source of secondary spherical wavelets — directly derives the laws of refraction, reflection, and rectilinear propagation. The key quantity is phase, which accumulates proportional to optical path length. In the short-wavelength limit, constructive interference from paths with stationary phase gives Fermat’s principle: geometric optics emerges from wave physics.
Huygens’ Principle#
Huygens’ Principle
Each point on a wavefront acts as a source of secondary wavelets (spherical waves). The wavefront at a later time is the envelope of all these secondary wavelets.
A wavefront is a surface of constant phase: \(\Phi(\boldsymbol{r}) = \text{const}\). Rays are perpendicular to wavefronts.
Phase and Optical Path Length#
Along a ray from \(\boldsymbol{r}_0\) to \(\boldsymbol{r}\), the phase accumulated is
Here the integral is taken along the actual ray path, and
\(\mathrm{d}s'\) is arc length element at \(\boldsymbol{r}'\) on the path,
\(n(\boldsymbol{r}')\) is the local refractive index,
\(k = \omega/c \) is the vacuum wave number, with \(\omega = 2\pi \nu\) the angular frequency.
Why does \(n\) appear?
For a monochromatic wave, frequency \(\omega\) is fixed across interfaces (the field must stay in phase at boundaries), while the phase speed \(v(\boldsymbol{r}) = c/n(\boldsymbol{r})\) changes. The local wave number is \(k_{\mathrm{loc}}(\boldsymbol{r}) = \omega/v(\boldsymbol{r}) = k \cdot n(\boldsymbol{r})\), so
Frequency is shared across media; wavelength is not. This is why the refractive index enters the optical path length.
Wave Derivation of the Three Laws#
Using Huygens’ principle and simple geometry, all three laws of geometric optics follow:
Rectilinear propagation. In a uniform medium, spherical wavelets from a flat wavefront form another flat wavefront. Rays are straight lines.
Reflection. Secondary wavelets from a reflective surface form a reflected wavefront with \(\theta_r = \theta_i\) by symmetry of the wavelet construction.
Refraction (Snell’s law). An incident wavefront in medium \(n_1\) hits a boundary with medium \(n_2\). By the envelope geometry:
because \(\lambda_i\propto v_i\propto 1/n_i\).
Derivation: Snell’s Law from Wavefronts
Point \(A\) on the wavefront hits the boundary at \(t = 0\). Point \(B\), at distance \(\ell\) along the wavefront, arrives at \(t_B = \ell n_1 / c\). In that time, the wavelet from \(A\) travels \(d_A = (c/n_2) t_B = \ell \, n_1/n_2\) into medium 2. The envelope gives \(\sin\theta_2 = d_A / \ell = (n_1/n_2)\sin\theta_1\).
Interference: Why Wave Theory is Needed#
Geometric optics cannot explain interference. In Young’s double-slit experiment, the wave amplitude is a superposition \(\psi = \psi_1 + \psi_2\), and the detected intensity
shows fringes from the path-length difference \(\Delta\Phi = k \, \Delta L\). The interpretation \(I = |\psi|^2\) as light intensity (photon probability density) foreshadows Born’s rule in quantum mechanics.
From Huygens to Fermat: Stationary Phase#
Huygens’ principle sums wavelets from all points; Fermat’s principle selects one stationary ray. The connection is stationary phase:
In the short-wavelength limit (\(k \to \infty\)), the phase \(\Phi = k L\) oscillates rapidly for paths far from an extremum. Only paths where \(\delta L = 0\) contribute constructively. All other paths cancel by destructive interference.
Limit |
Framework |
Picture |
|---|---|---|
Short wavelength |
Fermat’s principle |
Single extremal ray |
General wavelength |
Huygens’ superposition |
All wavelets contribute |
Geometric optics is the classical limit of wave optics.
Discussion
Young’s double-slit experiment shows interference fringes for light. What happens if you send electrons one at a time through a double slit? Would you still see fringes? What does the answer tell you about the nature of quantum particles versus classical waves?
Summary#
Huygens’ principle: wavefront points emit secondary wavelets; the envelope is the new wavefront.
Phase \(\Phi = k L\) accumulates proportional to optical path length. Frequency is constant across interfaces; wavelength changes by factor \(n\).
Three laws: rectilinear propagation, reflection, and Snell’s law all follow from wavefront geometry.
Interference (\(I = |\psi_1 + \psi_2|^2\)) is the decisive phenomenon that geometric optics cannot explain.
Stationary phase: Fermat’s principle emerges from Huygens’ superposition in the short-wavelength limit.
Homework#
1. Light of frequency \(f\) crosses from vacuum (\(n=1\)) into glass (\(n=1.5\)). Does the frequency change? Does the wavelength change? Compute the ratio \(\lambda_{\text{glass}}/\lambda_{\text{vacuum}}\) and explain why the refractive index appears in the optical path length.
2. Using Huygens’ principle and the wavefront envelope construction, derive the law of reflection for a plane wave hitting a flat mirror. Draw the incident and reflected wavefronts and label the angles.
3. A plane wavefront in medium 1 (\(n_1\)) arrives at a flat interface with medium 2 (\(n_2\)) at angle \(\theta_1\). Using the wavefront construction (secondary wavelets travel at speed \(c/n_2\) in medium 2), derive Snell’s law \(n_1\sin\theta_1 = n_2\sin\theta_2\) from geometry.
4. Two paths from \(A\) to \(B\) through a medium have optical path lengths \(L_1\) and \(L_2\). The phase difference is \(\Delta\Phi = k_0(L_1 - L_2)\) where \(k_0 = 2\pi/\lambda_0\). For what values of \(\Delta\Phi\) is the interference constructive? Destructive?
5. In the short-wavelength limit (\(\lambda \to 0\), so \(k_0 \to \infty\)), argue using the stationary phase idea that only paths near the one with extremal optical path length contribute significantly to the wave amplitude. How does this recover Fermat’s principle from the wave picture?
6. A slab of thickness \(d\) and refractive index \(n\) is inserted into one arm of an interferometer. By how much does the optical path length change compared to the same thickness of vacuum? Express your answer in terms of \(d\), \(n\), and \(\lambda_0\).