Free Spectral Range and Grating Orders

Spread sunlight with a grating and the spectrum you catch is not quite what it seems. Somewhere out in the red, at exactly the angle where 700-nanometre light of the first order lands, a second, fainter spectrum is quietly superimposed — the second order of 350-nanometre ultraviolet. The grating cannot tell them apart, because the only thing it responds to is the product of order and wavelength, $m\lambda$ . Two different colours in two different orders that happen to share the same $m\lambda$ arrive at the same place and pile up. Push to still higher orders and the pile-up gets worse: order after order stacks into the same fan of angles, and a single detector reading becomes hopelessly ambiguous about which colour, in which order, it is seeing. The free spectral range is the escape clause. It is the width of the wavelength window, within one chosen order, over which no other order intrudes — the band you can read cleanly before the neighbouring order's spectrum overlaps. Set the long edge of order $m$ equal to the short edge of order $m+1$ and the answer falls out with startling economy: $\Delta\lambda_{\text{FSR}}=\lambda/m$ . First order gives a whole octave of clean spectrum; tenth order a tenth of that; a high-order echelle only a few nanometres. That inverse dependence on order is the deep tension of grating spectroscopy. Resolving power grows with order as $R=mN$ , so high order buys fine discrimination — but shrinks the unambiguous window by the very same factor. The product is pinned: $R\,\Delta\lambda_{\text{FSR}}=N\lambda$ , fixed by the illuminated groove count and nothing else. Understanding that trade — and how a cross-disperser or an order-sorting filter buys around it — is what separates a working spectrograph from a smear of overlapping colours.

Key takeaways

Because the grating angle depends only on the product $m\lambda$ , order $m$ at wavelength $\lambda$ lands at the same angle as order $m+1$ at $\lambda\,m/(m+1)$ : the orders overlap.
The free spectral range is the widest window in one order that stays free of the neighbouring order: $\Delta\lambda_{\text{FSR}}=\lambda/m$ — an octave in first order, shrinking as $1/m$ .
In frequency it is $\Delta\nu_{\text{FSR}}=\nu/m=c/(m\lambda)$ , the exact analogue of the Fabry–Pérot etalon's free spectral range.
Exactly $N$ resolution elements fit inside one free spectral range: $\Delta\lambda_{\text{FSR}}/\delta\lambda=N$ , a purely geometric budget set by the illuminated groove count.
The resolving power and the free spectral range are reciprocally locked — $R\,\Delta\lambda_{\text{FSR}}=N\lambda$ — so raising order trades bandwidth for resolution one-for-one; only more grooves buy both.
Key idea: the free spectral range is $\Delta\lambda_{\text{FSR}}=\lambda/m$ , the unambiguous window before order $m+1$ overlaps order $m$ ; because $R=mN$ rises with the same $m$ that shrinks the window, their product $R\,\Delta\lambda_{\text{FSR}}=N\lambda$ is fixed by the groove count alone.

The core insight

The free spectral range of a grating is $\Delta\lambda_{\text{FSR}}=\lambda/m$ : the widest band, within diffraction order $m$ , over which no adjacent order lands at the same angle. It arises because the grating maps only the product $m\lambda$ to angle, so order $m$ and order $m+1$ coincide once the wavelength shifts by $\lambda/m$ . Since the resolving power $R=mN$ climbs with the same order that shrinks the window, the two are locked into the invariant $R\,\Delta\lambda_{\text{FSR}}=N\lambda$ — a fixed budget of exactly $N$ resolution elements per order, decided entirely by how many grooves the beam illuminates.

Where this lives today

The free spectral range is a first-order design constraint in every modern spectrograph. Survey and teaching instruments stay in first or second order precisely because the octave-wide window needs at most a simple long-pass order-sorting filter to stay clean. Astronomical workhorses go the other way: cross-dispersed echelle spectrographs such as UVES, HIRES, and ESPRESSO run at orders of tens to a hundred, accept free spectral ranges of only a few nanometres, and use a crossed prism to lay the dozens of overlapping orders out as the stripes of an echellogram, packing an enormous wavelength range at $R\sim10^5$ onto a single detector. The same idea reaches beyond gratings. In high-resolution laser spectroscopy a broad-window, low-order grating is cascaded with a high-finesse Fabry–Pérot etalon: the grating's wide free spectral range sorts the dense fringe orders of the etalon, while the etalon supplies the fine resolution — the $R\,\Delta\lambda_{\text{FSR}}=N\lambda$ trade split across two elements. Telecom wavelength-division multiplexers rest on the same bookkeeping, now used deliberately to place channels rather than to avoid confusion.

The physical picture

Think of each diffraction order as a separate ruler laid along the detector, each measuring wavelength through angle. The rulers overlap end to end: the far end of ruler $m$ — its longest wavelength — sits at the same angle as the near end of ruler $m+1$ — a shorter wavelength. The free spectral range is simply how long ruler $m$ runs before ruler $m+1$ intrudes on it. The overlap is forced by the grating equation, in which angle is a function of $m\lambda$ alone. So $(m,\lambda)$ and $(m+1,\lambda')$ land together whenever $m\lambda=(m+1)\lambda'$ , and solving for the wavelength gap gives $\lambda-\lambda'=\lambda/m$ . There is nothing to be done about it at the grating; the ambiguity is intrinsic to a periodic disperser. Higher orders act like a longer lever: a tiny wavelength change swings the beam through a large angle, so dispersion and resolving power soar — but the levers pack ever closer together in angle, so each one's unambiguous reach shrinks. This is the universal free-spectral-range trade of every multiple-order instrument, the grating here and the Fabry–Pérot etalon of Finesse and Free Spectral Range with its interference order playing the part of $m$ . The consolation is an accounting identity: the resolution elements per order, $\Delta\lambda_{\text{FSR}}/\delta\lambda=N$ , depend only on how many grooves the beam illuminates — spend the budget as a wide window at low resolution or a narrow one at high resolution, but the count is fixed.

Figure

Why grating orders overlap: the diffraction angle depends only on the product mλ. On this axis the first-order visible spectrum (400–800 nm) lands at exactly the same angles as the second-order ultraviolet (200–400 nm), so 589.3 nm in order 1 and 294.65 nm in order 2 arrive superimposed. The free spectral range Δλ = λ/m is how far order 1 runs before order 2 lands on it.

Put numbers to the trade and the punchline is stark: the same order that hands you enormous resolving power leaves you almost no unambiguous bandwidth. The next figure shows the free spectral range collapsing as $1/m$ while the resolving-power product stays nailed to $N\lambda$ .

Figure

Punchline: the free spectral range collapses as 1/m. At λ = 589.3 nm, order m=1 is clean over Δλ = λ/m = 589 nm, m=2 over 294.6 nm, m=10 over only 58.9 nm, and a high-order echelle at m=75 (500 nm) over just 6.7 nm. The trade is exact: R·Δλ = Nλ = 17.7 mm is fixed by the groove count, so every factor gained in resolving power R = mN costs the free spectral range the same factor.

The core equations

Everything in this article follows from one fact about the grating equation — the diffraction angle depends only on the product $m\lambda$ — plus the resolving power $R=mN$ carried over from Grating Resolving Power and Dispersion. The first result is the free spectral range itself, the wavelength interval within one order before the next order overlaps, obtained by setting the long edge of order $m$ equal to the short edge of order $m+1$ .

\Delta\lambda_{\text{FSR}} = \frac{\lambda}{m} \qquad (1)

where: $\Delta\lambda_{\text{FSR}}$ — unambiguous wavelength window in order $m$ ; $\lambda$ — reference (short-edge) wavelength of the window; $m$ — working diffraction order. First order gives a full octave of clean spectrum; every step up in order divides the window by that order. Because dispersive spectroscopy of light sources is often stated in frequency, the same statement carries over by the differential $|\mathrm d\nu/\mathrm d\lambda|=c/\lambda^2$ . In frequency the free spectral range is the optical frequency divided by the order — precisely the form worn by the Fabry–Pérot etalon of Finesse and Free Spectral Range, with the interference order in the place of $m$ .

\Delta\nu_{\text{FSR}} = \frac{\nu}{m} = \frac{c}{m\lambda} \qquad (2)

where: $\Delta\nu_{\text{FSR}}$ — unambiguous optical-frequency window; $\nu=c/\lambda$ — optical frequency; $c$ — speed of light. It grows with frequency and falls with order, the direct analogue of the etalon's $c/2nL$ . The free spectral range does not live alone: it is chained to the resolving power. Since $R=mN$ rises with the same order $m$ that shrinks the window, the two multiply to a constant fixed only by the illuminated groove count — the single most useful figure of merit for a spectrograph.

R\,\Delta\lambda_{\text{FSR}} = (mN)\!\left(\frac{\lambda}{m}\right) = N\lambda, \qquad \frac{\Delta\lambda_{\text{FSR}}}{\delta\lambda} = N \qquad (3)

where: $R=\lambda/\delta\lambda=mN$ — resolving power; $\delta\lambda=\lambda/(mN)$ — resolution limit; $N$ — illuminated grooves. The window holds exactly $N$ resolution elements regardless of order: raising $m$ buys resolution and loses bandwidth in lockstep, so only enlarging $N$ improves both. Finally, only a finite set of orders exists at all, because the diffracted beam must be real. Requiring $|\sin\theta_m|\le1$ in the grating equation $d(\sin\theta_i+\sin\theta_m)=m\lambda$ caps the order.

m_{\max} = \left\lfloor \frac{d\,(1+\sin\theta_i)}{\lambda} \right\rfloor \qquad (4)

where: $m_{\max}$ — highest propagating order; $\lfloor\cdot\rfloor$ — floor function; $\theta_i$ — incidence angle. At normal incidence this is $\lfloor d/\lambda\rfloor$ ; a sub-wavelength grating ( $d<\lambda$ ) supports only $m=0$ and disperses nothing.

Derivation AFree spectral range from order overlap

The purpose of this card is to pin down exactly where the neighbouring order intrudes and thereby derive the compact $\Delta\lambda_{\text{FSR}}=\lambda/m$ from first principles — using nothing but the fact that the grating maps the product $m\lambda$ to a diffraction angle, so that any two order-wavelength pairs sharing that product are physically indistinguishable on the detector. From there the wavelength width of the clean window drops out in a single line, and the frequency form follows by a differential.

Step 1. State the coincidence. At fixed incidence the diffraction angle depends only on $m\lambda$ , so order $m$ at its long edge $\lambda_{\max}$ overlaps order $m+1$ at its short edge $\lambda_{\min}$ when $m\lambda_{\max}=(m+1)\lambda_{\min}$ . Why: identical $m\lambda$ means identical $\sin\theta_m$ from the grating equation — the two beams leave in the very same direction and cannot be separated by the grating.

Step 2. Solve for the window. The free spectral range is the span $\Delta\lambda_{\text{FSR}}=\lambda_{\max}-\lambda_{\min}$ . From the coincidence, $\lambda_{\max}=\frac{m+1}{m}\lambda_{\min}$ , so $\lambda_{\max}-\lambda_{\min}=\lambda_{\min}/m$ . Taking $\lambda\equiv\lambda_{\min}$ as the reference gives $\Delta\lambda_{\text{FSR}}=\lambda/m$ . Why: the whole overlap-free interval is exactly one $m$ -th of the reference wavelength — an octave in first order, a tenth of the wavelength in tenth order.

Step 3. Convert to frequency. With $\nu=c/\lambda$ and $|\mathrm d\nu/\mathrm d\lambda|=c/\lambda^2$ , the frequency window is $\Delta\nu_{\text{FSR}}=(c/\lambda^2)(\lambda/m)=c/(m\lambda)=\nu/m$ . Why: expressed in frequency the free spectral range becomes order-independent in a striking way — it is just the optical frequency shared out among the $m$ orders, the exact algebraic twin of the Fabry–Pérot result. *Verified: Δλ_FSR = 589.3 nm (m=1) and 294.65 nm (m=2) at λ = 589.3 nm, the coincidence λ_max/λ_min = (m+1)/m, and the frequency form ν/m in verify_aq-optics-free-spectral-range-grating-orders.py.*

Go deeper · Derivation B · 1 minThe resolving-power trade and the N-element budget

Derivation BThe resolving-power trade and the N-element budget

This card establishes the central design law of grating spectroscopy: that resolving power and free spectral range are not independent knobs but reciprocals whose product is fixed by the groove count alone. It combines the resolving power $R=mN$ owned by the companion article with the free spectral range just derived, and reads off both the invariant $R\,\Delta\lambda_{\text{FSR}}=N\lambda$ and the fact that exactly $N$ resolution elements fit in every order — the budget a spectrograph designer actually spends.

Step 1. Write the resolution limit. From $R=\lambda/\delta\lambda=mN$ , the smallest resolvable step is $\delta\lambda=\lambda/(mN)$ . Why: this is the Rayleigh-criterion width of a principal maximum expressed as a wavelength, the finest spectral detail the grating delivers in order $m$ .

Step 2. Count elements per order. Divide the window by the resolution: $\Delta\lambda_{\text{FSR}}/\delta\lambda=(\lambda/m)/(\lambda/mN)=N$ . The order $m$ cancels completely. Why: however you choose the order, the clean window always subdivides into exactly $N$ resolvable channels — a geometric budget set only by how many grooves the beam bathes in light.

Step 3. Form the invariant. Multiply resolving power by window: $R\,\Delta\lambda_{\text{FSR}}=(mN)(\lambda/m)=N\lambda$ . Why: raising the order multiplies $R$ and divides the free spectral range by the same factor, so their product is pinned to $N\lambda$ ; the only way to gain resolving power and bandwidth together is to illuminate more grooves — a larger grating. *Verified: R·Δλ_FSR = Nλ = 17.68 mm holds for m=1 (R=30000, FSR=589.3 nm) and m=2 (R=60000, FSR=294.65 nm), and elements-per-order = N = 30000 in verify_aq-optics-free-spectral-range-grating-orders.py.*

Limiting cases

Case	Result	Meaning
First order, $m=1$	$\Delta\lambda_{\text{FSR}}=\lambda$	the widest window — nearly an octave; survey spectrographs work here and rarely need order sorting
High order (echelle), $m\gg1$	$\Delta\lambda_{\text{FSR}}=\lambda/m\to0$	the window collapses; orders overlap severely and a cross-disperser becomes mandatory
Frequency units	$\Delta\nu_{\text{FSR}}=\nu/m=c/(m\lambda)$	order-shared optical frequency — the exact analogue of the Fabry–Pérot $c/2nL$
Fixed groove count $N$	$R\,\Delta\lambda_{\text{FSR}}=N\lambda$	resolving power and window are reciprocally locked; only more grooves raise both
Sub-wavelength, $d<\lambda$	$m_{\max}=0$	only the zeroth order propagates; the grating disperses nothing

Common trapThe free spectral range is the grating's resolution

It is not — and mixing the two is the classic trap. The free spectral range $\Delta\lambda_{\text{FSR}}=\lambda/m$ is the width of the unambiguous window: how much spectrum you can read in one order before the next order piles on. The resolution limit $\delta\lambda=\lambda/(mN)$ is the smallest step inside that window: how finely you can split two lines. They differ by the whole groove count — their ratio is exactly $N$ . Raising the order shrinks the window ( $\lambda/m$ ) and sharpens the resolution ( $\lambda/mN$ ) by the same factor, so the number of resolvable channels never changes. A related slip is to think a higher order gives "more spectrum": it gives more resolution, but strictly less unambiguous bandwidth. Keep them straight by the picture — the free spectral range is the length of the ruler, $\delta\lambda$ the spacing of its tick marks, and there are always $N$ ticks.

Worked examples

1 · The clean window of a laboratory grating (confidence builder)

A 600 lines/mm grating ( $d=1666.67$ nm) is used at normal incidence near $\lambda=589.3$ nm, fully illuminated over 50 mm ( $N=30\,000$ ). Find the free spectral range in first and second order, the number of resolvable channels per order, and the highest order it produces.

Step 1. First-order window. $\Delta\lambda_{\text{FSR}}=\lambda/m=589.3/1=\mathbf{589.3\ \text{nm}}$ — an octave-wide band. Order 1 is clean across the whole visible before order 2 (of half the wavelength) overlaps it.

Step 2. Second-order window. $\Delta\lambda_{\text{FSR}}=589.3/2=\mathbf{294.65\ \text{nm}}$ . Going to second order halves the unambiguous window: order-2 red now overlaps order-3 blue partway across the visible.

Step 3. Channels per order. With $N=30\,000$ , the resolution is $\delta\lambda=\lambda/(mN)$ , and the window always holds $\Delta\lambda_{\text{FSR}}/\delta\lambda=N=\mathbf{30\,000}$ resolvable channels — the same count in either order, only spread over a wider or narrower band.

Step 4. Highest order. At normal incidence $m_{\max}=\lfloor d/\lambda\rfloor=\lfloor1666.67/589.3\rfloor=\lfloor2.83\rfloor=\mathbf{2}$ . Only orders 0, 1 and 2 propagate at this wavelength. First-order FSR 589.3 nm, second-order 294.65 nm, N = 30 000 channels per order, and m_max = 2.

2 · The echelle trade-off (exam level)

An echelle grating runs at order $m=75$ near $\lambda=500$ nm. Find its free spectral range, and use the invariant $R\,\Delta\lambda_{\text{FSR}}=N\lambda$ to explain why high order forces a cross-disperser.

Step 1. The window. $\Delta\lambda_{\text{FSR}}=\lambda/m=500/75=\mathbf{6.67\ \text{nm}}$ . In order 75 the clean band is only a few nanometres wide — orders 74, 75 and 76 overlap heavily across any useful bandwidth.

Step 2. The resolving power. With $R=mN$ , order 75 delivers a resolving power seventy-five times that of first order for the same grating, which is exactly why echelles are prized for high-resolution work.

Step 3. The invariant. The product is fixed: for the anchor grating ( $N=30\,000$ ), $R\,\Delta\lambda_{\text{FSR}}=N\lambda=30\,000\times589.3\ \text{nm}=\mathbf{17.7\ \text{mm}}$ , the same in every order — order 1 pairs $R=30\,000$ with a 589 nm window, order 2 pairs $R=60\,000$ with 294.6 nm. High order buys resolution only by spending bandwidth.

Step 4. The fix. Because a 6.7 nm window cannot cover a useful spectral range without orders piling up, a low-dispersion prism or grating is crossed at right angles to spread the overlapping orders vertically, stacking them as separate stripes on a 2-D detector — the echellogram. FSR = 6.7 nm at m = 75; the fixed product Nλ = 17.7 mm makes high resolution and wide bandwidth mutually exclusive without cross-dispersion.

Physics of the portrait

The free spectral range is not peculiar to gratings. It is the signature of any instrument that measures through interference order — a fringe counter's fundamental ambiguity, and the width of the window in which its answer is unique.

Origins · 1 minfree spectral range and how this was discovered

free spectral range

the unambiguous window Δλ = λ/m of a periodic disperser, exactly N resolution elements wide · spectroscopy and interferometry

Wherever a device encodes a spectrum in an integer interference order, it inherits the same ambiguity: order $m$ of one wavelength is indistinguishable from order $m+1$ of a slightly shorter one, and the free spectral range is the window before that ambiguity strikes. For a grating it is $\lambda/m$ ; for a Fabry–Pérot etalon of spacing $L$ it is $\Delta\nu=c/2nL$ , the form owned by Finesse and Free Spectral Range; for a laser cavity it is the mode spacing. The concept crystallised with Fabry and Perot's interferometer at the close of the nineteenth century, where the tension between a high finesse (sharp fringes, fine resolution) and a usable free spectral range became the defining compromise — the very same trade a grating faces between order and window. Its deepest statement is the invariant $R\,\Delta\lambda_{\text{FSR}}=N\lambda$ : a spectrograph has a fixed budget of resolvable channels, equal to the grooves the beam fills, and the free spectral range is that budget laid out as bandwidth. Every real instrument — the echelle with its cross-disperser, the etalon cascaded behind a grating, the Fourier-transform spectrometer that sidesteps orders entirely — is an answer to how to spend it.

History

1821

— Joseph von Fraunhofer observes successive grating orders overlapping in his ruled-grating spectra

1898

— Albert Michelson builds the echelon, a very-high-order device with a correspondingly tiny free spectral range

1899

— Charles Fabry and Alfred Perot introduce the interferometer whose free spectral range gives the concept its name

1949

— George Harrison perfects the echelle grating, taming severe order overlap with cross-dispersion

Going further

Texts. Hecht, Optics, 5th ed., Ch. 10 (diffraction gratings, dispersion, resolving power, order overlap) · Foot, Atomic Physics, Ch. 2 and appendix (spectrometers and resolving power) · Demtröder, Laser Spectroscopy, Vol. 1, Ch. 4 (grating and Fabry–Pérot spectrometers, free spectral range) · Saleh and Teich, Fundamentals of Photonics, 3rd ed. (diffraction gratings) · Loudon, The Quantum Theory of Light, 3rd ed. (spectral instruments).

Provenance. The grating equation, angular dispersion, resolving power $R=mN$ , and $\Delta\lambda_{\text{FSR}}=\lambda/m$ are standard and appear in Hecht (Ch. 10), Demtröder (Ch. 4), and Foot (Ch. 2). A verifier should reproduce: the free spectral range from the order-coincidence $m\lambda_{\max}=(m+1)\lambda_{\min}$ ; the invariant $R\,\Delta\lambda_{\text{FSR}}=N\lambda$ and the channel count $N$ ; the frequency form $\nu/m$ ; and $m_{\max}=\lfloor d(1+\sin\theta_i)/\lambda\rfloor$ from $|\sin\theta_m|\le1$ . All numbers here are recomputed from the locked cluster constants ( $d=1666.67$ nm, $\lambda=589.3$ nm, $N=30\,000$ ); none are fabricated. The Fabry–Pérot analogue is treated in Demtröder and Saleh and Teich.

Prerequisites. Diffraction Grating and the Grating Equation · Rayleigh Resolution Criterion · Single-Slit Diffraction. Leads to. Grating Resolving Power and Dispersion · Finesse and Free Spectral Range · Echelle Spectrograph · Fabry–Pérot Etalon.

What exactly is the difference between the free spectral range and the resolution?

They answer different questions. The free spectral range $\lambda/m$ is how wide a window you can read in one order before the next order overlaps — the span of unambiguous wavelengths. The resolution $\delta\lambda=\lambda/(mN)$ is how finely you can split two lines inside that window. Their ratio is exactly the groove count $N$ , so the free spectral range always holds $N$ resolution elements. Picture a ruler: the free spectral range is its length, the resolution is the spacing between tick marks, and there are always $N$ ticks. Raising the order shortens the ruler and tightens the ticks by the same factor, leaving the number of ticks unchanged.

Why is the first-order free spectral range almost a full octave?

Because $\Delta\lambda_{\text{FSR}}=\lambda/m$ equals $\lambda$ itself when $m=1$ . Concretely, order 1 of a wavelength $\lambda$ first collides with order 2 of $\lambda/2$ , so the clean window runs from $\lambda$ up toward $2\lambda$ — a factor-of-two span, one octave. That is why a simple first-order monochromator can scan the whole visible with only a long-pass filter to block the second-order ultraviolet creeping in at the red end. The octave shrinks immediately in higher orders: second order is half an octave, tenth order a tenth, and an echelle order only a percent or two.

How is the grating's free spectral range related to a Fabry–Pérot etalon's?

They are the same concept in different hardware. Both instruments encode the spectrum in an integer interference order, and both suffer the ambiguity that order $m$ of one wavelength overlaps order $m+1$ of a neighbour. In frequency the grating gives $\Delta\nu_{\text{FSR}}=\nu/m$ and the etalon gives $\Delta\nu=c/2nL$ ; writing the etalon's order as $m\approx2nL/\lambda$ makes the two expressions identical. The practical difference is scale: a grating typically works at low order with a broad free spectral range, an etalon at very high order with a narrow one. This is exactly why the two are cascaded — the grating's wide window sorts the etalon's dense orders, as detailed under Finesse and Free Spectral Range.

Why do echelle spectrographs need a cross-disperser?

Because they operate at high order, where the free spectral range $\lambda/m$ is only a few nanometres, so dozens of orders overlap across any useful bandwidth. The cross-disperser — a prism or low-dispersion grating set at right angles to the echelle — deflects each order by a slightly different amount in the perpendicular direction, separating the piled-up orders into distinct horizontal stripes on a 2-D detector. The result, the echellogram, lays out an enormous wavelength range at high resolution on one chip. Without cross-dispersion the overlapping orders would be an irretrievable jumble.

Does the free spectral range depend on the number of grooves?

No. The free spectral range $\lambda/m$ depends only on the wavelength and the order — a coarse two-groove grating and a fine million-groove one have the same first-order window. What the groove count $N$ sets is the resolution inside that window, $\delta\lambda=\lambda/(mN)$ , and hence the number of resolvable channels, $N$ . This clean separation is why the two figures multiply to the groove-count invariant $R\,\Delta\lambda_{\text{FSR}}=N\lambda$ : the free spectral range is pure geometry of orders, the resolution is set by how many grooves cooperate, and the illuminated groove count is the one lever that moves both.

Should I quote the free spectral range in wavelength or frequency?

Use whichever matches your source and detector, but the frequency form is often more revealing. In wavelength, $\Delta\lambda_{\text{FSR}}=\lambda/m$ varies across the band because $\lambda$ appears explicitly. In frequency, $\Delta\nu_{\text{FSR}}=\nu/m$ ties directly to the Fabry–Pérot analogue and to laser-comb spacing, where equal frequency intervals are what matter. For a broadband survey the wavelength window is intuitive; for high-resolution laser work near a fixed line the frequency free spectral range, in GHz or THz, is the natural currency. They convert through $\Delta\nu=(c/\lambda^2)\,\Delta\lambda$ .