Optical Path Difference and Phase Retardation

Two waves leave a source in step, take different routes, and meet again. Whether they arrive crest-on-crest or crest-on-trough — bright fringe or dark — is decided not by how far each travelled but by how many wave crests each accumulated along the way. In a denser medium the crests pack tighter ( $\lambda=\lambda_0/n$ ), so a centimetre of glass "contains" more crests than a centimetre of air. The bookkeeping that gets this right is the optical path length, $\mathrm{OPL}=\int n\,ds$ : the geometric distance re-expressed as the equivalent distance in vacuum, which is exactly the number of crests re-scaled to vacuum wavelengths. Interference lives entirely on the difference of these two counts, the optical path difference $\Lambda=\mathrm{OPL}_2-\mathrm{OPL}_1$ . It converts geometry and materials into a single phase, $\delta=(2\pi/\lambda_0)\Lambda$ , and that phase alone sets the fringe: bright when $\Lambda$ is a whole number of vacuum wavelengths, dark when it is a half-integer. A glass slab dropped into one beam lengthens that route's phase distance without moving its endpoints, sliding the fringes by $(n-1)t$ worth of wavelengths. A gas cell being filled does the same, more gently, at parts-per-million sensitivity. A thin film folds two reflections a distance $2n_f t\cos\theta_t$ apart. One subtlety no geometry can supply must ride along in the same ledger: reflection off a denser medium flips the field's sign, an instantaneous $\pi$ jump worth exactly half a wavelength of path. That half-wave is why a vanishingly thin soap film looks dark in reflection even though its two reflections travel no extra distance. Keep the path counts and the reflection phases in one account and the entire zoo of interference — Newton's rings, anti-reflection coatings, interferometric metrology, phase microscopy — reduces to reading $\Lambda$ in units of $\lambda_0$ .

Key takeaways

The optical path length is $\mathrm{OPL}=\int n\,ds$ — geometric distance weighted by refractive index, i.e. the vacuum-equivalent distance a wave "feels."
Phase follows optical, not geometric, path: $\delta=k_0\,\mathrm{OPL}=(2\pi/\lambda_0)\,\mathrm{OPL}$ , using the vacuum wavelength $\lambda_0$ because the index already sits inside the OPL.
Interference is decided by the optical path difference $\Lambda=\mathrm{OPL}_2-\mathrm{OPL}_1$ : bright when $\Lambda=m\lambda_0$ , dark when $\Lambda=(m+\tfrac12)\lambda_0$ , with integer order $m$ .
A slab of index $n$ , thickness $t$ , inserted normally into one beam adds $(n-1)t$ of OPD for a single pass and shifts the interference order by $(n-1)t/\lambda_0$ ; a double pass gives twice this OPD shift.
Reflection at a low-to-high index interface adds a $\pi$ phase shift worth $\lambda_0/2$ of OPD; this reflection term must be kept in the same ledger as the path term.
Key idea: phase follows optical path, $\delta=(2\pi/\lambda_0)\int n\,ds$ . Interference is set by the optical path difference $\Lambda$ via $\delta=2\pi\Lambda/\lambda_0$ , with $\Lambda=m\lambda_0$ bright and $\Lambda=(m+\tfrac12)\lambda_0$ dark — plus any reflection half-waves.

The core insight

The phase a wave accumulates depends on the optical path length $\mathrm{OPL}=\int n\,ds$ , not the geometric distance: $\delta=(2\pi/\lambda_0)\,\mathrm{OPL}$ . Interference between two routes is governed entirely by their optical path difference $\Lambda$ , through $\delta=2\pi\Lambda/\lambda_0$ , giving bright fringes at $\Lambda=m\lambda_0$ and dark fringes at $\Lambda=(m+\tfrac12)\lambda_0$ — after adding any $\lambda_0/2$ from reflection phase shifts.

Where this lives today

Optical path difference is the working variable of precision metrology. Interferometric refractometers still measure gas density, humidity, and plasma electron content by counting the fringes that sweep by as one arm's OPD changes; the same fringe-counting, now stabilized to a frequency comb, realizes displacement standards down to picometres in semiconductor lithography stages. Wavefront-testing interferometers such as Fizeau and Twyman–Green systems map residual optical path differences across polished mirrors and lenses to fractions of a wavelength, providing a key acceptance test for precision telescope and camera optics. In the life sciences the same physics makes the invisible visible. A transparent cell imposes an OPD $\Lambda=(n_{\text{cell}}-n_{\text{medium}})t$ across its thickness; phase-contrast and differential-interference-contrast microscopes convert that OPD into brightness, and quantitative phase imaging inverts the fringes to weigh a living cell's dry mass without any stain. And the grandest example is gravitational-wave detection: LIGO senses a passing wave as a differential OPD between its two 4-kilometre arms of order $10^{-18}\,\mathrm{m}$ — a ten-thousandth of a proton's width — read out as a fractional fringe shift.

The physical picture

A wave is a marching column of crests. What matters where two routes recombine is how many crests have passed along each — the accumulated phase. Because crests crowd together in a denser medium ( $\lambda=\lambda_0/n$ ), a given geometric length in glass holds more of them than the same length in air. The optical path length $\int n\,ds$ is precisely that crest count, re-expressed in vacuum wavelengths, which is why it — and not raw distance — controls interference.

Figure

Two routes to the same point: the lower ray crosses a slab (index

n

, thickness

t

), gaining extra optical path

(n−1)t

even though the geometric lengths match

Think of the OPD as a stopwatch difference. Two runners start together; when they meet, the OPD is the difference in optical path accumulated along the two routes. A whole number of wavelengths and they arrive in step (bright); a half-integer and they arrive in opposition (dark). Dropping a glass slab of thickness $t$ and refractive index $n$ into one route adds an optical path difference $(n-1)t$ for a single pass, shifting the fringe pattern by $(n-1)t/\lambda_0$ fringes. Because a single wavelength is sub-micrometre, counting even one fringe pins down index changes at the level of parts per million. The next figure shows this linear fringe shift directly as a function of slab thickness.

Figure

Punchline: fringe shift ΔN = (n−1)t/λ₀ grows linearly with slab thickness — glass n = 1.5 at λ = 550 nm gives 4.55 fringes at t = 5 μm

The core equations

The optical path length weights every step of a ray by the local index, so that a wave crossing many materials accumulates one honest, additive measure of "phase distance." It is defined by an integral because the index can vary along the ray.

\mathrm{OPL}=\int_A^B n(s)\,ds \qquad (1)

where: $\mathrm{OPL}$ — optical path length from A to B (vacuum-equivalent distance); $n(s)$ — refractive index along the ray, possibly position-dependent; $ds$ — element of geometric arc length. In a uniform medium this collapses to the product $\mathrm{OPL}=n\ell$ . The reason OPL matters is that phase advances at the local wavenumber $k=nk_0$ , so integrating the phase along the path pulls the index inside and leaves the vacuum wavenumber outside. The vacuum wavelength therefore sets the phase even deep inside glass.

\delta=k_0\,\mathrm{OPL}=\frac{2\pi}{\lambda_0}\int n\,ds \qquad (2)

where: $\delta$ — accumulated phase; $k_0=2\pi/\lambda_0$ — vacuum wavenumber; $\lambda_0$ — vacuum wavelength. The index lives inside the integral (in the OPL); dividing by $\lambda_0$ , not by the medium wavelength $\lambda=\lambda_0/n$ , avoids double-counting it. For two interfering routes only the difference of the two OPLs survives, and it maps directly to a phase difference and thence to the bright and dark conditions.

\Lambda=\mathrm{OPL}_2-\mathrm{OPL}_1,\qquad \delta=\frac{2\pi\Lambda}{\lambda_0} \qquad (3)

where: $\Lambda$ — optical path difference (OPD); $\delta$ — phase difference at recombination. Interference maxima occur at $\delta=2\pi m$ ( $\Lambda=m\lambda_0$ ) and minima at $\delta=(2m+1)\pi$ ( $\Lambda=(m+\tfrac12)\lambda_0$ ), before any reflection phase is added. Two special OPDs recur constantly: the slab (or gas cell) that changes one route's index over a length, and the thin film whose two reflections are separated by a round trip through the layer.

\Lambda_{\text{slab}}=(n-1)t,\qquad \Lambda_{\text{film}}=2n_f t\cos\theta_t\;\Big(\pm\tfrac{\lambda_0}{2}\Big) \qquad (4)

where: $t$ — slab or film thickness; $n,n_f$ — slab and film indices; $\theta_t$ — refraction angle inside the film; $\pm\lambda_0/2$ — reflection half-wave shift when exactly one of the two reflections is at a low-to-high index interface. The slab term is measured relative to vacuum; the film term is a physical round trip.

Derivation APhase follows the optical path, measured in vacuum wavelengths

The aim of this derivation is to prove the single most useful fact about interference bookkeeping: the phase a wave accumulates is the optical path length times the vacuum wavenumber, so that all the index-dependence is carried by the OPL and none of it by the wavelength you divide by. We track a monochromatic wave from vacuum into a medium, using the continuity of frequency and the slowing of the phase velocity.

Step 1. Across any interface the temporal frequency is continuous, $\omega_{\text{vac}}=\omega_{\text{med}}=\omega$ , because the fields must match at all times. Why: a boundary condition that held only at one instant would be violated a moment later, so the oscillation rate cannot jump — only the spatial structure can.

Step 2. A non-magnetic dielectric slows the phase velocity to $v=c/n$ , so with $\omega$ fixed the wavelength shrinks to $\lambda=v/f=\lambda_0/n$ and the wavenumber rises to $k=2\pi/\lambda=nk_0$ . Why: fewer metres travelled per second at the same beat rate means the crests sit closer together, and the spatial phase rate is exactly that crest density.

Step 3. The differential phase advance along the ray is $d\delta=k\,ds=nk_0\,ds$ , so integrating gives $\delta=\int nk_0\,ds=k_0\int n\,ds=k_0\,\mathrm{OPL}$ . Why: pulling the constant $k_0$ out of the integral leaves the index inside — the OPL — and shows the phase is the OPL measured in units of the vacuum wavenumber.

Step 4. Hence for a fixed geometric length in a uniform medium, $\delta=nk_0\ell=k_0(n\ell)=(2\pi/\lambda_0)\,\mathrm{OPL}$ ; the medium wavelength never appears in the final phase. Why: this is precisely why OPD is divided by $\lambda_0$ and not $\lambda$ — the index has already done its work inside the OPL, and dividing by $\lambda$ would count it twice.

Go deeper · Derivation B · 1 minFringe shift from an inserted slab, and single versus double pass

Derivation BFringe shift from an inserted slab, and single versus double pass

This derivation delivers the working tool of interferometric refractometry: the number of fringes that sweep past a fixed point when a transparent slab or gas cell is introduced into one beam. The result is the added OPD in vacuum wavelengths, with a crucial factor of two depending on whether the beam crosses the sample once or twice. We replace a length of vacuum in one route by a medium of index $n$ .

Step 1. Before insertion the route traverses $\mathrm{OPL}_{\text{before}}=1\cdot t$ of vacuum; after insertion it traverses $\mathrm{OPL}_{\text{after}}=n\,t$ of the medium, so the change in OPD is $\Delta\Lambda=(n-1)t$ . Why: only the excess of the index over vacuum changes the count of crests; the geometric length is untouched, so nothing but the $(n-1)$ factor enters.

Step 2. The pattern shifts by one order for every $\lambda_0$ of added OPD: $\Delta N=\Delta\Lambda/\lambda_0=(n-1)t/\lambda_0$ (single pass). Why: each $2\pi$ of phase carries the cosine through one full cycle, moving each fringe to where its neighbour stood — one order per vacuum wavelength.

Step 3. In a Michelson interferometer the sample sits in an arm the light crosses twice (out and back), so the OPD change doubles: $\Delta N=2(n-1)t/\lambda_0$ , and inverting gives $n-1=\Delta N\,\lambda_0/(2t)$ . Why: a round trip through the arm passes through the sample on the way out and again on the way back, so the index excess is accumulated twice — miss this and you infer half the true refractivity.

Step 4. The sensitivity is extreme: with $\lambda_0\sim 0.5\,\mu\mathrm{m}$ and $t\sim 0.1\,\mathrm{m}$ , a single fringe in a single-pass measurement corresponds to $\Delta n\sim\lambda_0/t\sim 5\times10^{-6}$ ; in a double-pass measurement the sensitivity improves by a factor of two, to $\Delta n\sim\lambda_0/(2t)\sim 2.5\times10^{-6}$ . Why: dividing a sub-micrometre wavelength by a decimetre baseline turns a countable fringe into a parts-per-million index change — which is how gas density and plasma refractivity are measured.

Go deeper · Derivation C · 1 minThe reflection half-wave belongs in the same ledger as the path

Derivation CThe reflection half-wave belongs in the same ledger as the path

Path length is not the whole story: a reflection can add phase all by itself, and this derivation shows that the $\pi$ shift picked up on reflection from a denser medium enters the interference condition exactly like an extra $\lambda_0/2$ of optical path. Keeping it in the same account is what makes a vanishingly thin film look dark. We compare the two reflections that produce thin-film fringes.

Step 1. The wave reflected from the top surface (air to film, low to high index) suffers a sign flip of the field, a phase jump of $\pi$ ; the wave reflected from the bottom surface (film to air, high to low) does not. Why: the boundary conditions force the reflected field to invert only when the wave meets a slower medium, so exactly one of the two reflections carries the extra $\pi$ .

Step 2. Convert that $\pi$ into path units: a phase $\pi$ equals $2\pi\cdot(\lambda_0/2)/\lambda_0$ , i.e. a half-wave, so the reflection contributes $\lambda_0/2$ of effective OPD. Why: phase and path are interchangeable through $\delta=2\pi\Lambda/\lambda_0$ , so a bare $\pi$ is worth precisely half a vacuum wavelength of path.

Step 3. The total OPD between the two reflected waves is therefore $\Lambda=2n_f t\cos\theta_t+\lambda_0/2$ , and constructive reflection requires $\Lambda=m\lambda_0$ , i.e. $2n_f t\cos\theta_t=(m-\tfrac12)\lambda_0$ . Why: the physical round trip and the reflection half-wave add in the same ledger, so the bright condition on the path term picks up a half-integer offset.

Step 4. Take $t\to 0$ : the path term vanishes but the $\lambda_0/2$ remains, so $\Lambda=\lambda_0/2$ — the destructive condition — and the film is dark in reflection. Why: this is the famous dark centre of a wedge fringe or Newton's rings at contact, produced entirely by the reflection phase with no path difference at all.

Limiting cases

Case	Result	Meaning
uniform medium	$\mathrm{OPL}=n\ell$	the integral collapses to a product — most problems are $n\ell$ bookkeeping
vacuum or air	$n\to 1,\ \mathrm{OPL}\to\ell$	OPL equals geometric length; $\lambda=\lambda_0$ , phase is $k_0\ell$
thin gas slab	$\Lambda=(n-1)L$	OPD relative to vacuum; fringe count gives $n-1$ to parts in $10^7$
near-normal thin film	$\Lambda\approx 2n_f t\ (\pm\lambda_0/2)$	round-trip OPD; add the half-wave if one reflection is low-to-high
film at contact ( $t\to 0$ )	$\Lambda\to\lambda_0/2$	reflection half-wave alone — dark centre, no path difference
dispersive medium	$\mathrm{OPL}(\omega)=n(\omega)\ell$	each colour sees its own OPL; group delay uses $n_g=n+\omega\,dn/d\omega$

Common trapDividing the OPD by the wrong wavelength — and forgetting the reflection half-wave

Two classic errors sink OPD calculations. First, converting OPD to phase with the medium wavelength $\lambda=\lambda_0/n$ instead of the vacuum wavelength $\lambda_0$ : the OPD already carries the index inside $\int n\,ds$ , so you must divide by $\lambda_0$ — using $\lambda$ counts the index twice and gives a phase too large by a factor $n$ . Second, omitting the $\pi$ (half-wave, $\lambda_0/2$ ) shift on reflection from a low-to-high index interface: it flips the bright and dark conditions and is why a thin soap film is dark, not bright, at its thinnest. Path phase and reflection phase belong in one ledger; leave either out and every fringe count is wrong.

Worked examples

1 · A mica sheet shifts Young's fringes (confidence builder)

A thin mica sheet of index $n=1.58$ and thickness $t=5\,\mu\mathrm{m}$ is laid over one slit of a double-slit experiment illuminated at $\lambda_0=589\,\mathrm{nm}$ . The light crosses the sheet once. By how many fringes does the pattern shift?

Step 1. The added optical path is the index excess times the thickness: $\Lambda=(n-1)t=(1.58-1)(5\,\mu\mathrm{m})=0.58\times 5\,\mu\mathrm{m}=\mathbf{2.90\,\mu m}$ .

Step 2. Convert to a fringe shift by dividing by the vacuum wavelength: $\Delta N=\Lambda/\lambda_0=2.90\,\mu\mathrm{m}/0.589\,\mu\mathrm{m}=\mathbf{4.92}$ orders.

Step 3. Interpret the sign: the covered slit's route is lengthened optically, so the central (zero-OPD) fringe moves toward that slit, dragging the whole pattern with it by almost five fringe spacings. Measuring this shift with a known $t$ yields $n$ ; with a known $n$ it yields $t$ — the working principle of the classic mica-sheet determination of thin-film thickness. The pattern shifts by 4.92 fringes toward the covered slit.

2 · Weighing air with a Michelson interferometer (exam level)

A $10\,\mathrm{cm}$ gas cell in one arm of a Michelson interferometer is slowly evacuated at $\lambda_0=633\,\mathrm{nm}$ , and $N=90$ fringes are counted passing a fixed point. Find the refractive index of the air that was in the cell.

Step 1. In a Michelson the arm is traversed twice, so the OPD change on removing gas of index $n$ over length $L$ is a double pass: $\Delta\Lambda=2(n-1)L$ , and the fringe count is $N=\Delta\Lambda/\lambda_0=2(n-1)L/\lambda_0$ .

Step 2. Solve for the refractivity: $n-1=\dfrac{N\lambda_0}{2L}=\dfrac{90\times 633\times 10^{-9}\,\mathrm{m}}{2\times 0.10\,\mathrm{m}}=\dfrac{5.697\times10^{-5}}{0.20}=\mathbf{2.85\times10^{-4}}$ , so $n=\mathbf{1.000285}$ .

Step 3. Check plausibility and the factor of two: standard air at $633\,\mathrm{nm}$ has $n-1\approx 2.7\times10^{-4}$ , so $2.85\times10^{-4}$ is right for slightly dense or humid air — and had we forgotten the double pass we would have reported $1.4\times10^{-4}$ , badly wrong. Counting a single fringe here already resolves $\Delta n=\lambda_0/(2L)\approx 3\times10^{-6}$ . n − 1 = 2.85 × 10⁻⁴, i.e. n = 1.000285, with the factor of two from the round trip essential.

Physics of the portrait

Optical path length is the quantity that organizes the whole subject — the vacuum-equivalent distance accumulated by a ray, in terms of which Fermat's principle can be expressed and interference fringes are counted. It is the bridge between ray optics, wave optics, and coherence.

Origins · 1 minoptical path length and how this was discovered

optical path length

the currency of interference, from Fermat 1662 · n times geometric distance, measured in vacuum wavelengths

Optical path length, $\mathrm{OPL}=\int n\,ds$ , is the single quantity that unifies geometric and wave optics. Pierre de Fermat stated his principle of least time in 1662 as a condition on this quantity — a real ray renders the OPL stationary, $\delta\int n\,ds=0$ — and the entire edifice of ray optics, from image formation to aberration theory, is written in OPLs. A lens forms a sharp image precisely by equalising the OPL of every ray from object to image point: the central rays travel less distance but through more glass, the marginal rays more distance through less glass, and the design goal is that all arrive with the same accumulated phase and interfere constructively. In wave language the OPL from a reference surface is the eikonal $S(\mathbf r)$ , satisfying $|\nabla S|=n$ , and surfaces of constant $S$ are the wavefronts — so OPL is literally the phase clock of the field. Coherence theory then adds the final layer: coherence simply tells us over what range of OPD the phase relationship, and hence the fringes, survive. From Fermat's stopwatch to the metre defined by a krypton wavelength, the OPL is the ruler optics measures with.

History

1662

— Fermat states the principle of least time as a condition on optical path length

1801

— Young interprets interference as a difference in optical path, explaining fringe colours

1851

— Fizeau measures the drag of light in moving water from the fringe shift of a split beam

1881

— Michelson builds the interferometer that reads optical path differences to a fraction of a wavelength

1896

— Rayleigh develops the interference refractometer, measuring gas refractivity from fringe shifts

Going further

Texts. Hecht, Optics, 5th ed., Ch. 4 (propagation, optical path length) and Ch. 9 (interference) · Saleh & Teich, Fundamentals of Photonics, 3rd ed., Ch. 1 (ray optics, Fermat) and Ch. 2 (wave optics) · Born & Wolf, Principles of Optics, 7th ed., §3.1 (eikonal) and Ch. 7 (interference) · Foot, Atomic Physics, appendix on interferometry · Demtröder, Laser Spectroscopy, Ch. 4 (interferometers). The core relations — $\mathrm{OPL}=\int n\,ds$ , $\delta=k_0\,\mathrm{OPL}$ , $\delta=2\pi\Lambda/\lambda_0$ , the bright/dark conditions $\Lambda=m\lambda_0$ / $(m+\tfrac12)\lambda_0$ , the slab OPD $(n-1)t$ , the film OPD $2n_f t\cos\theta_t\pm\lambda_0/2$ , and the eikonal $|\nabla S|=n$ — are standard results of wave optics. The worked numbers confirm these: mica single-pass $\Delta N=4.92$ , Michelson double-pass gas result $n-1=2.85\times10^{-4}$ , the $\pi\leftrightarrow\lambda_0/2$ equivalence, and the dark-at-contact film condition. Historical dates: Fermat 1662, Michelson interferometer 1881, Rayleigh refractometer 1896.

Prerequisites. Superposition Principle of Electromagnetic Waves · Two-Wave Interference and the Interference Term. Leads to. Conditions for Observable Interference · Thin-Film Interference · Temporal Coherence and Coherence Length · Newton's Rings.

Why divide the optical path difference by the vacuum wavelength and not the wavelength in the medium?

Because the optical path difference $\Lambda=\int n\,ds$ already has the index folded in — it is a vacuum-equivalent length. The phase is $\delta=k_0\Lambda=2\pi\Lambda/\lambda_0$ , using the vacuum wavenumber. If you instead divided by the medium wavelength $\lambda=\lambda_0/n$ , you would multiply by $n$ a second time and overstate the phase by a factor $n$ . The clean way to remember it: put all index dependence into the OPL, then measure that OPL in vacuum wavelengths and nowhere else.

Does the frequency or wavelength of light change when it enters glass?

The frequency does not change — it is fixed by the source and forced to be continuous across the interface by the boundary conditions. What changes is the phase velocity ( $v=c/n$ ) and therefore the wavelength ( $\lambda=\lambda_0/n$ ): the crests bunch closer together. A red laser stays red in glass because colour is set by frequency, not by the local wavelength. This is exactly why the optical path length uses $n\,ds$ : the same geometric length holds more crests in the medium, and OPL counts those crests in vacuum units.

Why does inserting a slab shift the fringes rather than just dimming them?

A slab adds a fixed optical path $(n-1)t$ to one route, which slides that route's phase by $2\pi(n-1)t/\lambda_0$ everywhere on the screen. Since the fringe pattern is a map of OPD across the screen, a constant added OPD simply translates the whole map sideways by $(n-1)t/\lambda_0$ orders — the zero-OPD (central) fringe moves toward the lengthened arm. The contrast is unchanged as long as the slab is uniform and the light stays coherent; only the position shifts. Counting that shift is how the slab's index or thickness is measured.

How can one fringe correspond to a nanometre-scale index change?

Because a fringe is a full vacuum wavelength of OPD, and in refractometry that OPD is spread over a long path. Over a $10\,\mathrm{cm}$ double-pass cell, one fringe means $\Delta\Lambda=\lambda_0\approx 633\,\mathrm{nm}$ of OPD change, i.e. $\Delta n=\lambda_0/(2L)\approx 3\times10^{-6}$ . The long baseline amplifies a tiny index change into a countable, whole-fringe event. Stabilize the source and interpolate within a fringe and refractometers reach parts in $10^8$ , which is how air's dispersion is tabulated for length metrology.

Is the reflection half-wave shift a path difference or something else?

It is a genuine phase shift, not a path, but it enters the interference condition exactly as if it were an extra $\lambda_0/2$ of optical path — because phase and path are interchangeable through $\delta=2\pi\Lambda/\lambda_0$ . Physically it comes from the boundary conditions forcing the reflected field to flip sign when the wave meets a higher-index medium. You keep it in the same ledger as the geometric OPD, adding $\lambda_0/2$ whenever exactly one of the two interfering reflections is low-to-high. When both (or neither) reflect low-to-high, the two half-waves cancel and no net shift remains.

How does Fermat's principle relate to optical path difference?

Fermat's principle says a real ray makes the optical path length stationary, $\delta\int n\,ds=0$ . The deep reason, from wave optics, is interference: contributions from the bundle of nearby paths all share nearly the same OPL near a stationary path, so they add in phase, while off-stationary paths have rapidly varying OPL and cancel. A perfect image is the extreme case — every ray from object to image point is arranged to have the same OPL, so their OPDs are all zero and they interfere fully constructively. Optical path difference is thus the quantity Fermat's principle is quietly minimizing the spread of.