Lesson 3.5 · 3. Electromagnetism

Wave Optics

In the previous lessons on electromagnetism, we established that light is an electromagnetic wave. But what happens when light encounters obstacles, slits, or thin films? The wave nature of light produces phenomena that no particle theory can explain: interference and diffraction. These effects, invisible in everyday life because the wavelength of visible light is so small, reveal themselves spectacularly when we look carefully. Wave optics is the study of light in situations where its wave character cannot be ignored.

Huygens' Principle

Every point on a wavefront is a source

Huygens' principle states that every point on a wavefront acts as a source of secondary spherical wavelets. The new wavefront at a later time is the envelope (the tangent surface) of all these wavelets. This geometric construction, proposed by Christiaan Huygens in 1678, explains reflection, refraction, and diffraction without any reference to rays.

Consider a plane wave traveling to the right. At time $t$, every point on the wavefront emits a spherical wavelet. After a time interval $\Delta t$, each wavelet has radius $v \Delta t$, where $v$ is the wave speed. The envelope of all these wavelets is a new plane, displaced by $v \Delta t$ to the right. Huygens' principle recovers the trivial result that a plane wave remains a plane wave in free space.

The power of the principle emerges when the wavefront encounters an obstacle or an aperture. Only the portions of the wavefront that are not blocked emit secondary wavelets, and the resulting envelope is no longer a simple plane. The wave bends around obstacles and spreads after passing through openings. This is diffraction.

Huygens' principle applied to a slit: each point in the opening emits a spherical wavelet. The resulting wave spreads into the geometric shadow region, demonstrating diffraction.

Why don't we notice diffraction in everyday life?

Diffraction becomes significant when the size of the aperture or obstacle is comparable to the wavelength. Visible light has wavelengths of $400$-$700$ nm, roughly a thousand times smaller than a millimeter. Everyday objects (doors, windows, furniture) are millions of wavelengths across, so diffraction effects are negligibly small. But pass light through a narrow slit or look at a distant streetlight through a fine mesh, and diffraction becomes visible.

Young's Double-Slit Experiment

In 1801, Thomas Young performed one of the most important experiments in the history of physics. He passed light through two narrow slits and observed a pattern of bright and dark bands on a distant screen. This pattern, called an interference pattern, provided definitive evidence that light is a wave.

Setup and Geometry

Consider two narrow slits separated by a distance $d$, illuminated by monochromatic light of wavelength $\lambda$. A screen is placed at a large distance $L$ from the slits ($L \gg d$). We want to find the intensity pattern on the screen.

Young's double-slit setup: light from two slits S₁ and S₂ arrives at point P with a path difference that depends on the angle θ. The right edge shows the resulting bright and dark fringes.

Deriving the Interference Pattern

Let $r_1$ and $r_2$ be the distances from slits $S_1$ and $S_2$ to a point P on the screen. The key quantity is the path difference $\Delta r = r_2 - r_1$.

Step 1: Path difference. If the point P is at position $y$ on the screen (measured from the central axis), the angle $\theta$ satisfies $\tan\theta = y/L$. For small angles ($y \ll L$), we have $\sin\theta \approx \tan\theta \approx y/L$. The path difference is: $$\Delta r = d \sin\theta \approx \frac{dy}{L}$$

Step 2: Phase difference. A path difference $\Delta r$ corresponds to a phase difference $\delta = \frac{2\pi}{\lambda}\Delta r = \frac{2\pi d \sin\theta}{\lambda}$.

Step 3: Superposition. The electric fields from the two slits at point P are: $$E_1 = E_0 \cos(\omega t), \qquad E_2 = E_0 \cos(\omega t + \delta)$$ The total field is $E = E_1 + E_2$. Using the identity $\cos A + \cos B = 2\cos\left(\frac{A-B}{2}\right)\cos\left(\frac{A+B}{2}\right)$: $$E = 2E_0 \cos\left(\frac{\delta}{2}\right)\cos\left(\omega t + \frac{\delta}{2}\right)$$

Step 4: Intensity. The intensity is proportional to the time-averaged square of the field: $$I = I_0 \cos^2\left(\frac{\delta}{2}\right) = I_0 \cos^2\left(\frac{\pi d \sin\theta}{\lambda}\right)$$ where $I_0 = 4 \langle E_0^2 \rangle$ is four times the single-slit intensity (constructive interference doubles the amplitude, quadrupling the intensity).

Double-slit interference conditions

Bright fringes (constructive interference) occur when the path difference is a whole number of wavelengths:

$$d\sin\theta = m\lambda, \qquad m = 0, \pm 1, \pm 2, \ldots$$

Dark fringes (destructive interference) occur when the path difference is a half-integer number of wavelengths:

$$d\sin\theta = \left(m + \tfrac{1}{2}\right)\lambda$$

The fringe spacing on the screen is $\Delta y = \lambda L / d$.

Notice an important and initially counterintuitive feature: $\Delta y = \lambda L / d$ means that narrower slit spacing produces wider fringes. This inverse relationship between spatial scale and diffraction angle is a hallmark of wave phenomena and connects directly to the Heisenberg uncertainty principle in quantum mechanics.

From Classical Waves to Quantum Mechanics

Young's experiment is not just about light. In Lesson 6 (Quantum Mechanics Foundations), we saw that electrons, neutrons, and even molecules produce the same interference pattern when sent through a double slit, one particle at a time. The wave nature of matter and the wave nature of light are unified by quantum mechanics: both are described by probability amplitudes that interfere.

Interactive: Double Slit Experiment

Slit distance 40

Wavelength 20

Adjust slit distance and wavelength to see how the interference pattern changes.

Single-Slit Diffraction

Even a single slit produces a pattern on a distant screen, not just a uniform blob, but a central bright maximum flanked by progressively weaker subsidiary maxima separated by dark minima. This is single-slit diffraction, and it arises from the interference of wavelets emitted from different parts of the slit itself.

Deriving the Single-Slit Pattern

Consider a slit of width $a$ illuminated by a plane wave of wavelength $\lambda$. We divide the slit into infinitely many infinitesimal sources and add their contributions at a distant point P.

Step 1: Divide the slit. Consider a thin strip of width $dy$ at position $y$ within the slit ($-a/2 \leq y \leq a/2$). This strip contributes a wavelet with amplitude proportional to $dy$ and phase $\frac{2\pi}{\lambda} y \sin\theta$ relative to the center of the slit.

Step 2: Integrate. The total amplitude at angle $\theta$ is: $$E(\theta) \propto \int_{-a/2}^{a/2} e^{i k y \sin\theta} \, dy = \frac{e^{i k (a/2)\sin\theta} - e^{-i k (a/2)\sin\theta}}{ik\sin\theta}$$ Using Euler's formula: $$E(\theta) \propto a \cdot \frac{\sin\left(\frac{\pi a \sin\theta}{\lambda}\right)}{\frac{\pi a \sin\theta}{\lambda}}$$

Step 3: Define the sinc function. Introducing $\beta = \frac{\pi a \sin\theta}{\lambda}$, the amplitude is $E \propto a \cdot \text{sinc}(\beta)$ where $\text{sinc}(x) = \sin(x)/x$.

Step 4: Intensity. The intensity pattern is: $$I(\theta) = I_0 \left[\frac{\sin\beta}{\beta}\right]^2, \qquad \beta = \frac{\pi a \sin\theta}{\lambda}$$

Single-slit diffraction minima

The intensity is zero (dark fringes) when $\sin\beta = 0$ but $\beta \neq 0$, which gives:

$$a \sin\theta = m\lambda, \qquad m = \pm 1, \pm 2, \pm 3, \ldots$$

The central maximum has angular half-width $\theta_1 \approx \lambda/a$ (for small angles). The central maximum contains about 84% of the total diffracted power.

The single-slit diffraction pattern: a broad central maximum flanked by much weaker subsidiary maxima. The first zeros occur at sinθ = ±λ/a.

Why does the single slit give minima at $a\sin\theta = m\lambda$ while the double slit gives maxima at $d\sin\theta = m\lambda$?

For the double slit, the condition $d\sin\theta = m\lambda$ means the two beams are exactly in phase: constructive interference. For the single slit, $a\sin\theta = m\lambda$ means we can divide the slit into pairs of strips separated by $a/2$ whose contributions exactly cancel. The physics is the same (path-difference-based interference), but the geometry differs.

Diffraction Gratings

A diffraction grating is an array of $N$ equally spaced slits (or reflecting grooves) with spacing $d$. It is one of the most important optical instruments because it separates light into its component wavelengths with extraordinary precision.

The intensity pattern from $N$ slits is:

$$I(\theta) = I_{\text{single}} \left[\frac{\sin(N\alpha)}{\sin(\alpha)}\right]^2, \qquad \alpha = \frac{\pi d \sin\theta}{\lambda}$$

where $I_{\text{single}}$ is the single-slit diffraction envelope. The factor $[\sin(N\alpha)/\sin(\alpha)]^2$ produces extremely sharp principal maxima at the same positions as the double-slit maxima:

$$d\sin\theta = m\lambda, \qquad m = 0, \pm 1, \pm 2, \ldots$$

but now with angular width proportional to $1/N$. Between adjacent principal maxima, there are $N-2$ secondary maxima and $N-1$ minima. As $N$ increases, the principal maxima become sharper and the secondary maxima become negligible.

Resolving power of a grating

The resolving power $R$ of a grating measures its ability to distinguish two nearby wavelengths $\lambda$ and $\lambda + \Delta\lambda$:

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

where $m$ is the diffraction order and $N$ is the total number of slits. A grating with 10,000 slits working in second order ($m = 2$) can resolve wavelengths differing by as little as one part in 20,000. This is why diffraction gratings are the backbone of spectroscopy.

Thin-Film Interference

Some of the most beautiful optical effects in nature arise from thin-film interference: the iridescent colors of soap bubbles, oil slicks on wet pavement, and the wings of certain butterflies. These colors arise because light reflects from both the front and back surfaces of a thin film, and the two reflected beams interfere.

The Physics of Thin Films

Consider a thin film of thickness $t$ and refractive index $n$ surrounded by air (index $n_0 = 1$). Light striking the film at near-normal incidence reflects from both surfaces. The beam reflected from the back surface has traveled an extra optical path $2nt$ (twice through the film). However, there is a subtlety.

Phase change upon reflection

When light reflects from a medium with a higher refractive index, the reflected wave undergoes a phase shift of $\pi$ (equivalent to half a wavelength). When it reflects from a medium with a lower refractive index, there is no phase shift. This is analogous to a wave on a string reflecting from a fixed end (phase flip) versus a free end (no flip).

For a soap film (air-film-air), the first reflection (air to film, low to high index) picks up a $\pi$ phase shift, while the second reflection (film to air, high to low) does not. The total phase difference between the two reflected beams is:

$$\delta = \frac{2\pi}{\lambda}(2nt) + \pi$$

Constructive interference (bright reflection) occurs when $\delta = 2m\pi$:

$$2nt = \left(m - \frac{1}{2}\right)\lambda, \qquad m = 1, 2, 3, \ldots$$

Destructive interference (no reflection, meaning the light is transmitted) occurs when:

$$2nt = m\lambda, \qquad m = 0, 1, 2, \ldots$$

When white light illuminates a thin film, different wavelengths satisfy the constructive condition for different thicknesses or viewing angles. This is why soap bubbles display swirling rainbow colors as their thickness varies across the surface, and why the colors change as the bubble thins and eventually pops.

Why do very thin soap films appear black just before they pop?

When the film thickness approaches zero ($t \to 0$), the path difference $2nt$ vanishes. The only phase difference comes from the $\pi$ shift at the first surface, giving destructive interference for all wavelengths. No light is reflected, so the film appears black. This is a striking confirmation of the phase-change rule.

Coherence

Interference requires that the combining waves maintain a stable phase relationship. This property is called coherence. There are two types.

Temporal Coherence

Temporal coherence and coherence length

Temporal coherence describes how well a wave maintains a consistent phase over time. A perfectly monochromatic wave (single frequency) has infinite temporal coherence. A real light source emits wave trains of finite duration $\tau_c$ (the coherence time). The coherence length is $\ell_c = c\tau_c$.

Two portions of a beam can interfere only if their path difference is less than $\ell_c$. The coherence length is related to the spectral bandwidth $\Delta\nu$ by:

$$\ell_c \approx \frac{c}{\Delta\nu} \approx \frac{\lambda^2}{\Delta\lambda}$$

An ordinary incandescent bulb has $\ell_c \sim 1$ $\mu$m (a few wavelengths), which is why everyday light sources do not easily produce visible interference fringes. A laser, by contrast, can have $\ell_c$ ranging from centimeters to kilometers, enabling interference over macroscopic distances.

Spatial Coherence

Spatial coherence and the van Cittert-Zernike theorem

Spatial coherence describes the correlation between the wave at two different points in space, measured at the same time. An extended source (like the Sun) produces spatially incoherent light: the phase at one point on the wavefront is unrelated to the phase at a distant point. A point source, or a source very far away, produces spatially coherent light.

The van Cittert-Zernike theorem states that the spatial coherence of light from an incoherent source of angular size $\alpha$ has a coherence radius:

$$r_c \approx \frac{\lambda}{\alpha}$$

This is why starlight (tiny angular size) is spatially coherent enough to produce interference.

Young's double-slit experiment requires spatial coherence across both slits. If the source has spatial coherence radius $r_c < d$ (where $d$ is the slit separation), the fringe visibility degrades. This is why Young used a single narrow source slit to improve spatial coherence before the double slit.

Coherence and Quantum Mechanics

The concept of coherence extends far beyond optics. In quantum mechanics (Lessons 6-11), a quantum system in a superposition state $|\psi\rangle = \alpha|0\rangle + \beta|1\rangle$ is "coherent" when the off-diagonal terms of the density matrix are nonzero. Decoherence, the loss of these off-diagonal terms through interaction with the environment, is the quantum analog of losing interference fringe visibility. The mathematics is strikingly similar.

Exercises

Exercise 1: Double-Slit Fringe Spacing

A double slit with separation $d = 0.25$ mm is illuminated with light of wavelength $\lambda = 550$ nm. The screen is $L = 1.5$ m away. Calculate (a) the fringe spacing, (b) the angle to the third bright fringe, and (c) how the pattern changes if the experiment is performed underwater ($n = 1.33$).

(a) Fringe spacing: $$\Delta y = \frac{\lambda L}{d} = \frac{(550 \times 10^{-9})(1.5)}{0.25 \times 10^{-3}} = 3.3 \times 10^{-3} \text{ m} = 3.3 \text{ mm}$$

(b) Third bright fringe ($m = 3$): $$\sin\theta_3 = \frac{3\lambda}{d} = \frac{3 \times 550 \times 10^{-9}}{0.25 \times 10^{-3}} = 6.6 \times 10^{-3}$$ $$\theta_3 = 0.38°$$

(c) Underwater: The wavelength in the medium is $\lambda' = \lambda/n = 550/1.33 = 413$ nm. The fringe spacing decreases to $\Delta y' = \lambda' L / d = 2.5$ mm.

Exercise 2: Resolving Power

A diffraction grating has 6000 lines/cm and is 3 cm wide. (a) What is its resolving power in second order? (b) Can it resolve the sodium doublet at $\lambda = 589.0$ nm and $\lambda = 589.6$ nm?

(a) Total number of slits: $N = 6000 \times 3 = 18{,}000$. Resolving power: $R = mN = 2 \times 18{,}000 = 36{,}000$.

(b) Required resolving power: $R_{\text{req}} = \lambda / \Delta\lambda = 589.0 / 0.6 = 982$. Since $36{,}000 \gg 982$, the grating easily resolves the sodium doublet. In fact, this grating could resolve wavelength differences as small as $\Delta\lambda = 589/36{,}000 \approx 0.016$ nm.

Exercise 3: Thin-Film Colors

A soap film ($n = 1.33$) is illuminated with white light at normal incidence. What is the minimum film thickness that produces a bright reflection for green light ($\lambda = 520$ nm)? What other visible wavelengths are strongly reflected at this thickness?

Minimum thickness: For constructive interference (with the extra $\pi$ phase shift): $$2nt = \left(m - \frac{1}{2}\right)\lambda$$ The minimum thickness corresponds to $m = 1$: $$t = \frac{\lambda}{4n} = \frac{520}{4 \times 1.33} = 97.7 \text{ nm}$$

Other wavelengths: At this thickness, the constructive condition $2nt = (m - 1/2)\lambda$ gives $\lambda = 2nt/(m - 1/2) = 260/(m - 1/2)$. For $m = 1$: $\lambda = 520$ nm (green). For $m = 2$: $\lambda = 173$ nm (ultraviolet, not visible). So at this minimum thickness, only green light is strongly reflected. At larger thicknesses, multiple visible wavelengths can satisfy the condition simultaneously, producing mixed colors.

Exercise 4: Coherence Length

A helium-neon laser emits light at $\lambda = 632.8$ nm with a spectral linewidth of $\Delta\nu = 1.5$ GHz. (a) What is the coherence length? (b) A sodium lamp emits the D-line at $\lambda = 589.3$ nm with $\Delta\lambda = 0.6$ nm. What is its coherence length? (c) Which source can produce interference fringes in a Michelson interferometer with a path difference of 10 cm?

(a) Laser: $$\ell_c = \frac{c}{\Delta\nu} = \frac{3 \times 10^8}{1.5 \times 10^9} = 0.20 \text{ m} = 20 \text{ cm}$$

(b) Sodium lamp: $$\ell_c = \frac{\lambda^2}{\Delta\lambda} = \frac{(589.3 \times 10^{-9})^2}{0.6 \times 10^{-9}} = 5.8 \times 10^{-4} \text{ m} = 0.58 \text{ mm}$$

(c) The laser ($\ell_c = 20$ cm) can produce fringes at 10 cm path difference. The sodium lamp ($\ell_c = 0.58$ mm) cannot, since the path difference far exceeds its coherence length.

Key Takeaways

Huygens' principle explains how waves propagate and diffract: every point on a wavefront is a source of secondary wavelets.
Young's double-slit experiment produces an intensity pattern $I = I_0 \cos^2(\pi d\sin\theta/\lambda)$, with fringe spacing $\Delta y = \lambda L/d$.
Single-slit diffraction gives a sinc-squared pattern $I \propto [\sin(\beta)/\beta]^2$ with minima at $a\sin\theta = m\lambda$.
Diffraction gratings sharpen the interference maxima and achieve resolving power $R = mN$, enabling precision spectroscopy.
Thin-film interference produces colorful reflections due to the interplay of optical path differences and phase shifts upon reflection.
Coherence (temporal and spatial) determines the ability of light to produce interference; lasers have much higher coherence than thermal sources.

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