Optics · Chapter 18

Wave Optics

When the scale of optical elements approaches the wavelength of light, geometric optics breaks down. Wave optics explains interference, diffraction, and the fundamental resolution limit of every optical instrument.

Prerequisites

Wave properties (Ch. 8) \cdot Interference (Ch. 11) \cdot Geometric optics (Ch. 17

Learning Goals

Predict bright and dark fringe positions in a double-slit experiment using the path difference condition.
$Calculate fringe spacing \Deltay = \lambdaL/d and infer wavelength from observed patterns$ .
Apply the Rayleigh criterion to find the diffraction-limited resolution of an optical instrument.
Determine constructive and destructive reflection conditions for thin films, accounting for phase shifts.
Explain how anti-reflection coatings exploit destructive interference to minimize reflection losses.

18.1 The Wave Nature of Light

Light is an electromagnetic wave with wavelength roughly 380–780 nm. Its wave nature becomes apparent only when it interacts with objects or apertures comparable in size to its wavelength. For everyday objects (centimeters), λ/object ≈ 10⁻⁵ and light behaves as a ray. For a 500 nm aperture, wave effects dominate entirely.

The principle of superposition governs wave optics: at any point, the total electric field is the sum of contributions from all sources. Constructive interference (crests meet crests) produces bright fringes; destructive interference (crests meet troughs) produces dark fringes.

18.2 Young's Double-Slit Experiment

In 1801, Thomas Young demonstrated light's wave nature by shining monochromatic light through two narrow slits separated by distance d and observing alternating bright and dark bands on a screen at distance L. The path difference from the two slits to a point at height y on the screen is Δ = d sin θ ≈ dy/L for small angles.

Theorem 18.1 — Double-Slit Interference Conditions

Bright fringes (constructive): \qquad d sin \theta = m\lambda \qquad (m = 0, \pm1, \pm2, \cdots

Dark fringes (destructive): \qquad d sin \theta = (m + \frac{1}{2})\lambda

Fringe spacing on screen: \qquad \Deltay = \lambdaL/d \qquad (small angle

The fringe spacing Δy = λL/d reveals the wavelength: smaller d or larger L spreads the fringes. This is how wavelengths of light were first measured with precision.

Figure 18.1. Double-slit in

terference pattern. The colored band on the right is the intensity at the screen. Change \lambda

ontrols fringe color), d (slit separation controls fringe spacing), and a (slit width controls the single-slit envelope that modulates the pattern). Notice how narrower slits spread the envelope.

Example 18.1 — Fringe Spacing in Young's Experiment

Light of wavelength 589 nm passes through two slits 0.25 mm apart. A screen is 1.2 m away. Find the distance between adjacent bright fringes.

Formula:

\Deltay = \lambdaL/d

Substitute:

\Deltay = (589\times10^{-9} m \times 1.2 m) / (0.25\times10^{-3} m

Calculate:

\Deltay = 7.068\times10^{-7} / 2.5\times10^{-4} = 2.83\times10^{-3} m =

2.83 mm

18.3 Single-Slit Diffraction

Even a single slit of finite width a produces a diffraction pattern. Each point within the slit acts as a secondary wave source (Huygens' principle). The intensity pattern is:

I(\theta) = I_{0} [sin(\beta/2) / (\beta/2)]^{2} \qquad where \beta = (2\pia/\lambda) sin \theta

(18.1)

This sinc² function gives a central maximum flanked by minima at a sin θ = mλ (m = ±1, ±2, …). The central maximum has width 2λ/a — a narrower slit creates awider diffraction pattern. This reciprocal relationship between object size and diffraction spread is a fundamental feature of wave physics (and of Fourier transforms).

Definition 18.1 — Rayleigh Criterion — Resolution Limit

Two point sources are just resolved when the central maximum of one falls on the first minimum of the other. For a circular aperture of diameter D:

\theta_min = 1.22 \lambda/D

This is

the diffraction-limited resolution of a telescope, microscope, or eye. Larger aperture D \to

resolution. This formula ends the dream of perfect optical images — all real optical instruments are resolution-limited by diffraction.

Example 18.2 — Resolution of the Human Eye

The pupil di $ameter in bright light is about 2 mm. What is the angular resolution of the human eye at \lambda$

Rayleigh criterion:

\theta_min = 1.22 \times \lambda/D = 1.22 \times 550\times10^{-9} / 2\times10^{-3}

Calculate:

\theta_min = 1.22 \times 2.75\times10^{-4} = 3.35\times10^{-4} rad \approx 1.2 arcminutes

Context:

At arm's length (60 cm), this resolves features separated by 60 cm \times 3.35\times10^{-4} = 0.2 mm. T

he retina's cone spacing (\approx3 \mum at the fovea) matches this limit

18.4 Thin Film Interference

When light reflects off the top and bottom surfaces of a thin film (soap bubble, oil slick, lens coating), the two reflected waves interfere. The path difference is approximately 2t (where t is the film thickness), but reflections at boundaries where n increases introduce an extra half-wavelength phase shift.

Theorem 18.2 — Thin Film Conditions (film of index n, in air)

Light reflects with phase shift at air\tofilm boundary (lower\tohigher n), but not at film\toair

Constructive reflection: \qquad 2nt = (m + \frac{1}{2})\lambda \qquad (m = 0, 1, 2, \cdots

Destructive reflection: \qquad 2nt = m\lambda

The wavelength in the film is \lambda/n, so effective path is scaled by n

This is why soap bubbles show rainbow colors: different wavelengths constructively interfere at different film thicknesses. Anti-reflection coatings on camera lenses and eyeglasses are designed to cancel reflected light: choose thickness t = λ/(4n) so 2nt = λ/2, producing destructive reflection for the design wavelength.

Definition 18.2 — Common Traps

Use path difference for phase: bright and dark conditions depend on relative phase, not absolute distance.
Single-slit diffraction envelopes double-slit fringes: real slits are not point sources.
Rayleigh resolution is an angular limit: convert to linear separation only after using the distance to the image or object.
Thin-film phase flips matter: reflection from a higher-index boundary adds a half-cycle phase shift.

Exercises — 18.1–18.4 Wave Optics

Light of 589 nm passes through two slits 0.25 mm apart, screen at 1.2 m. Find the fringe spacing.

Straightforward

Find the diffraction-limited angular resolution of a 2 mm pupil at \lambda = 550 nm

rad

Straightforward

An anti-reflection coating (n = 1.38) is applied for \lambda = 800 nm. Find the minimum thicknes

s for destructive reflection.

Intermediate

4.A soap bubble appears bright in the red (600 nm) at what minimum film thickness? What color does a very thin soap film (approaching zero thickness) appear, and why?

Intermediate

5.C

alculate the diffraction-limited angular resolution of (a) the Hubble Space Telescope (D =

2.4 m, \lambda = 550 nm), (b) the James Webb Space Telescope (D = 6.5 m, \lambda = 2 \mum), (c) the ELT

ground telescope (D = 39 m). Express in arcseconds

Challenging

Key Takeaways

$Wave optics governs when aperture or object size \approx \lambda; geometric optics fails in this regim$ e.
$Double-slit: bright fringes at d sin\theta = m\lambda; fringe spacing \Deltay = \lambdaL/d on a distant screen$ .
$Single-slit diffraction: I \propto sinc^{2}(\pia sin\theta/\lambda); narrower slit \to wider central maximum$ .
$Rayleigh criterion: \theta_min = 1.22\lambda/D — diffraction limits resolution of all optical instrum$ ents.
$Thin film: constructive reflection when 2nt = (m+\frac{1}{2})\lambda (accounting for phase shifts at bound$ aries).
Anti-reflection coatings use destructive interference to eliminate reflection at a design wavelength.