Thin-Film Interference

The colours that swirl across a soap bubble, the purple bloom on a camera lens, the oily rainbow on a wet road — all are the same effect, and it is interference of the most tangible kind. A transparent film a few hundred nanometres thick reflects light twice: once at its front surface, once at its back. Those two reflected waves are amplitude-split copies of one incident ray, so they are mutually coherent, and they recombine having travelled paths that differ by $2n_f t\cos\theta_t$ — twice through the film, corrected for the slant. Whether they crest together or cancel depends on how many wavelengths fit into that extra path, and because the count depends on $\lambda$ , white light is sorted into colour. That is the whole phenomenon; everything else is bookkeeping. The bookkeeping has one famous subtlety, and it is electromagnetic. Reflection off an optically denser medium flips the wave's sign — a $\pi$ phase shift, an extra half-wave — exactly as a pulse on a rope inverts when it reflects from a fixed end. This article recalls that fact by name from the electromagnetic article aq-em-fresnel-equations, whose reflection coefficient goes negative ( $r<0$ , a $180^\circ$ shift) when light enters a higher-index medium; the phase-change-on-reflection is itself owned within our track by the Stokes relations. For a film in air only the front reflection flips, and that single extra half-wave is why a soap film thinner than a quarter-wavelength looks black in reflection — the two beams cancel even though their path difference is nearly zero. Get the half-wave bookkeeping right and thin films explain themselves: the shifting bands of a draining bubble, the quarter-wave layer that erases a lens's glare, and the stacked dielectric mirror that reflects better than any metal.

Key takeaways

A thin film reflects light at both surfaces; the two amplitude-split reflections are coherent and differ in optical path by $\Delta=2n_f t\cos\theta_t$ (twice through the film, corrected for angle).
One reflection — the one from a low-to-high refractive-index interface — carries a $\pi$ phase shift (an extra $\lambda/2$ ), as follows from the Fresnel reflection coefficient.
For a film in air (one $\pi$ shift) a bright reflected fringe needs $2n_f t\cos\theta_t=(m+\tfrac12)\lambda$ ; the thinnest bright film at normal incidence is $t=\lambda/(4n_f)$ .
Because the condition depends on $\lambda$ , white light gives colour: the iridescence of soap, oil, and oxide layers, sweeping through the spectrum as the film thins.
Anti-reflection coating: a quarter-wave layer of index $n_f=\sqrt{n_0 n_s}$ makes the two reflections equal and antiphase, cutting glass reflectance from 4% toward zero (MgF₂ gives ≈1.3%).
Key idea: two amplitude-split reflections separated by $2n_f t\cos\theta_t$ interfere; adding the single $\pi$ shift at the low-to-high reflection gives the bright/dark conditions, and the $\lambda$ -dependence turns white light into colour.

The core insight

A thin film splits an incident ray by amplitude into a front- and a back-surface reflection, coherent copies separated by optical path $2n_f t\cos\theta_t$ . A single $\pi$ phase shift occurs at the reflection entering the denser medium (recalled from the Fresnel coefficient turning negative), so for a film in air the reflected maxima are $2n_f t\cos\theta_t=(m+\tfrac12)\lambda$ . The wavelength dependence sorts white light into the colours of soap films and enables anti-reflection coatings.

Where this lives today

Thin-film interference is a multi-billion-dollar technology hiding in plain sight. Every lens in a camera, telescope, or pair of binoculars wears an anti-reflection coating — usually a stack of several layers, not just one, optimized across the visible band — because a modern zoom lens has a dozen or more air-glass surfaces, and 4% lost at each would leave a dim, flare-ridden image. Stack many quarter-wave layers of alternating high and low index and the partial reflections add in phase instead, building dielectric mirrors that reflect more than 99.99% of a chosen wavelength: these are the mirrors of every laser cavity, far better than any polished metal, and the same principle makes the narrow-band interference filters that isolate a single spectral line for astronomy or fluorescence microscopy. The everyday and the industrial meet in the same physics. The iridescence of a butterfly's wing or a beetle's shell is structural colour — layered nanostructures doing thin-film interference with no pigment at all — and it is being copied into non-fading paints and anti-counterfeiting inks. In the semiconductor fab, the colour of an oxide or photoresist layer on a silicon wafer is read as a thickness gauge to within nanometres, and ellipsometry refines the same interference into a sub-Ångström measurement of film thickness and index. From the bloom on your glasses to the mirror in a gravitational-wave detector, the two reflections of a thin film are quietly at work.

The physical picture

Follow one ray onto a soap film. Part of it bounces off the front surface; the rest dives in, crosses the film, reflects off the back, and climbs back out parallel to the first. The second ray has made a round trip through the film — a geometric path $2t$ at normal incidence, an optical path $2n_f t$ — and whether it emerges in step with the front reflection or out of step depends on how that extra path compares to the wavelength. Slide across a film of varying thickness and the two reflections drift in and out of phase, painting bright and dark bands. Now the twist that makes thin films behave against naive intuition. Reflecting off a denser medium inverts the wave — a $\pi$ shift — while reflecting off a rarer medium does not. For a film in air the front reflection (air into film, low into high index) flips; the back reflection (film into air, high into low) does not. So the two reflections start life already half a wavelength out of step, before any path difference. At the very top of a draining soap film, where the membrane has thinned below a quarter-wavelength and the path difference is almost nothing, that built-in half-wave makes the two beams cancel — the film goes black an instant before it bursts, a small, reliable piece of theatre you can watch on any bubble. Everywhere else, the path difference $2n_f t\cos\theta_t$ competes with the half-wave, and since the winning wavelength depends on thickness and angle, white light is combed into the moving colours of oil slicks and bubbles.

Figure

A thin film splits one ray into a front and a back reflection separated by optical path 2n_f t cosθ_t; the front reflection (into the denser film) gains a π half-wave shift

Because the winning colour depends on thickness, a single wavelength drawn against film thickness already tells the whole story. The next figure plots how a soap film's reflected brightness at one colour rises and falls as the film thickens — and marks the thinnest film that shines, the quarter-wave point every coating engineer knows by heart.

Figure

Reflected intensity of a soap film

(n_f = 1.33)

\lambda = 550

nm versus thickness: black at

t \to 0

(the

\pi

shift), first bright at

t = \lambda/(4n_f) = 103

The zero-thickness black is the fingerprint of the half-wave shift — no path-difference argument alone would predict it. Move right and the reflected intensity oscillates with a period of $\lambda/(2n_f)$ in thickness: bright at $103$ nm, dark again at $207$ nm, bright at $310$ nm. Change the wavelength and the whole curve rescales, so at a fixed thickness different colours peak — which is exactly why a wedge of soap or a slick of oil shows bands of colour rather than uniform grey.

The core equations

The physics starts with the geometric path difference between the two reflected rays. Tracing the ray that dives through the film and reflects off the back, against the ray reflected directly off the front, and projecting both onto a common wavefront, the slant terms cancel by Snell's law and leave a compact result.

\Delta = 2\,n_f\,t\cos\theta_t \qquad (1)

where: $\Delta$ — optical path difference of the two reflections; $n_f$ — film index; $t$ — film thickness; $\theta_t$ — refraction angle inside the film. At normal incidence ( $\theta_t=0$ ) this is simply $\Delta=2n_f t$ : twice through the film in optical units. Converting the path to phase and adding the interfacial half-wave gives the total phase difference, from which the bright and dark conditions follow. The $\pi$ appears because exactly one of the two reflections — the front one, entering the denser film — inverts the wave.

\delta = \frac{2\pi}{\lambda_0}\,2n_f t\cos\theta_t + \pi \qquad\Longrightarrow\qquad 2n_f t\cos\theta_t = \left(m+\tfrac12\right)\lambda_0 \ \ (\text{bright, film in air}) \qquad (2)

where: $\delta$ — total phase difference; $\lambda_0$ — vacuum wavelength; $m=0,1,2,\dots$ — integer order; the $+\pi$ is the single half-wave shift. Bright reflection therefore needs a half-integer number of waves in the path; dark reflection needs a whole number, $2n_f t\cos\theta_t=m\lambda_0$ . At normal incidence the thinnest bright film is $t=\lambda_0/(4n_f)$ — the quarter-wave thickness. The $\pi$ shift itself is not a thin-film postulate but an electromagnetic result: the Fresnel amplitude reflection coefficient at an interface changes sign when the second medium is denser.

r_{12}=\frac{n_1\cos\theta_i-n_2\cos\theta_t}{n_1\cos\theta_i+n_2\cos\theta_t}\ \ \Longrightarrow\ \ n_2>n_1\Rightarrow r_{12}<0\ (\pi\ \text{shift}) \qquad (3)

where: $r_{12}$ — amplitude reflection coefficient (s-polarization) from medium 1 into medium 2. A negative amplitude is a $180^\circ$ phase shift; this sign, recalled from aq-em-fresnel-equations, is the entire origin of the half-wave. For the anti-reflection problem, cancelling the two reflections needs them equal in amplitude and antiphase, which fixes both the film index and thickness.

n_f=\sqrt{n_0\,n_s},\qquad t=\frac{\lambda_0}{4n_f},\qquad R=\left(\frac{n_f^2-n_0 n_s}{n_f^2+n_0 n_s}\right)^2 \qquad (4)

where: $n_0$ — incident medium (air, $\approx1$ ); $n_s$ — substrate index; $n_f$ — coating index; $R$ — residual reflectance of a real quarter-wave coating whose index misses the ideal $\sqrt{n_0 n_s}$ . For glass ( $n_s=1.52$ ) the ideal coating index is $\sqrt{1.52}=1.23$ ; the practical MgF₂ value $1.38$ leaves a small residual.

Derivation AOptical path difference and the half-wave bookkeeping

The goal is to pin down the two ingredients of every thin-film calculation: the geometric optical path difference between the front and back reflections, and the single extra half-wave that reflection off the denser medium adds — the piece that flips bright and dark and blackens a vanishingly thin film. We trace the two rays to a common wavefront and then account for the interfacial phase from the Fresnel sign.

Step 1. The refracted ray crosses the film twice, each leg of length $t/\cos\theta_t$ , for an optical path $2n_ft/\cos\theta_t$ ; the front-reflected ray, meanwhile, runs ahead in air by $\Delta d=2t\tan\theta_t\sin\theta_i$ to reach the shared wavefront. Why: interference compares the two beams at a common wavefront perpendicular to the emerging parallel rays, so we must subtract the head start the front ray gets.

Step 2. Subtract, using Snell's law $n_0\sin\theta_i=n_f\sin\theta_t$ : the removed path is $n_0\cdot2t\tan\theta_t\sin\theta_i=2n_f t\,\sin^2\theta_t/\cos\theta_t$ , so $\Delta=\frac{2n_f t}{\cos\theta_t}(1-\sin^2\theta_t)=2n_f t\cos\theta_t$ . Why: Snell's law converts the external head start into the film's own variables, and $1-\sin^2=\cos^2$ collapses the two terms into the clean angle-dependent result.

Step 3. Add the interfacial phase. By the Fresnel coefficient (recalled from aq-em-fresnel-equations), reflection at the front surface (air→film, $n_f>n_0$ ) gives $r<0$ : a $\pi$ shift. The back surface (film→air, $n_f>n_0$ going the other way) gives $r>0$ : no shift. Exactly one $\pi$ , so $\delta=(2\pi/\lambda_0)\,2n_f t\cos\theta_t+\pi$ . Why: only the low-to-high reflection inverts the wave, so the two reflections of a film-in-air are born half a wavelength apart before any path difference — this is the owned bookkeeping.

Step 4. Read off the conditions. Bright reflection ( $\delta=2\pi m$ ) requires $2n_f t\cos\theta_t=(m+\tfrac12)\lambda_0$ ; dark requires $2n_f t\cos\theta_t=m\lambda_0$ . As $t\to0$ , $\delta\to\pi$ : destructive, so a film far thinner than a wavelength reflects almost nothing — it looks black. Why: this is the owned result, and the $t\to0$ black film is the crisp, checkable signature of the half-wave that a path-only account would get exactly backwards. *Quick checks: $\Delta=2n_f t\cos\theta_t$ reduces to $2n_f t$ at normal incidence; the first bright soap film ( $n_f=1.33$ , $550$ nm) appears at $t=\lambda_0/(4n_f)=103$ nm; and as $t\to0$ the result is correctly destructive.*

Go deeper · Derivation B · 1 minThe quarter-wave anti-reflection coating and its residual

Derivation BThe quarter-wave anti-reflection coating and its residual

Here the aim is to design a single-layer coating that makes a surface reflect nothing at a chosen wavelength, and to compute what a real coating — whose index is not the ideal value — actually leaves behind. Complete cancellation needs the two partial reflections equal in amplitude and exactly out of phase, which fixes both the coating index and its thickness, and any index mismatch feeds a clean residual-reflectance formula.

Step 1. Antiphase condition: put the two reflections half a wavelength apart in the round trip, $2n_f t=\lambda_0/2$ , i.e. the quarter-wave thickness $t=\lambda_0/(4n_f)$ . Here both reflections (air→film and film→glass) are low-to-high, so both carry a $\pi$ shift and the shifts cancel — the phase offset comes purely from the path. Why: two equal $\pi$ shifts leave no net interfacial phase, so the quarter-wave path is what sets the reflections in opposition.

Step 2. Equal-amplitude condition: the two reflectances match, $\dfrac{n_0-n_f}{n_0+n_f}=\dfrac{n_f-n_s}{n_f+n_s}$ , which cross-multiplies to $n_f^2=n_0 n_s$ , so the ideal index is the geometric mean $n_f=\sqrt{n_0 n_s}$ . Why: only if the two reflected amplitudes are equal can antiphase addition cancel them completely — the geometric mean is exactly the index that balances them.

Step 3. For a real coating of index $n_f\neq\sqrt{n_0 n_s}$ at the quarter-wave point, the two reflections still oppose but no longer fully cancel; summing them gives residual reflectance $R=\big(\tfrac{n_f^2-n_0 n_s}{n_f^2+n_0 n_s}\big)^2$ . Why: this is the owned result — it measures how far a practical coating misses perfection because no durable material has exactly the ideal index.

Step 4. Numbers for glass ( $n_0=1$ , $n_s=1.52$ ): the ideal index is $\sqrt{1.52}=1.23$ , unavailable in a hard coating; magnesium fluoride ( $n_f=1.38$ ) gives $R=\big(\tfrac{1.38^2-1.52}{1.38^2+1.52}\big)^2=0.0126$ , cutting a bare surface's $4.3\%$ reflection to about $1.3\%$ . Why: this is why lens blooms are faint purple, not perfectly black — MgF₂ is the durable compromise, not the ideal geometric-mean glass. *Quick checks: ideal $n_f=\sqrt{1.52}=1.23$ ; MgF₂ quarter-wave thickness $t=\lambda_0/(4n_f)=99.6$ nm at $550$ nm; residual $R=1.26\%$ against bare glass at $4.3\%$ .*

Limiting cases

Case	Result	Meaning
vanishing thickness $t\to0$	$\delta\to\pi$ , destructive	the $\pi$ shift dominates — a very thin soap film reflects almost nothing (black before it bursts)
normal incidence	$\Delta=2n_f t$ , bright at $t=(m+\tfrac12)\lambda_0/2n_f$	thinnest bright film at $t=\lambda_0/(4n_f)$ (quarter-wave)
low reflectance (soap, oil)	$R\approx r_1^2+r_2^2+2r_1r_2\cos\delta$	pure two-wave cosine fringes — the everyday colour regime
high reflectance (coatings)	Airy peaks, $F=4R/(1-R)^2$	multiple reflections sharpen into narrow bands — dielectric mirrors, filters
thick plate ( $2n_f t\gg\ell_c$ )	no fringes	path difference exceeds the coherence length; reflections add incoherently (a window pane has no colour)

Common trap“Both reflections get the half-wave shift, so the shifts cancel and you can ignore them”

A reflection from a low-to-high refractive-index interface carries a $\pi$ phase shift, while a reflection from a high-to-low interface does not. For the commonest case — a film in air — the front surface flips and the back surface does not, so there is one net half-wave, and it swaps the bright/dark conditions relative to a path-only argument. The shifts cancel only in the symmetric case where both reflections happen to be low-to-high (as in a coating on glass) or both high-to-low. Apply the rule mechanically: check each interface, count the low-to-high reflections, and add $\lambda/2$ for each. Get it wrong and you predict a bright film where it is black — the single most common thin-film error, and the reason the top of a soap film goes dark is precisely this one uncancelled half-wave.

Worked examples

1 · The thinnest bright soap film (confidence builder)

A soap film ( $n_f=1.33$ ) in air is viewed in reflected green light, $\lambda_0=550$ nm, at normal incidence. What is the thinnest film that appears bright, and what happens as the film drains thinner still?

Step 1. Film in air means one $\pi$ shift, so a bright reflection needs $2n_f t=(m+\tfrac12)\lambda_0$ . The thinnest bright film is $m=0$ : $2n_f t=\tfrac12\lambda_0$ , i.e. $t=\lambda_0/(4n_f)$ .

Step 2. Evaluate: $t=550\times10^{-9}/(4\times1.33)=550/5.32$ nm $=\mathbf{103\ nm}$ . A quarter-wave of green light — barely a hundred atoms thick — is the thinnest film that shines green in reflection.

Step 3. As the film drains below $103$ nm the reflected intensity falls, and at $t\to0$ the path difference vanishes but the uncancelled half-wave makes the two reflections destructive: the top of the film goes black just before it pops. Between $103$ nm and zero the film sweeps through the colours as different wavelengths hit their quarter-wave point at different thicknesses — the shifting iridescence of a bubble. Thinnest bright film t = 103 nm; the film blackens as t → 0.

2 · A magnesium-fluoride anti-reflection coating (exam level)

A camera lens of glass ( $n_s=1.52$ ) is coated with magnesium fluoride ( $n_f=1.38$ ) for $\lambda_0=550$ nm. Find the quarter-wave thickness, the residual reflectance, and compare to the ideal coating and to bare glass.

Step 1. Quarter-wave thickness: $t=\lambda_0/(4n_f)=550\times10^{-9}/(4\times1.38)=550/5.52$ nm $=\mathbf{99.6\ nm}$ . This sets the two reflections (both low-to-high, both $\pi$ -shifted, so the shifts cancel) half a wavelength apart in path — antiphase.

Step 2. Ideal index would be the geometric mean $n_f=\sqrt{n_0 n_s}=\sqrt{1\times1.52}=1.23$ , which would cancel the reflection completely — but no durable coating has that index. MgF₂ at $1.38$ misses it, leaving $R=\big(\tfrac{1.38^2-1.52}{1.38^2+1.52}\big)^2=\big(\tfrac{1.904-1.52}{1.904+1.52}\big)^2=\big(\tfrac{0.384}{3.424}\big)^2=\mathbf{0.0126}=1.3\%$ .

Step 3. Bare glass reflects $R_0=\big(\tfrac{n_s-1}{n_s+1}\big)^2=\big(\tfrac{0.52}{2.52}\big)^2=0.043=4.3\%$ per surface. The single MgF₂ layer cuts that to $1.3\%$ — a factor of three, and across the many air-glass surfaces of a camera lens the difference between a hazy, flare-prone image and a crisp one. The faint purple-green bloom you see on a coated lens is the residual reflection: the coating is optimized for green, so red and blue leak slightly and tint the ghost. t = 99.6 nm, residual R = 1.3% (vs 4.3% bare, 0% ideal at n_f = 1.23).

Physics of the portrait

Thin-film colours were among the first optical phenomena anyone tried to explain quantitatively — and the man who catalogued them under a new microscope got tantalizingly close, a generation before the wave theory could finish the job.

Origins · 1 minRobert Hooke and how this was discovered

Robert Hooke

1635–1703 · London (the Royal Society)

Robert Hooke was the Royal Society's first Curator of Experiments and one of the most inventive experimentalists of the seventeenth century — a polymath who gave his name to the law of springs, coined the biological term "cell", drew the first detailed maps of the Moon, and quarrelled famously with Newton over almost everything. In his 1665 masterpiece Micrographia, alongside the celebrated engravings of a flea and a fly's eye, Hooke recorded the "fantastical colours" of thin transparent bodies: the films of mica he split into ever-thinner sheets, the coloured rings in soap bubbles, the tarnish on heated steel. He saw clearly that the colour depended on the thickness of the film, and he proposed that it arose from reflections at the two surfaces combining — a strikingly modern intuition. What he lacked was the wave and the wavelength: without them he could not turn "reflections combining" into a quantitative condition. Newton took up the same rings a few decades later (they bear Newton's name, not Hooke's, though Hooke saw them first), measured them meticulously in the Opticks of 1704, yet his corpuscular theory forced an awkward machinery of "fits of easy reflection and transmission" to explain the periodicity. Only with Young and Fresnel, a century after Hooke, did the wave theory supply the missing piece — the half-wave shift and the path condition $2n_f t\cos\theta_t=(m+\tfrac12)\lambda$ — that makes Hooke's coloured films the clean textbook effect they are today.

History

1665

— Hooke describes the colours of thin plates and soap films in Micrographia

1704

— Newton publishes precise measurements of the rings in Opticks

1801

— Young explains the colours by wave interference and the phase change on reflection

1817

— Young and Fresnel establish the half-wave shift on reflection from a denser medium

1886

— Lord Rayleigh shows that a thin, low-index layer can reduce reflection from glass

1935

— Smakula patents practical anti-reflection coatings for optical glass

Going further

Texts. Hecht, Optics, 5th ed., §9.6–9.9 (thin films, coatings) · Born & Wolf, Principles of Optics, 7th ed., §1.6 (stratified media, Airy reflectance) · Saleh & Teich, Fundamentals of Photonics, 3rd ed., Ch. 7 (resonators, multilayers) · Loudon, The Quantum Theory of Light, 3rd ed., Ch. 3. The optical path difference $\Delta=2n_f t\cos\theta_t$ follows from Snell's law; the $\pi$ phase shift corresponds to the Fresnel reflection coefficient satisfying $r_{12}<0$ (a $180^\circ$ shift) for $n_2>n_1$ , recalled from aq-em-fresnel-equations. The anti-reflection design gives $n_f=\sqrt{n_0 n_s}$ , quarter-wave thickness, and residual $R=\big(\tfrac{n_f^2-n_0 n_s}{n_f^2+n_0 n_s}\big)^2$ . Key numbers: soap first-bright $t=103$ nm; MgF₂ quarter-wave $99.6$ nm; residual $1.3\%$ vs bare $4.3\%$ ; ideal $n_f=1.23$ .

Prerequisites. Superposition Principle of Electromagnetic Waves · Two-Wave Interference and the Interference Term · Optical Path Difference and Phase Retardation · Conditions for Observable Interference · Temporal Coherence and Coherence Length. Leads to. Fringes of Equal Inclination and Equal Thickness · Newton's Rings · Wavefront-Splitting and Amplitude-Splitting Interferometers · Fringe Visibility and Contrast.

Why does exactly one reflection get the π phase shift and not both?

Because the two reflections happen at interfaces crossed in opposite senses. For a film in air, the front reflection is air→film — light meeting a denser medium — and the Fresnel coefficient goes negative there, a $\pi$ shift, like a rope pulse reflecting off a fixed end. The back reflection is film→air — light meeting a rarer medium — where the coefficient stays positive, no shift, like a rope pulse reflecting off a free end. So there is one net half-wave. In a symmetric situation where both reflections meet a denser medium (a coating between air and glass, both low-to-high) the two $\pi$ shifts cancel and you drop them. Always count the low-to-high reflections one interface at a time.

How thick can a film be and still show interference colours?

Only up to about the coherence length of the illuminating light. The two reflections are separated by an optical path $2n_f t$ , and they interfere coherently only while that stays within the source's coherence length $\ell_c=\lambda^2/\Delta\lambda$ . For sunlight $\ell_c\approx1\,\mu$ m, so films up to roughly a micrometre thick show colour, but a millimetre-thick window has a path difference thousands of times the coherence length — the two reflections arrive uncorrelated and simply add their intensities, giving no fringes and no colour. This is why oil slicks and bubbles are iridescent but plate glass is not, even though both have two surfaces.

Why is a single MgF₂ coating only "purple" rather than perfectly clear?

Two reasons, both visible in the numbers. First, MgF₂'s index (1.38) is not the ideal geometric mean (1.23 for glass), so even at the design wavelength a residual $\approx1.3\%$ reflection survives. Second, the quarter-wave condition is exact at only one wavelength — usually chosen in the green, where the eye is most sensitive — so red and blue are slightly off the null and reflect a little more. The leftover reflection is therefore purple-magenta (red + blue), the familiar bloom on a coated lens. Multilayer "broadband" coatings stack several films to flatten the reflectance across the whole visible band, which is why high-end lenses look almost black in reflection rather than purple.

Is thin-film interference the same effect as Newton's rings?

Yes — Newton's rings are one special geometry of thin-film interference. The "film" is the thin wedge of air trapped between a convex lens and a flat plate; its thickness grows as $r^2$ with distance from the contact point, so the equal-thickness contours are circles. Each ring is a locus where $2t=(m+\tfrac12)\lambda$ (with the single air-gap half-wave), and the centre, where the thickness is zero, is dark for exactly the reason a thin soap film is black. What changes from a soap film is only the shape of the thickness profile; the interference condition $2n_f t\cos\theta_t\pm\lambda/2$ is identical. Fringes of equal thickness and equal inclination are the general framework of which Newton's rings are the circular case.

Why does increasing the surface reflectance sharpen the fringes?

At low reflectance only the first two reflections matter and the pattern is a gentle two-wave cosine. As the surfaces become more reflective, the light bounces back and forth many times before escaping, and each extra round trip adds another partial wave to the sum. Summing that geometric series gives the Airy function, whose bright bands narrow as the reflectance rises — quantified by the finesse $F=4R/(1-R)^2$ . Physically, the many beams reinforce sharply only in a narrow band of phase and cancel elsewhere, so the broad cosine collapses into sharp spikes. This is the continuous bridge from a soap film (two beams, broad fringes) to a Fabry–Pérot etalon (many beams, razor-sharp peaks) — the same film, driven to high reflectance.

Does the film have to be in air for the half-wave rule to apply?

No — the rule is completely general, you just apply it at each interface separately. What matters at every reflection is whether the second medium is denser (low-to-high, $\pi$ shift) or rarer (high-to-low, no shift). A film of oil on water, for instance, has air→oil (say $n$ up, $\pi$ shift) and oil→water (which way the index steps depends on the oil), so you must check the actual indices. When both reflections are low-to-high, or both high-to-low, their $\pi$ shifts cancel and you use the plain path condition $2n_f t\cos\theta_t=m\lambda$ for bright. The single most reliable habit is to never memorize "bright = half-integer" but to write down each interface, tag it with a $\pi$ or not, and add up.