Thin Films, Wedges & Newton's Rings

Set a film of index n and thickness t in air and send in light of vacuum wavelength λ₀ at near-normal incidence. Part reflects at the top face; part enters, reflects at the bottom face and comes back out, having travelled an extra 2t in a medium where the wavelength is λ₀/n. That second wave has advanced through 2t ÷ (λ₀/n) = 2nt/λ₀ extra cycles: its phase contribution is 2π(2nt)/λ₀, and 2nt, not 2t, is the optical path difference. At angle of refraction θₜ inside the film the internal path lengthens to 2t/cos θₜ, while the top-reflected wave gains an air path 2t tan θₜ sin θᵢ; with Snell's law sin θᵢ = n sin θₜ the difference collapses to Δ = 2nt cos θₜ, which is within 1 % of 2nt for θₜ below 8°. A 400 nm layer of n = 1.50 at normal incidence gives Δ = 1200 nm: two full wavelengths of 600 nm light but 2.4 wavelengths of 500 nm light. The path is fixed by geometry; what it means depends on the colour.

Superposition needs the phase of each reflected wave, and reflection itself can supply a phase. At normal incidence the reflected amplitude is r = (n₁ − n₂)/(n₁ + n₂), where n₁ is the medium the wave arrives from. Air to glass gives r = (1.00 − 1.52)/2.52 = −0.206: a negative amplitude, which for a sinusoid is a π phase shift. The field reverses on reflection from a denser medium, like a pulse on a string meeting a fixed end. Glass to air gives r = +0.206, the same 4.3 % reflectance and no reversal. Only the difference between the two reflections enters the budget. A soap film in air: top face flips, bottom does not, net π. An air gap between glass plates: top face (glass to air) does not flip, bottom (air to glass) does, net π. A MgF₂ coating (1.38) on glass (1.52): both faces flip, net zero. Oil (1.45) on water (1.33): top flips, bottom does not, net π. Write these two entries down before touching 2nt; most wrong answers in this topic go wrong here.

With both entries the total phase is δ = 4πnt/λ₀ + φᵣ, with φᵣ = 0 or π, and the reflected intensity follows the two-beam rule I = I₁ + I₂ + 2√(I₁I₂) cos δ. Maxima sit at δ = 2πm. With one flip that reads 2nt = (m + ½)λ₀ for bright and 2nt = mλ₀ for dark; with no flip, or two, the conditions swap. The rule assumes exactly two beams, and the Fresnel numbers say why that is safe. For a glass plate in air the second beam leaves with amplitude (1 − r²)r ≈ 0.96r, having crossed the top face twice; the third beam, which makes another round trip, carries (1 − r²)r³ ≈ 0.008 relative to the incident field, only 4 % of the second. Its cross terms move the peak intensity by about 4 %, so the fringes are the smooth cos² of two-beam interference. Silver the faces to r ≈ 0.95 and the neglected beams become the whole story: the sum turns into the Airy function and the fringes sharpen into the narrow rings of a Fabry–Pérot. The two-beam formula is a weak-reflection limit, and its own coefficients say so.

An antireflection layer uses the budget backwards. A film of lower index than the substrate flips both reflections, so the reflection phases cancel, and the two beams then cancel when 2n_f t = λ₀/2: a quarter-wave layer, t = λ₀/(4n_f). MgF₂ on crown glass at 550 nm needs t = 550/(4 × 1.38) = 99.6 nm. Cancellation is complete only if the amplitudes match, (1 − n_f)/(1 + n_f) = (n_f − nₛ)/(n_f + nₛ), which solves to n_f = √nₛ = 1.233; no durable material has it, and MgF₂'s 1.38 leaves r₁ = −0.160 against r₂ = −0.048, which the two crossings of the top face trim to −0.047, so the residual is (0.160 − 0.047)² ≈ 1.3 % against 4.3 % for bare glass. Away from the design wavelength δ leaves π: at 400 nm, δ = 4π(1.38)(99.6)/400 = 4.32 rad and R ≈ 2.2 %; at 700 nm, δ = 2.47 rad and R ≈ 1.6 %. Red and blue leak more than green, which is why a coated lens looks faintly purple by reflection. Multilayer stacks widen the notch; they are the same budget summed over more faces.

Let the thickness vary as t = αx, an air wedge between two flats touching along x = 0. The reflection budget is one flip (glass to air, then air to glass), so dark fringes lie where 2αxₘ = mλ₀: straight lines parallel to the contact edge, the m = 0 dark fringe on the edge itself, spaced Δx = λ₀/(2α). Each fringe is a contour of constant thickness — a Fizeau fringe — and crossing one means the gap grew by λ₀/2. Fill the wedge with a liquid of index n and the spacing shrinks to λ₀/(2nα), because the optical path is what counts. The scale is ferocious. A 25 μm strip under one end of 100 mm flats gives α = 2.5 × 10⁻⁴ rad, and 546 nm light lays down fringes every 1.09 mm; 2t/λ₀ = 92 fringes cross the plate. Lay a reference flat over a surface under test and a straight fringe reports a flat surface, while a fringe that bows by half a spacing reports a bump or hollow of λ₀/4, about 140 nm, seen by eye with nothing but a lamp.

Rest a convex lens of radius of curvature R on a flat and the air gap at distance r from the contact is a sagitta: R² = (R − t)² + r² gives t = r²/(2R) once t² is dropped, an error of two parts in 10⁶ for R = 2.00 m and r = 5 mm. One flip again, so dark rings sit at 2t = mλ₀, that is rₘ = √(mλ₀R), and bright rings at r² = (m + ½)λ₀R. The centre, where t = 0, is dark: a zero path difference producing cancellation is the reflection phase made visible, and Young used exactly this spot to argue for it. With R = 2.00 m and λ₀ = 589 nm the first dark ring has radius √(1.178 × 10⁻⁶ m²) = 1.09 mm and the tenth 3.43 mm; the rings crowd together because dr/dm = λ₀R/(2r) falls as r grows. Because rₘ² is linear in m, a plot of rₘ² against m has slope λ₀R and an intercept that absorbs any dust-induced offset in the gap, so the radius of a lens can be measured to three figures without knowing which ring is which. Filling the gap with a liquid divides every rₘ² by n.