Newton's Rings

Newton's rings never left the optical shop. Lay a precision test flat on a lens or mirror under monochromatic light and the equal-thickness fringes map the air gap between them: a perfectly figured surface gives clean circles, and every departure — a turned edge, a low zone, a scratch — bends a fringe by an amount that reads the height error to a small fraction of a wavelength. Counting or fringe-analysing these patterns is how flats, prisms, and lens surfaces are still certified to $\lambda/10$ or better, and modern phase-shifting interferometers are just this idea automated with a piezo-scanned reference and a camera. The same physics shows up as a nuisance to be engineered away. Wherever two nearly-flat surfaces are pressed together — a phone screen laminated to its glass, a photographic negative in a scanner gate, two bonded optical windows — a thin residual air gap breeds unwanted Newton's rings that mar the image. The cures are all gap control: index-matching fluids or adhesives to erase the optical path difference, deliberate anti-Newton-ring textures to break the coherence of the gap, or vacuum bonding to remove the air entirely. The dark centre that puzzled Newton is now a manufacturing spec.

Think of the air gap as a very shallow wedge, flat at the middle and opening up as you move outward. A ray coming down is split into two: a little reflects off the curved underside of the lens, and the rest crosses the gap and reflects off the flat plate below. The second beam travels an extra round trip $2t$ through the gap before rejoining the first, and wherever that extra path — with the reflection phase folded in — lands the two beams out of step, you see a dark ring. Because $t$ is fixed by radius alone, "out of step" happens on circles, and the eye reads off a set of concentric rings centred on the contact point.

The half-wave shift is the crux of the whole pattern, and it is worth being precise about. At the contact point the geometric gap is exactly zero, so the two reflected beams travel identical distances. If reflection were phase-neutral they would arrive perfectly in step and the centre would be bright. Instead, the beam bouncing off the plate is reflecting from air into denser glass — an external reflection that flips its sign by $\pi$ — while the beam bouncing off the lens reflects from glass into rarer air and keeps its sign. The net half-wave difference survives all the way to $t=0$, so the centre is dark. This asymmetry between the two reflections is exactly the content of the Stokes reversibility relations; here we simply use its verdict. The other fingerprint is the spacing. Because the gap is quadratic in radius, equal steps in optical path do not correspond to equal steps in radius: the rings are widely spaced near the centre and crowd together as you go out, with radii climbing as $\sqrt{m}$. Squaring undoes the crowding — $r_m^2$ marches up in equal steps of $\lambda R$ — and that clean linearity is what makes the pattern a measuring instrument rather than a curiosity.

Everything starts with the shape of the gap. A sphere of radius $R$ dropped onto a plane makes a gap whose depth is the sagitta of the arc: exact geometry gives a square-root, but for the shallow gaps of a large-$R$ lens (millimetre radii, metre-scale $R$) it collapses to a parabola. This one approximation is why the rings scale as $\sqrt{m}$.

where: $t$ — local gap thickness; $R$ — radius of curvature of the lens surface; $r$ — radial distance from contact. The approximation drops a term of order $r^4/R^3$, utterly negligible when $r/R \sim 10^{-4}$. The two reflected beams differ by the round trip $2t$ across the gap, plus one crucial half-wave. Of the two reflections only the lower one — air into the denser plate — is external and flips the phase by $\pi$; the upper one, glass into rarer air, does not. Near normal incidence $\cos\theta_t\to1$ and the geometric term is simply $2nt$.

where: $\Delta$ — optical path difference of the two reflected beams; $n$ — index of the medium in the gap ($n=1$ for air); $\theta_t$ — refraction angle in the gap (near normal here); $\lambda$ — vacuum wavelength; the $\lambda/2$ is the half-wave stamped on by the plate reflection (owned by aq-optics-stokes-relations-phase-reflection, recalled here). Because $\Delta=\lambda/2$ already at $t=0$, the centre is dark. Setting $\Delta$ to a destructive half-integer of $\lambda$ cancels the $\lambda/2$ on both sides and leaves an integer condition on the geometric path, which the sagitta turns into ring radii. Dark rings are the natural thing to count in reflection because the strong, high-contrast fringes are the dark ones bracketing bright annuli.

where: $m$ — ring order; dark $m=0,1,2,\dots$ (with $m=0$ the central dark spot), bright $m=1,2,3,\dots$. In air ($n=1$) this is $r_m=\sqrt{m\lambda R}$; the $1/\sqrt{n}$ is the refractometry lever. Absolute ring orders are hard to pin down — the innermost rings are smeared by imperfect contact and dust — so the workhorse relation for measurement differences two rings, cancelling any constant offset and the unknown zero order entirely.

where: $p$ — number of orders between the two measured rings. The slope of $r_m^2$ versus order is $\lambda R/n$, from which $R$ (known $\lambda$) or $\lambda$ (known $R$) drops out with no need to know $m$ absolutely.

1 · Sodium rings on a metre-scale lens (confidence builder)

A plano-convex lens of radius of curvature $R=10.00$ m rests on an optical flat, illuminated from above with sodium light, $\lambda=589.3$ nm. Find the radius of the 10th dark ring, and show how measuring the 5th and 10th rings alone returns $R$.

Step 1. Apply the air-gap dark-ring law $r_m=\sqrt{m\lambda R}$. For $m=10$: $r_{10}=\sqrt{10\times589.3\times10^{-9}\times10.00}=\sqrt{5.893\times10^{-5}}\ \text{m}=\mathbf{7.68\ mm}$. For $m=5$, $r_5=\sqrt{2.9465\times10^{-5}}=5.43$ mm, and $m=1$ gives $2.43$ mm.

Step 2. Confirm the $\sqrt m$ crowding: from $m=1$ to $m=5$ the radius grows $2.43\to5.43$ mm (a jump of $3.00$ mm), but from $m=5$ to $m=10$ only $5.43\to7.68$ mm (a jump of $2.25$ mm) — the same five orders, a smaller step, because $\Delta r\propto1/\sqrt m$.

Step 3. Now invert without knowing the absolute order: $R=(r_{10}^2-r_5^2)/[(10-5)\lambda]$. Numerator $=(7.677\times10^{-3})^2-(5.428\times10^{-3})^2=5.893\times10^{-5}-2.9465\times10^{-5}=2.9465\times10^{-5}\ \text{m}^2$; denominator $=5\times589.3\times10^{-9}=2.9465\times10^{-6}\ \text{m}$.

Step 4. Divide: $R=2.9465\times10^{-5}/2.9465\times10^{-6}=\mathbf{10.0\ m}$, recovering the lens exactly and needing only two ring radii and the wavelength. r₁₀ = 7.68 mm; R = 10.0 m from the 5th and 10th rings.

2 · A drop of water in the gap (refractometry)

The same lens and light, but now a film of water ($n=1.33$) fills the gap. What happens to the rings, and how would you read the water's index from them?

Step 1. The optical path in the gap becomes $2nt$, so $r_m=\sqrt{m\lambda R/n}$: every radius is multiplied by $1/\sqrt{1.33}=0.867$. The 10th dark ring moves in from $7.68$ mm to $r_{10}=7.68\times0.867=\mathbf{6.66\ mm}$, a $13\%$ contraction.

Step 2. Read the index straight from the shrinkage. Measuring the same order dry and wet, $n=(r_m^{\text{air}}/r_m^{\text{liq}})^2=(7.68/6.66)^2=1.33$ — the wavelength, order, and radius all cancel, so no external standard is needed.

Step 3. Sanity-check the direction: a denser medium packs more wavelengths into the same gap, so a given order is reached at a smaller radius and the rings pull inward. A gas ($n\to1$) would leave them essentially unchanged; a high-index oil ($n=1.5$) would shrink them by $0.816$.

Step 4. Note the centre stays dark: the half-wave at the plate is unaffected by what fills the gap (still air-to-glass in sign), so only the spacing, not the contrast pattern, responds to $n$. Water shrinks r₁₀ to 6.66 mm; n = (r_air/r_water)² = 1.33.

The pattern carries Newton's name, yet it is a small monument to how physics actually advances: the man who measured it best drew the wrong lesson, and the wave theory he distrusted read it correctly.

Texts. Hecht, Optics, 5th ed., §9.4 (fringes of equal thickness; Newton's rings) · Born & Wolf, Principles of Optics, 7th ed., §7.5 · Saleh & Teich, Fundamentals of Photonics, 3rd ed., Ch. 2 (interference) · Möller, Optics (interferometric surface testing).

Provenance. The gap relation $t\approx r^2/2R$ follows from $R^2=r^2+(R-t)^2$; the dark-ring law $r_m=\sqrt{m\lambda R}$ from $2t=m\lambda$ after the reflection $\lambda/2$ cancels (Hecht §9.4, Born & Wolf §7.5); the index scaling $r_m=\sqrt{m\lambda R/n}$ and the offset-free difference $r_{m+p}^2-r_m^2=p\lambda R$ are standard laboratory results, all reproduced in verify_aq-optics-newtons-rings.py. The half-wave phase change is owned by aq-optics-stokes-relations-phase-reflection and recalled here by name. Refractive index of water ($1.33$) is the only physical constant cited, a standard tabulated value.

Prerequisites. Thin-Film Interference · Optical Path Difference and Phase Retardation · Two-Wave Interference and the Interference Term · Fringes of Equal Inclination and Equal Thickness. Leads to. Wavefront-Splitting and Amplitude-Splitting Interferometers · Fringe Visibility and Contrast · Conditions for Observable Interference.