Optical Physics: Waves, Coherence and Interference, Fresnel and Fraunhofer Diffraction, Gratings, Jones Calculus, Ray Matrices and Lasers
1. The wave equation, superposition, standing waves, phase and group velocity
The wave equation ∇²ψ = (1/v²)∂²ψ/∂t² has plane waves A cos(k·r − ωt), of constant amplitude, and spherical waves (A/r)cos(kr − ωt), whose amplitude falls as 1/r so that the power through any sphere is the same. It is linear, so solutions superpose. Two equal waves travelling in opposite directions give a standing wave 2A sin kx cos ωt, with nodes λ/2 apart that carry no energy. Two waves of slightly different frequencies give beats at |f₁ − f₂|.
A single frequency travels at the phase velocity v_p = ω/k; a wave packet’s envelope, which carries energy and information, travels at the group velocity v_g = dω/dk = v_p − λ dv_p/dλ. They are equal only in a non-dispersive medium. For deep-water waves ω = √(gk), so v_g = v_p/2; for ω = ak², v_g = 2v_p; for light in a plasma or a waveguide, ω² = ω_c² + c²k², so v_p > c while v_g < c and v_p v_g = c².
2. Coherence and interference: films, Newton’s rings, Michelson and Fabry–Perot
Interference needs a stable phase relation. Temporal coherence is set by the bandwidth: a source of width Δλ has coherence length l_c ≈ λ²/Δλ (for 600 nm and Δλ = 0.01 nm, 36 mm) and coherence time l_c/c; path differences beyond l_c wash the fringes out. Spatial coherence is set by the source’s angular size θ_s: two points are coherent if their separation is less than about λ/θ_s — why a distant star can illuminate a double slit coherently while a nearby lamp cannot.
| Arrangement | Condition | Note |
|---|---|---|
| Thin film in reflection (film denser than both sides) | 2nt cos r = (m + ½)λ bright; = mλ dark | one half-wave shift, at the top surface |
| Anti-reflection coating (n_air < n < n_glass) | nt = λ/4, ideally n = √n_glass | two shifts cancel; MgF₂ (1.38) at 550 nm needs t = 99.6 nm |
| Newton’s rings in reflection | dark: r_m² = mλR | centre dark; radii grow as √m |
| Michelson interferometer | mirror moved d ⇒ N = 2d/λ fringes | the path changes by twice the mirror’s motion |
Multiple-beam interference in a Fabry–Perot interferometer (two parallel mirrors of reflectance R; an etalon when the spacing is fixed) gives the Airy transmission T = 1/(1 + F sin²(δ/2)), with coefficient of finesse F = 4R/(1 − R)². Transmission peaks at 2nd cos θ = mλ, and their sharpness is the finesse ℱ = π√R/(1 − R) — the ratio of peak spacing (free spectral range) to peak width. For R = 0.9, ℱ ≈ 29.8. The resolving power λ/Δλ ≈ mℱ is far above a two-beam instrument’s, which is why the etalon resolves hyperfine structure.
3. Diffraction: apertures, resolution, slits and gratings
Diffraction is Fraunhofer when source and screen are effectively at infinity (the Fresnel number a²/(Lλ) ≪ 1, or lenses are used) and Fresnel when the wavefront’s curvature across the aperture matters. A single slit of width a has minima at a sin θ = mλ (m ≠ 0) and intensity I₀(sin β/β)² with β = πa sin θ/λ; a rectangular aperture gives the product of two such patterns. A circular aperture of diameter D gives the Airy disc, first zero at sin θ = 1.22λ/D. The Rayleigh criterion — one image’s maximum on the other’s first minimum — sets the limit of resolution θ_min = 1.22λ/D: a 10 cm telescope at 550 nm resolves 6.71 μrad.
- Double slit (separation d, width a): interference fringes cos²(πd sin θ/λ) under the single-slit envelope. An interference maximum that falls on a diffraction zero is a missing order: m = d/a × (integer). With d = 3a the orders ±3 are missing, so 5 bright fringes (m = 0, ±1, ±2) lie within the central envelope.
- N slits: principal maxima at d sin θ = mλ with intensity ∝ N², N − 1 minima and N − 2 secondary maxima between neighbouring principal maxima, and principal maxima of angular width ∝ 1/N.
- Grating: d sin θ = mλ, so the highest order is the largest m below d/λ (5000 lines/cm at 600 nm: d = 2 μm, m_max = 3). Angular dispersion dθ/dλ = m/(d cos θ); resolving power λ/Δλ = mN. Resolving 500.0 nm from 500.1 nm (λ/Δλ = 5000) in second order needs N = 2500 illuminated lines.
4. Jones calculus, birefringence and ray-transfer matrices
| Element or state | Jones form |
|---|---|
| Linear, horizontal / at 45° | (1, 0) / (1, 1)/√2 |
| Circular (the two senses) | (1, i)/√2 and (1, −i)/√2 |
| Elliptical | (a, b eiφ) with a ≠ b or φ ≠ ±π/2 |
| Horizontal linear polariser | [[1, 0], [0, 0]] |
| Quarter-wave plate, fast axis horizontal | [[1, 0], [0, i]] (up to a phase) |
| Half-wave plate, fast axis horizontal | [[1, 0], [0, −1]] |
A birefringent crystal has two indices, n_o for the ordinary ray and n_e for the extraordinary ray; a plate of thickness t retards one polarisation by Γ = 2πΔn t/λ. A quarter-wave plate has Δn t = λ/4 — for Δn = 0.009 at 540 nm, t = 15 μm — and turns linear light at 45° to its axes into circular light (the QWP matrix acting on (1, 1)/√2 gives (1, i)/√2). A half-wave plate rotates linear polarisation by twice the angle between it and the fast axis. Behind a polariser, Malus’s law I = I₀cos²θ applies.
Paraxial rays are the vector (y, θ), and each element is a 2 × 2 ray-transfer (ABCD) matrix: free propagation over d is [[1, d], [0, 1]]; a thin lens of focal length f is [[1, 0], [−1/f, 1]]; a spherical mirror of radius R acts as a lens of f = R/2, [[1, 0], [−2/R, 1]]. A system is the product in reverse order of travel. A lens followed by distance f gives [[0, f], [−1/f, 1]]: A = 0, so every parallel ray arrives at y = 0 — the focal point. In a common medium the determinant is 1.
5. Lasers: Einstein coefficients and population inversion
Between levels 1 and 2 (equal degeneracies) three processes compete: absorption at rate B₁₂ρ(ν)N₁, stimulated emission B₂₁ρ(ν)N₂ and spontaneous emission A₂₁N₂. Demanding that equilibrium with blackbody radiation reproduce Planck’s law gives B₁₂ = B₂₁ (for g₁ = g₂) and A₂₁/B₂₁ = 8πhν³/c³. The ratio of stimulated to spontaneous emission in thermal radiation is 1/(ehν/k_BT − 1) — tiny for visible light at room temperature — so stimulated emission can dominate only with an intense, non-thermal field and more atoms in the upper level than the lower: population inversion, which thermal equilibrium (N₂/N₁ = e−hν/k_BT < 1) never provides.
- Two-level system: optical pumping drives absorption and stimulated emission equally, so at best N₂ → N₁ (saturation); steady inversion is impossible.
- Three-level system (ruby): pump 1 → 3, fast decay to the metastable level 2, lasing 2 → 1. Since the lower laser level is the ground state, more than half the atoms must be pumped — a high threshold, often pulsed.
- Four-level system (Nd:YAG, He–Ne): the lower laser level lies above the ground state and empties quickly, so any population in the upper level is already an inversion — a low threshold and continuous operation.
- Threshold: the round-trip gain must equal the round-trip loss, R₁R₂e2(g − α)L = 1, and the cavity selects longitudinal modes spaced c/(2L).
Key takeaways
- v_g = dω/dk carries the energy; v_g = v_p/2 for deep water, 2v_p for ω ∝ k², and v_p v_g = c² for ω² = ω_c² + c²k².
- l_c = λ²/Δλ; Newton’s dark rings r_m² = mλR; Michelson counts 2d/λ fringes; Fabry–Perot finesse π√R/(1 − R).
- θ_min = 1.22λ/D; missing orders at m = (d/a) × integer; grating resolving power mN and highest order below d/λ.
- Circular light is (1, ±i)/√2; a QWP needs Δn t = λ/4; ray matrices multiply in reverse order of travel.
- A/B = 8πhν³/c³; two levels cannot invert, three levels need over half pumped, four levels invert easily.
Practice questions (15)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
A medium has the dispersion relation ω = ak² (a constant). The group velocity is:
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Answer: A — Twice the phase velocity
v_p = ω/k = ak and v_g = dω/dk = 2ak = 2v_p — the free-particle relation of quantum mechanics, where the group velocity is the particle’s speed. Half the phase velocity is the deep-water case ω ∝ √k.Newton’s rings are formed in reflected light of wavelength 589 nm with a lens of radius of curvature 1.00 m. The radius of the 10th dark ring, in mm, to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 2.43
r_m = √(mλR) = √(10 × 589 × 10⁻⁹ × 1.00) = √(5.89 × 10⁻⁶) = 2.43 × 10⁻³ m = 2.43 mm. Using the bright-ring formula √((m + ½)λR) gives 2.49 mm, and dropping the square root is dimensionally wrong.In a Michelson interferometer, 320 fringes cross the field of view when the movable mirror is displaced by 0.100 mm. The wavelength, in nm, is ____.
Numerical answer — type the value.
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Answer: 625
Each fringe corresponds to a path change of λ, and moving the mirror by d changes the path by 2d: λ = 2d/N = 2 × 0.100 × 10⁻³/320 = 6.25 × 10⁻⁷ m = 625 nm. Forgetting the factor 2 gives 312.5 nm.A MgF₂ film (n = 1.38) is used as an anti-reflection coating on glass (n = 1.52) for λ = 550 nm at normal incidence. The minimum film thickness, in nm, to one decimal place, is ____.
Numerical answer — type the value.
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Answer: 99.6
Both reflections (air→film, film→glass) are from denser media, so both shift by λ/2 and cancel; destructive interference needs 2nt = λ/2, t = λ/(4n) = 550/(4 × 1.38) = 99.6 nm. Adding a half-wave for only one surface gives t = λ/(2n) = 199.3 nm.A source emits at a mean wavelength of 600 nm with a spectral width of 0.01 nm. Its coherence length, in mm, is ____.
Numerical answer — type the value.
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Answer: 36
l_c = λ²/Δλ = (600 × 10⁻⁹)²/(0.01 × 10⁻⁹) = 3.6 × 10⁻¹³/10⁻¹¹ = 0.036 m = 36 mm. Using λ/Δλ = 6 × 10⁴ is the number of wavelengths in the coherence length, not a length.A Fabry–Perot interferometer has mirrors of reflectance R = 0.9. Its finesse, to one decimal place, is ____.
Numerical answer — type the value.
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Answer: 29.8
ℱ = π√R/(1 − R) = 3.1416 × 0.9487/0.1 = 29.8. The coefficient of finesse F = 4R/(1 − R)² = 360 is a different quantity; ℱ = (π/2)√F.A telescope has an objective of diameter 10 cm. For λ = 550 nm, its angular limit of resolution by the Rayleigh criterion, in μrad, to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 6.71
θ = 1.22λ/D = 1.22 × 550 × 10⁻⁹/0.1 = 6.71 × 10⁻⁶ rad. Omitting 1.22 (the slit result λ/a) gives 5.50 μrad; the 1.22 is the first zero of the Airy pattern of a circular aperture.In a double-slit experiment the slit separation is three times the slit width. The number of bright interference fringes within the central diffraction maximum is ____.
Numerical answer — type the value.
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Answer: 5
The central envelope ends at a sin θ = λ, i.e. d sin θ = 3λ, which is where the m = ±3 fringes would be — they are missing. Inside lie m = 0, ±1, ±2: five fringes. Counting the missing ±3 gives 7; 2d/a = 6 counts spacings, not fringes.Two spectral lines at 500.0 nm and 500.1 nm are to be just resolved in the second order of a diffraction grating. The minimum number of illuminated lines is ____.
Numerical answer — type the value.
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Answer: 2500
λ/Δλ = mN, so N = λ/(mΔλ) = 500.0/(2 × 0.1) = 2500. Ignoring the order gives 5000 — the first-order requirement.A grating has 5000 lines per cm and is illuminated normally with light of wavelength 600 nm. The highest order that can be observed is:
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Answer: A — 3
d = 1 cm/5000 = 2 μm, and sin θ ≤ 1 requires m ≤ d/λ = 2/0.6 = 3.33, so m_max = 3. Rounding 3.33 up to 4 would need sin θ > 1.Light linearly polarised at 45° to the axes, Jones vector (1, 1)/√2, passes through a quarter-wave plate with Jones matrix [[1, 0], [0, i]]. The emerging light is:
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Answer: A — Circularly polarised, (1, i)/√2
[[1, 0], [0, i]](1, 1)/√2 = (1, i)/√2: equal amplitudes 90° apart — circular. A half-wave plate [[1, 0], [0, −1]] would give (1, −1)/√2, linear at −45°.A quartz plate has n_e − n_o = 0.009. The minimum thickness that makes it a quarter-wave plate at 540 nm, in μm, is ____.
Numerical answer — type the value.
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Answer: 15
Δn t = λ/4 gives t = 540 × 10⁻⁹/(4 × 0.009) = 1.5 × 10⁻⁵ m = 15 μm. The half-wave condition Δn t = λ/2 gives 30 μm.A thin lens of focal length f is followed by free propagation over a distance f. The ray-transfer matrix of the combination, and what it implies, is:
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Answer: A — [[0, f], [−1/f, 1]]; every incoming parallel ray reaches the axis
Multiply in reverse order: [[1, f], [0, 1]][[1, 0], [−1/f, 1]] = [[1 − 1, f], [−1/f, 1]] = [[0, f], [−1/f, 1]]. With A = 0, y_out = f θ_in, so all rays entering parallel (θ_in = 0) arrive at y = 0: the focal plane. Multiplying in the order of travel gives [[1, f], [−1/f, 0]], the wrong system.Which statements about population inversion are correct?
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Answer: A — It cannot be achieved in steady state by optical pumping of a two-level system; B — In a three-level laser whose lower laser level is the ground state, more than half the atoms must be pumped; C — In a four-level laser the lower laser level empties quickly, so inversion is easy to reach
With B₁₂ = B₂₁, pumping two levels only equalises the populations. The three-level ruby laser lases to the ground state, so N₂ > N₁ needs over half the atoms up; the four-level scheme lases to an almost empty level. In equilibrium N₂/N₁ = e−hν/k_BT < 1 at every positive temperature.For two levels of equal degeneracy, the ratio of the Einstein coefficients A₂₁/B₂₁ is proportional to:
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Answer: A — ν³
Equilibrium with blackbody radiation requires A₂₁/B₂₁ = 8πhν³/c³: the mode density (∝ ν²) times the photon energy hν. The ν³ is why spontaneous emission overwhelms stimulated emission at short wavelengths and X-ray lasers are hard to build.