3.1.2 Physical Optics

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#

In §3.1.1, we saw how the particle theory of light explains reflection and refraction via Fermat’s principle. The wave theory offers an alternative foundation for the same laws, based on Huygens’ principle and the propagation of wave fronts.

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

\[ \boxed{\Phi(\boldsymbol{r}_0 \to \boldsymbol{r}) = k \, L(\boldsymbol{r}_0 \to \boldsymbol{r}) = k \int_{\boldsymbol{r}_0}^{\boldsymbol{r}} n(\boldsymbol{r}') \, \mathrm{d}s'} \]

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

\[ \Phi = \int k_{\mathrm{loc}}(\boldsymbol{r}) \, \mathrm{d}s = k \int n(\boldsymbol{r}) \, \mathrm{d}s = k \, L \]

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. In the figure, let \(\lambda_{\parallel}\) denote the distance along the mirror between two neighboring wavefront-intersection points. For a plane wave of wavelength \(\lambda\) meeting the mirror at angle \(\theta_i\) (measured from the normal), this projected spacing is

\[ \lambda_{\parallel}=\frac{\lambda}{\sin\theta_i}. \]

For the reflected wave, the same boundary spacing \(\lambda_{\parallel}\) must be reproduced by the outgoing wavefront pattern, so

\[ \lambda_{\parallel}=\frac{\lambda}{\sin\theta_r}. \]

In reflection the wave stays in the same medium, so \(\lambda\) is unchanged. Therefore

(60)#\[ \frac{\lambda}{\sin\theta_i}=\frac{\lambda}{\sin\theta_r} \quad\Rightarrow\quad \theta_r=\theta_i. \]

Refraction (Snell’s law). Again let \(\lambda_{\parallel}\) be the spacing along the interface between neighboring phase-matched points. The incident and transmitted wavefronts must share this same interface spacing:

\[ \lambda_{\parallel}=\frac{\lambda_1}{\sin\theta_1}=\frac{\lambda_2}{\sin\theta_2}. \]

Now medium change matters: frequency is continuous across the boundary, but wave speed changes from \(v_1=c/n_1\) to \(v_2=c/n_2\), so

\[ \lambda_i=\frac{v_i}{\nu}=\frac{c}{n_i\nu}\propto\frac{1}{n_i}. \]

Substituting into the \(\lambda_{\parallel}\) relation gives

(61)#\[ n_1\sin\theta_1=n_2\sin\theta_2. \]

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

\[ I = |\psi_1 + \psi_2|^2 = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos(\Delta\Phi) \]

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: electron double-slit interference

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?

Poll: Crossing an interface

A monochromatic light wave crosses from glass (\(n=1.5\)) into air (\(n=1.0\)). Which quantity stays the same across the interface?

(A) Wavelength \(\lambda\).

(B) Wave speed \(v\).

(C) Frequency \(\nu\).

(D) Local wave number \(k_{\mathrm{loc}}\).

Just as Huygens’ principle in wave optics and Fermat’s principle in geometric optics are two faces of the same physics, so too matter exhibits this duality. This dual description—wave and particle—is the subject of §3.1.3, which unifies both pictures.

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.

See Also

Homework#

1. Refraction at an interface. 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. Reflection from wavefronts. 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. Layered medium ray tracing. A light ray enters a stack of three parallel slabs with refractive indices \(n_{1} = 1.0\) (above), \(n_{2} = 1.5\), \(n_{3} = 1.2\), \(n_{4} = 1.0\) (below), at incidence angle \(\theta_{1} = 60^{\circ}\) from the surface normal in \(n_{1}\).

(a) Apply Snell’s law at each interface to compute \(\theta_{2}, \theta_{3}, \theta_{4}\).

(b) Show that \(\theta_{4}\) depends only on \(n_{1}, n_{4}\) and \(\theta_{1}\) — independent of the intermediate \(n_{2}, n_{3}\). Explain why the intermediate layers cannot change the exit direction.

(c) Find the entry angle \(\theta_{1}^{\text{crit}}\) above which total internal reflection occurs somewhere in the stack, and identify which interface fails first.

4. Interference of two paths. 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. Width of the Fermat band. In Huygens’ superposition, the amplitude integrates contributions from all paths weighted by \(\mathrm{e}^{\mathrm{i}k_{0}L(\boldsymbol r)}\) with \(k_{0} = 2\pi/\lambda_{0}\). Near the stationary path, expand to second order: \(L(\Delta) \approx L_{0} + \tfrac{1}{2}L''\,\Delta^{2}\).

(a) Constructive interference holds when the phase \(k_{0}L''\Delta^{2}/2\) stays below order unity. Show that this gives a Fermat-band width \(\Delta_{\text{Fermat}}\sim\sqrt{\lambda_{0}/(2\pi\,\vert L''\vert)}\).

(b) For visible light \(\lambda_{0} = 500\,\text{nm}\) and a typical lens curvature \(\vert L''\vert\sim 1/\mathrm{mm}\), estimate \(\Delta_{\text{Fermat}}\) in nm. Comment on whether geometric optics is a reliable approximation at this wavelength and lens scale.

(c) Repeat for X-rays \(\lambda_{0} = 1\,\text{Å}\). What does the answer say about why X-ray crystallography requires careful diffraction analysis rather than simple ray tracing?

6. Slab in an interferometer. 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 \(\lambda_{0}\).