Fringes of Equal Inclination and Equal Thickness

Reflect light off a transparent plate and you get two coherent beams — one from each surface — whose interference is set by a single phase, $\delta=(2\pi/\lambda_0)\,2n_f d\cos\theta_t\;(\pm\pi)$ . That phase holds two independent geometric quantities: the plate thickness $d$ and the internal ray angle $\theta_t$ . Hold one fixed and let the other sweep across the field of view, and the same film produces two visually and conceptually distinct families of fringes. Freeze the thickness and vary the angle — a strictly plane-parallel plate lit by a broad diffuse source — and each fringe is a contour of constant inclination: concentric circles, the Haidinger fringes, localized far away at infinity. Freeze the angle and vary the thickness — collimated near-normal light on a wedge, an air gap, or a soap film — and each fringe is a contour of constant thickness: a topographic map of the gap, the Fizeau fringes, localized right on the film. This one-phase, two-families picture is the organizing idea behind a surprising amount of practical optics. Fizeau's equal-thickness fringes are how an optician certifies that a surface is flat to a fraction of a wavelength: press it against a reference flat, and the straightness and spacing of the fringes in the thin air wedge read out the surface figure directly, one dark band per half-wavelength of gap change. Haidinger's equal-inclination rings are what you see peering into a Fabry–Pérot etalon or a Michelson interferometer set off zero — the circular ring system whose collapse toward the centre, one ring per half-wavelength of spacing change, measures wavelengths and cavity lengths. Newton's rings, the most famous fringes in optics, are simply the equal-thickness family for the circular air gap under a lens. Learn to ask which variable is sweeping — $\theta_t$ or $d$ — and every localized-fringe pattern in the laboratory sorts itself into one of these two boxes.

Key takeaways

Two amplitude-split reflections from a film share one phase $\delta=(2\pi/\lambda_0)\,2n_f d\cos\theta_t\;(\pm\pi)$ ; which of $d$ or $\theta_t$ varies across the field decides the fringe family.
Fix $d$ , vary $\theta_t$ (plane-parallel plate, extended source) ⟹ fringes of equal inclination (Haidinger): concentric circles localized at infinity, highest order $m_0=2n_f d/\lambda_0$ at the centre.
Fix $\theta_t$ , vary $d$ (collimated near-normal light, varying thickness) ⟹ fringes of equal thickness (Fizeau): contours of constant $d$ localized on the film.
Air-wedge (Fizeau) fringe spacing is $\Delta x=\lambda_0/(2n_f\alpha)$ for wedge angle $\alpha$ ; the total number of fringes across a gap of maximum thickness $D$ is $2D/\lambda_0$ .
Newton's rings are the equal-thickness family for a circular air gap ( $d\propto r^2$ ), with radii $\propto\sqrt{p}$ ; Fabry–Pérot and Michelson show equal-inclination rings.
Key idea: one phase governs both families — hold thickness fixed and vary the angle for Haidinger circles at infinity, or hold the angle fixed and vary the thickness for Fizeau contours on the film. The families differ only in which variable sweeps, and in where the fringes localize.

The core insight

The two reflections from a film obey one phase, $\delta=(2\pi/\lambda_0)\,2n_f d\cos\theta_t\;(\pm\pi)$ . Holding the thickness $d$ fixed and letting the inclination $\theta_t$ vary gives fringes of equal inclination (Haidinger): concentric rings localized at infinity. Holding the angle fixed and letting $d$ vary gives fringes of equal thickness (Fizeau): contours of constant thickness localized on the film. Same physics, opposite variable — and the localization follows the geometry.

Where this lives today

Fizeau's equal-thickness fringes are the daily language of precision optics manufacturing. Every telescope mirror, laser window, and lithography lens is certified by pressing it against a reference flat or a matched test plate and reading the fringes in the thin air gap: straight, evenly spaced bands mean a good surface, and any bow or twist is the surface error, measured in half-wavelengths, without ever touching the glass. Laser Fizeau interferometers automate this into a full surface map with nanometre resolution, and the same equal-thickness idea gauges the thickness of semiconductor films and the flatness of hard-disk platters. Haidinger's equal-inclination rings are what you look at inside a spectroscopic instrument. A Fabry–Pérot etalon fed with a spectral lamp shows the ring system, and the radii of the rings — or their motion as the spacing is scanned — measure wavelengths and hyperfine splittings to extraordinary precision; the same rings, in a Michelson set slightly off zero path difference, let you find and set the white-light fringe. Newton's rings, the circular equal-thickness special case, still serve as a quick check of a lens's radius of curvature and as the textbook demonstration of the half-wave phase shift, their dark centre a small daily proof that reflection off a denser medium flips the wave. In every case the instrument is reading the one phase this article names, with either the angle or the thickness doing the sweeping.

The physical picture

Write the whole business as one dial: $\delta=(2\pi/\lambda_0)\,2n_f d\cos\theta_t$ . A bright fringe is wherever this dial passes a multiple of $2\pi$ , so a fringe is a curve in the field of view along which $\delta$ is constant. Everything comes down to which variable — $d$ or $\theta_t$ — moves as your eye scans across that field. Freeze the thickness of a perfectly uniform plate and light it with a broad diffuse lamp: now $\delta$ depends only on the angle $\theta_t$ , and all the rays reaching your eye at a given inclination lie on a cone about the plate normal, which maps to a circle. Since $\cos\theta_t$ shrinks as you look further off-axis, the order drops outward, so the rings are highest-order and most crowded at the centre — the equal-inclination, or Haidinger, pattern. Now freeze the angle instead — collimate the light near normal — and let the thickness be a landscape. The bright condition $2n_f d=m\lambda$ becomes a level-set on $d$ : bright bands trace equal-thickness contours, exactly like elevation lines on a map. A wedge gives straight, parallel bands; a spherical air gap gives circles (Newton's rings). And the two families localize in different places for a physical reason. For a plane-parallel plate the two reflected rays that reach a given point on your retina are parallel, so they only recombine at infinity — you focus a telescope on infinity to see Haidinger rings sharply. For a thin wedge the two reflections effectively leave from the same point on the film, so the fringes are localized on the surface — you focus a microscope on the film to see Fizeau fringes. One phase, one film; the geometry decides both the shape and the place.

Figure

Two fringe families from one film: (top) a wedge under collimated light gives straight equal-thickness (Fizeau) fringes on the film; (bottom) a plane-parallel plate with an extended source gives circular equal-inclination (Haidinger) rings at infinity

The Fizeau family is the one you can turn into a ruler with a single number. The next figure takes a shallow air wedge — two optical flats with a thin wire trapped at one edge — and shows how the fringe spacing and the total fringe count read out the wedge angle and the wire's diameter directly.

Figure

Air-wedge Fizeau fringes: spacing Δx = λ/(2n_f α). A 25-μm wire at 5 cm gives α = 5×10⁻⁴ rad, so at λ = 589 nm the fringes are Δx = 0.589 mm apart — and 85 of them span the gap (2D/λ)

Eighty-five fringes from a wire a third the width of a human hair — that is the leverage interference gives you: the pattern amplifies a nanometre-scale height into a millimetre-scale band you can count by eye. Bend or dimple one of the flats and the straight fringes curve; the amount they wander from straight lines, in units of the fringe spacing, is the surface error in half-wavelengths. This is the everyday magic of an optical shop, and it is nothing more than the equal-thickness reading of the one phase.

The core equations

Both families descend from the single phase difference of the two reflections. Tracing the front-surface reflection against the ray that crosses the film and reflects off the back, and projecting onto a common wavefront, the geometry (via Snell's law) collapses to a compact angle-dependent path difference.

\delta = \frac{2\pi}{\lambda_0}\,2n_f d\cos\theta_t\;(\pm\pi),\qquad \Delta = 2n_f d\cos\theta_t \qquad (1)

where: $\delta$ — phase difference of the two reflections; $\Delta$ — their optical path difference; $n_f$ — film index; $d$ — local thickness; $\theta_t$ — internal (refraction) angle; $\pm\pi$ — the half-wave shift present when exactly one reflection is external (low-to-high index). Which variable sweeps decides the family. For the equal-inclination (Haidinger) family, $d$ is fixed, so $\delta$ depends on $\theta_t$ alone; each order is a cone of constant inclination, imaged as a ring. The centre carries the highest order, and the ring radii grow as the square root of the ring number.

m_0 = \frac{2n_f d}{\lambda_0}, \qquad \theta_{i,p}\approx\sqrt{\frac{n_f\lambda_0}{d}\,(p-\varepsilon)} \qquad (2)

where: $m_0$ — order at the centre ( $\theta_t=0$ ), generally non-integer; $p=1,2,\dots$ — ring number counted outward; $\theta_{i,p}$ — external angle of the $p$ -th ring; $\varepsilon$ — fractional part of $m_0$ . The $\sqrt{p}$ spacing means the rings crowd together as you move outward — the visual signature of an equal-inclination pattern. For the equal-thickness (Fizeau) family, $\theta_t$ is fixed (collimated, near-normal, $\cos\theta_t\approx1$ ), so $\delta$ depends on $d$ alone; each fringe is a contour of constant thickness. On a wedge of small angle $\alpha$ the contours are straight and evenly spaced.

\Delta x = \frac{\lambda_0}{2n_f\alpha}, \qquad N = \frac{2D}{\lambda_0}\ \ (\text{fringes across a gap of max thickness } D) \qquad (3)

where: $\Delta x$ — spacing between adjacent bright (or dark) bands; $\alpha$ — wedge angle; $N$ — total number of fringes over a gap that grows from $0$ to $D$ . One fringe corresponds to a thickness change of $\lambda_0/(2n_f)$ — half a wavelength of optical gap — which is why fringe-counting measures thickness to a fraction of a wavelength.

Derivation AFizeau fringes on a wedge — spacing and fringe count

The aim is to show that a shallow wedge under collimated near-normal light produces straight, equally spaced equal-thickness fringes, to find their spacing in terms of the wedge angle, and to count how many span a gap of given maximum thickness — the quantitative basis of optical-flat testing. We fix the angle, let the thickness grow linearly across the wedge, and impose the interference condition.

Step 1. With collimated near-normal light, $\cos\theta_t\approx1$ , so the phase depends only on the local thickness: $\delta(x)=(2\pi/\lambda_0)\,2n_f d(x)\;(\pm\pi)$ , and for a wedge $d(x)=\alpha x$ grows linearly with position $x$ from the contact line. Why: freezing the angle turns the two-variable phase into a function of thickness alone, and a wedge makes thickness a simple linear ramp.

Step 2. A dark (or bright) fringe sits wherever the order is an integer: $2n_f d(x_m)=2n_f\alpha x_m=m\lambda_0$ (dropping the constant $\pm\pi$ , which shifts all orders by half). So $x_m=m\lambda_0/(2n_f\alpha)$ , and successive fringes are equally spaced, $\Delta x=x_{m+1}-x_m=\lambda_0/(2n_f\alpha)$ . Why: because thickness is linear in position, equal steps in order map to equal steps in position — straight, evenly spaced Fizeau bands.

Step 3. Count the fringes over a wedge whose thickness runs from $0$ at contact to $D$ at the far edge: the order there is $m=2n_f D/\lambda_0$ , so the number of fringes is $N=2n_f D/\lambda_0$ , i.e. $N=2D/\lambda_0$ for an air wedge ( $n_f=1$ ). Why: this is the owned result — each fringe marks another $\lambda_0/(2n_f)$ of thickness, so the total count is the gap measured in those half-wave units.

Step 4. For a spacer of diameter $D$ at distance $L$ from the contact line, the wedge angle is $\alpha=D/L$ , giving $\Delta x=\lambda_0/(2n_f\alpha)=\lambda_0 L/(2n_f D)$ . A gauge is born: read $\Delta x$ (or count $N$ ) and you have $D$ to a fraction of a wavelength. Why: this turns the abstract fringe spacing into a direct micrometre-free measurement of a small gap or a thin wire. *Check: an air wedge with $D=25\,\mu$ m at $L=5$ cm gives $\alpha=5\times10^{-4}$ rad; at $\lambda_0=589$ nm, $\Delta x=0.589$ mm and $N=2D/\lambda_0=84.9\to85$ fringes.*

Go deeper · Derivation B · 2 minHaidinger rings — circular geometry and √p spacing

Derivation BHaidinger rings — circular geometry and √p spacing

Here the goal is to prove that a strictly plane-parallel plate under an extended source produces concentric circular equal-inclination fringes, to find the order at the centre, and to show the ring radii grow as the square root of the ring number — the tell-tale outward crowding. We fix the thickness, expand the phase near the axis, and map angle to radius.

Step 1. For constant $d$ , the bright condition $2n_f d\cos\theta_t=m\lambda_0$ depends on $\theta_t$ only. A fixed order $m$ picks a single $\theta_t$ — a cone about the plate normal — and by rotational symmetry that cone images to a circle, so the fringes are concentric rings. The centre ( $\theta_t=0$ ) has the largest $\cos\theta_t=1$ and hence the highest order $m_0=2n_f d/\lambda_0$ . Why: with thickness fixed, only inclination matters, and the axial symmetry of a flat plate turns constant-inclination cones into circles.

Step 2. Expand near the axis, $\cos\theta_t\approx1-\tfrac12\theta_t^2$ , so the order drops from the centre as $m_0-m\approx(n_f d/\lambda_0)\,\theta_t^2$ . Writing $m_0-m=p-\varepsilon$ for the $p$ -th ring (with $\varepsilon$ the fractional part of $m_0$ ) gives $\theta_t^2=(\lambda_0/n_f d)(p-\varepsilon)$ . Why: the small-angle expansion converts the cosine condition into a quadratic in angle, which is what makes the rings scale as $\sqrt{p}$ .

Step 3. Convert to the external angle with small-angle Snell's law $\theta_i\approx n_f\theta_t$ , so $\theta_{i,p}\approx\sqrt{(n_f\lambda_0/d)(p-\varepsilon)}$ ; a lens of focal length $f$ then images each ray angle to a radius $r_p=f\theta_{i,p}\propto\sqrt{p}$ . Why: this is the owned result — the $\sqrt{p}$ law means the rings pack closer as $p$ grows, the visual fingerprint distinguishing Haidinger rings from evenly spaced patterns.

Step 4. The emergent rays for a parallel plate are parallel, so they meet only at infinity: Haidinger fringes are localized at infinity and are seen sharply with the eye relaxed or a telescope focused far. Why: localization at infinity is the practical companion to the geometry — it tells you where to focus, and why these are the rings you see in a Fabry–Pérot or an off-zero Michelson. *Check: the centre order $m_0=2n_f d/\lambda_0=9478.7$ for a $2$ mm glass plate ( $n_f=1.5$ ) at $633$ nm, and ring radii scale as $\sqrt{p}$ .*

Limiting cases

Case	Result	Meaning
equal inclination ( $d$ fixed, $\theta_t$ varies)	$2n_f d\cos\theta_t=m\lambda$	Haidinger circles at infinity; order maximal at centre, rings crowd outward as $\sqrt{p}$
equal thickness ( $\theta_t\approx0$ , $d$ varies)	$2n_f d=m\lambda$	Fizeau contours of constant $d$ localized on the film; wedge $\Rightarrow$ straight bands
uniform film, normal incidence	$2n_f d=m\lambda$ (one tint)	no spatial fringes — a single colour, the origin of thin-film tint
one external reflection	$2n_f d\cos\theta_t=(m+\tfrac12)\lambda$ (bright)	the $\pi$ shift swaps bright/dark — hence the dark centre of Newton's rings

Common trap“Equal-inclination and equal-thickness fringes are localized in the same place”

They are not, and focusing in the wrong plane is why beginners "can't find the fringes." Equal-inclination (Haidinger) fringes live at infinity: the two reflected rays from a plane-parallel plate emerge parallel and only cross at infinity, so you must relax your eye or focus a telescope on infinity to see the rings sharply. Equal-thickness (Fizeau) fringes live on the film: the two reflections effectively originate from the same point of the thin gap, so you focus a microscope on the surface. A second, independent trap is forgetting the half-wave shift when exactly one reflection is external — it flips bright and dark, which is why the point of zero thickness (the centre of Newton's rings, the contact edge of a wedge) is dark, not bright. Diagnose a localized-fringe pattern by asking two questions: which variable sweeps ( $\theta$ or $d$ ), and how many reflections are external?

Worked examples

1 · Testing an optical flat with an air wedge (confidence builder)

Two glass flats are separated at one edge by a wire of diameter $D=25\,\mu$ m, forming an air wedge; the contact line is $L=5$ cm away. Under collimated sodium light ( $\lambda_0=589$ nm) at near-normal incidence, find the wedge angle, the fringe spacing, and the total number of fringes across the gap.

Step 1. Wedge angle from the geometry: $\alpha=D/L=25\times10^{-6}/0.05=5\times10^{-4}$ rad. A shallow wedge, half a milliradian.

Step 2. Fizeau fringe spacing (air, $n_f=1$ ): $\Delta x=\dfrac{\lambda_0}{2n_f\alpha}=\dfrac{589\times10^{-9}}{2\times1\times5\times10^{-4}}=5.89\times10^{-4}$ m $=\mathbf{0.589\ mm}$ . The bands sit about six-tenths of a millimetre apart — easily resolved and counted by eye.

Step 3. Total fringes across the gap: each fringe marks another $\lambda_0/2$ of thickness, so $N=\dfrac{2D}{\lambda_0}=\dfrac{2\times25\times10^{-6}}{589\times10^{-9}}=84.9\to\mathbf{85}$ fringes. Counting the bands between contact and the wire directly measures the wire diameter to a fraction of a wavelength; if the flats are not truly flat, the bands bow, and their departure from straight lines (in units of $\Delta x$ ) is the surface error in half-waves. α = 5×10⁻⁴ rad, Δx = 0.589 mm, N = 85 fringes.

2 · The highest-order Haidinger ring of a glass plate (exam level)

A plane-parallel glass plate, $d=2$ mm thick, index $n_f=1.5$ , is viewed in reflected HeNe light ( $\lambda_0=633$ nm) from an extended source. Find the order at the centre of the Haidinger ring pattern, and describe how the rings are spaced.

Step 1. Centre order: $m_0=\dfrac{2n_f d}{\lambda_0}=\dfrac{2\times1.5\times2\times10^{-3}}{633\times10^{-9}}=\dfrac{6\times10^{-3}}{633\times10^{-9}}=\mathbf{9478.7}$ . Nearly ten thousand wavelengths fit the round trip, so the pattern is very high order and the centre is generally between two integer orders (the fractional part $0.7$ sets where the first ring falls).

Step 2. Ring radii scale as $\theta_{i,p}\propto\sqrt{p-\varepsilon}$ , so the first few rings sit at relative radii $\sqrt{1}:\sqrt{2}:\sqrt{3}=1:1.41:1.73$ — the classic $\sqrt{p}$ crowding, rings packing tighter as you look outward.

Step 3. For fringes to be visible at all, the source's coherence length must exceed the round-trip path $2n_f d=6$ mm; a broad white lamp ( $\ell_c\sim1\,\mu$ m) shows nothing at this thickness, which is why Haidinger rings from a thick plate demand a narrow spectral line like a sodium or HeNe source. This coherence requirement, and the localization at infinity, are exactly why the same rings appear in a Fabry–Pérot etalon fed by monochromatic light. m₀ = 9478.7; rings crowd outward as √p; needs a source with ℓc ≳ 6 mm.

Physics of the portrait

Interference fringes are not a person but a phenomenon — bright and dark bands that turn an invisible optical path difference into something you can see and count — and their two great families carry the names of the men who first mapped them.

Origins · 1 mininterference fringes and how this was discovered

interference fringes

the visible signature of a path difference · Fizeau & Haidinger 1849 · from optical-flat testing to gravitational-wave readout

Interference fringes are the observable output of two-beam interference: a pattern of bright and dark bands, each a locus of constant phase difference, that converts a path difference of a fraction of a wavelength into a countable, measurable displacement. Their two localized families were charted in the same year, 1849. Hippolyte Fizeau — better remembered for the toothed-wheel measurement of the speed of light — studied the fringes formed in a thin film of varying thickness, the equal-thickness bands that now bear his name and that trace a contour map of any gap; they became, and remain, the workhorse of optical-shop metrology, certifying the flatness of every precision surface. Wilhelm Karl von Haidinger, an Austrian mineralogist with a gift for subtle optical observation (he also discovered "Haidinger's brush", the faint pattern the human eye sees in polarized light), described the complementary family: the concentric equal-inclination rings from a plane-parallel plate, localized at infinity, that appear whenever monochromatic light passes through an etalon. The two families are the same physics — one phase, $2n_f d\cos\theta_t$ — read with either the thickness or the angle held fixed. From certifying a telescope mirror to reading out the mirror motion of a kilometre-scale gravitational-wave interferometer, counting interference fringes is still how optics measures length against the wavelength of light.

History

1665

— Hooke records the colours of thin plates, the thickness-dependent fringes

1704

— Newton measures the circular equal-thickness rings that bear his name

1849

— Fizeau describes fringes of equal thickness in a varying-thickness film

1849

— Haidinger describes the circular equal-inclination fringes of a plane-parallel plate

1899

— Fabry and Pérot sharpen the equal-inclination rings with a high-reflectance cavity

Going further

Texts. Hecht, Optics, 5th ed., §9.4 (amplitude-splitting interferometers; Haidinger and Fizeau fringes) · Born & Wolf, Principles of Optics, 7th ed., §7.6 (fringes of equal inclination and equal thickness) · Saleh & Teich, Fundamentals of Photonics, 3rd ed., Ch. 2 · Demtröder, Laser Spectroscopy, Vol. 1, Ch. 4. The master phase $\delta=(2\pi/\lambda_0)\,2n_f d\cos\theta_t\;(\pm\pi)$ and the path difference $\Delta=2n_f d\cos\theta_t$ are derived from Snell's law. The Haidinger ring radii $\theta_{i,p}\approx\sqrt{(n_f\lambda_0/d)(p-\varepsilon)}$ and centre order $m_0=2n_f d/\lambda_0$ , the Fizeau wedge spacing $\Delta x=\lambda_0/(2n_f\alpha)$ , and the fringe count $N=2D/\lambda_0$ are the standard results of two-beam film interference. The dark-centre / half-wave shift follows the Fresnel sign convention. Key numbers: air wedge $\alpha=5\times10^{-4}$ rad, $\Delta x=0.589$ mm, $N=85$ ; Haidinger centre order $m_0=9478.7$ .

Prerequisites. Two-Wave Interference and the Interference Term · Optical Path Difference and Phase Retardation · Thin-Film Interference · Conditions for Observable Interference. Leads to. Newton's Rings · Wavefront-Splitting and Amplitude-Splitting Interferometers · Temporal Coherence and Coherence Length · Fringe Visibility and Contrast.

What actually decides whether I see equal-thickness or equal-inclination fringes?

Which of the two variables in $2n_f d\cos\theta_t$ is allowed to change across your field of view. If the film has a fixed thickness and you illuminate it with a broad source so many angles reach your eye, the angle sweeps and you see equal-inclination (Haidinger) rings. If instead you use collimated light (one angle) on a film whose thickness varies, the thickness sweeps and you see equal-thickness (Fizeau) contours. The same plate can show either, depending on how you light it and where you look — a plane-parallel plate under a diffuse lamp gives rings, a wedge under a collimated beam gives bands. Ask "what is changing as I scan the field?" and the family follows.

Why are Fizeau fringes on the film but Haidinger fringes at infinity?

It comes down to where the two interfering rays actually meet. For a plane-parallel plate, the two reflections of a given incident ray emerge exactly parallel — they never cross at any finite distance, so their interference is only defined at infinity, and you focus there (or use a lens) to see Haidinger rings. For a thin film of varying thickness, the two reflections come from very nearly the same point on the film — the gap is too thin to separate them — so they interfere right at the surface, and you focus a microscope on the film to see Fizeau fringes. Localization is not an optional detail: aim your focus at the wrong plane and the fringes blur away.

Are Newton's rings equal-thickness or equal-inclination fringes?

Equal-thickness — they are the Fizeau family for a circular air gap. The "film" is the wedge of air between a convex lens and a flat plate, whose thickness grows as $d\propto r^2$ with distance from the contact point, so the contours of constant thickness are circles and the ring radii scale as $\sqrt{p}$ (like $\sqrt{m}$ ). Because one reflection (at the glass-to-air top of the gap versus air-to-glass at the bottom) is external, there is a single half-wave shift, which makes the contact point — zero thickness — dark. So Newton's rings share the geometry of Haidinger rings (concentric circles) but belong to the opposite family: they are localized on the gap, not at infinity, and their order is lowest at the centre, the reverse of Haidinger.

How does counting fringes measure a surface to a fraction of a wavelength?

Because each fringe marks a change in gap thickness of exactly $\lambda_0/(2n_f)$ — half a wavelength of optical path. In an air gap that is about $300$ nm per fringe for green light. When you lay a test surface on a reference flat, a perfectly flat surface gives perfectly straight, evenly spaced fringes; any hill or valley on the surface bends the local gap and therefore bends the fringes. Measuring how far a fringe wanders from straight, as a fraction of the fringe spacing, gives the surface height error in units of $\lambda_0/2$ — so a fringe that bows by a quarter of its spacing signals a surface departure of about an eighth of a wavelength, tens of nanometres, read off by eye without any contact.

Why do the Haidinger rings crowd together as I look outward?

Because the order changes quadratically with angle near the axis. The bright condition $2n_f d\cos\theta_t=m\lambda$ expands to $m_0-m\approx(n_f d/\lambda_0)\theta_t^2$ , so the drop in order from the centre grows as the square of the angle. Inverting, the $p$ -th ring sits at $\theta\propto\sqrt{p}$ , which means the radii go as $\sqrt1,\sqrt2,\sqrt3,\dots=1,1.41,1.73,\dots$ — the gaps between successive rings shrink as $p$ increases. This $\sqrt{p}$ crowding is the instant visual signature of an equal-inclination pattern (and, with the same algebra, of Newton's rings), distinguishing it from the evenly spaced bands of a wedge.

Do I need a laser, or will any monochromatic lamp show these fringes?

Any sufficiently monochromatic source works, provided its coherence length exceeds the round-trip path in the film, $2n_f d$ . For a thick plate that can be demanding: a $2$ mm glass plate needs $\ell_c\gtrsim6$ mm, which rules out a broadband lamp ( $\ell_c\sim1\,\mu$ m) but is easily met by a sodium or mercury spectral line, and comfortably by a laser. For a thin film — a soap membrane, an air wedge near contact — the round-trip path is only a wavelength or two, so even white light shows a few coloured fringes near zero thickness. That is exactly the pattern of temporal coherence: the thicker the gap you want to see fringes across, the narrower the spectral line you need to light it.