
Physics III course sequence · Interactive lesson
Diffraction Gratings, Dispersion & Resolving Power
A grating is a thousand slits voting in phase. Learn why adding slits never moves a bright order but sharpens it, how that sharpness becomes an instrument's power to tell two wavelengths apart, and why the number on the spec sheet, mN, is a ceiling the real grating never quite reaches.
UPIII-05.07 · Lesson 7 of 9 in this unit. Position is not a completion score.Build the model
N slits vote in phase: the orders stay put, sharpen as 1/N, and resolve as mN.
The key idea
A diffraction grating is a periodic array of N slits or grooves of spacing d whose far-field interference throws light of wavelength λ into sharp principal maxima at d sin θ = mλ, each of angular half-width λ/(N d cos θ).
Example. A 600 lines mm⁻¹ grating (d = 1.667 μm) lit across 25 mm has N = 15 000, so in second order R = mN = 30 000 and it separates lines 0.020 nm apart at 589 nm; the sodium D pair, 0.597 nm apart, is split thirty times over.
The complete physical model and assumptions
Illuminate N identical slits of spacing d with a plane monochromatic wave and look far away. Each slit contributes a phasor of the same length, and neighbours differ in phase by δ = 2π d sin θ/λ, so the field is a geometric series whose sum has magnitude E₀ sin(Nδ/2)/sin(δ/2). Wherever δ is a whole number of turns the N phasors line up and the intensity is N² times a single slit's: the principal maxima, at d sin θ = mλ, which do not depend on N at all. What N controls is everything between them — N − 1 zeros, N − 2 faint secondary maxima, and a principal maximum whose half-width in angle is λ/(Nd cos θ), shrinking as 1/N while its height grows as N². Differentiating the grating equation gives the angular dispersion m/(d cos θ), and setting the dispersed separation of two lines equal to that half-width gives the Rayleigh limit Δλ = λ/(mN): the chromatic resolving power is R = mN, which is also W sin θ/λ, the illuminated width in wavelengths. Higher orders buy resolution at the price of a shorter free spectral range λ/m before orders overlap. The model is Fraunhofer, scalar and perfectly periodic: it cannot say how energy divides between orders, which is the groove profile's business and, for grooves near λ, needs vector theory; and a real ruling with position errors, or a beam that fills only part of it, resolves less than mN.
1Add N equal phasors and sum the series
Send a plane wave at normal incidence onto N slits of spacing d and collect the light at angle θ, far enough away that the rays are parallel. Slit j is d sin θ further from the screen than slit j − 1 for every j, so its field lags by the same δ = 2π d sin θ/λ. The total is E₀(1 + e(iδ) + e(2iδ) + … + e(i(N−1)δ)), a geometric series with sum E₀ (e(iNδ) − 1)/(e(iδ) − 1); factoring a half-angle out of each bracket leaves the magnitude E₀ sin(Nδ/2)/sin(δ/2). Square it: I = I₀ sin²(Nδ/2)/sin²(δ/2). At δ = 2πm both sines vanish and the ratio tends to N, so the principal maxima reach N² I₀ — N phasors laid end to end in a straight line. Between them the numerator vanishes N − 1 times, at δ = 2πm + 2πp/N with p = 1 … N − 1, where the phasors close into a regular polygon; between those zeros sit N − 2 secondary maxima. For N = 5 the first secondary maximum, near δ = 104°, is 1.56 I₀ against a principal 25 I₀, some 6%; for large N it settles at (2/(3π))² = 4.5%. Check N = 2: sin²δ/sin²(δ/2) = 4 cos²(δ/2), the double-slit result.
2More slits: the orders stay put and sharpen
The position of a principal maximum, d sin θ = mλ, has no N in it: a grating and a double slit of the same spacing throw their orders to the same angles. What N buys is width. The first zero beside the mth maximum sits at δ = 2πm + 2π/N, a phase step of 2π/N. Since dδ = (2π d cos θ/λ) dθ, that is an angle Δθ = λ/(N d cos θ) = λ/(W cos θ), with W = Nd the width of grating actually lit. Take d = 2.00 μm and λ = 500 nm: the first order lies at θ = 14.5°, and with N = 2 the half-width is 0.129 rad, or 7.4°, a fringe rather than a line. Light 20 mm of the same pitch, N = 10 000, and it shrinks to 2.6 × 10⁻⁵ rad, about 5 arcseconds, while the peak intensity has grown by (10 000/2)² = 2.5 × 10⁷. Height × width goes as N² × 1/N = N, in step with the light N slits admit: the sharpening costs no energy, it only pulls light out of the gaps between orders.
3The grating equation and how far apart two colours land
Hold m fixed and differentiate d sin θ = mλ: d cos θ dθ = m dλ, so the angular dispersion is D = dθ/dλ = m/(d cos θ). Substitute m/d = sin θ/λ and it becomes tan θ/λ — the dispersion depends only on where the line lands, not on the pitch or order that put it there. A 600 lines mm⁻¹ grating, d = 1.667 μm, sends 589 nm to 20.7° in first order, where D = 1/(1.667 μm × cos 20.7°) = 0.641 mrad nm⁻¹; in second order the line sits at 45.0° and D = 1.70 mrad nm⁻¹. The sodium D pair, 0.597 nm apart, is therefore separated by 0.38 mrad and 1.01 mrad, which a 300 mm focusing lens turns into 0.115 mm and 0.304 mm at the detector. Third order does not exist here: m cannot exceed d/λ = 2.83. Tilt the incident beam by θᵢ and every order shifts according to d(sin θᵢ + sin θₘ) = mλ, which is how reflection gratings reach orders that normal incidence cannot; the dispersion formula is unchanged.
4Resolving power: the Rayleigh criterion turns width into Δλ
Two lines at λ and λ + Δλ are resolved, by Rayleigh's convention, when the peak of one falls on the first zero of the other. Dispersion sets their angular separation, (m/(d cos θ)) Δλ; the half-width of a line is λ/(N d cos θ). Equate them and cos θ cancels: Δλₘᵢₙ = λ/(mN). The chromatic resolving power is R = λ/Δλₘᵢₙ = mN, order times illuminated slit count, and the summed profile of two lines at that limit dips to 8/π² = 81% of the peaks. Rewrite it with the grating equation: mN = N d sin θ/λ = W sin θ/λ. Resolving power is the illuminated width measured in wavelengths, scaled by sin θ ≤ 1, so W/λ is a ceiling no choice of pitch or order can beat; oblique incidence, where mλ = d(sin θᵢ + sin θₘ), only stretches it to 2W/λ. The 600 lines mm⁻¹ grating lit across 25 mm has N = 15 000 and R = 15 000 in first order, Δλₘᵢₙ = 0.039 nm at 589 nm: the D pair is fifteen times over resolved, but two lines 0.02 nm apart need second order, R = 30 000, or a wider beam.
5Free spectral range: where the orders start to overlap
Each order is a full spectrum, and they fan out together. Order m of wavelength λ + Δλ lands where order m + 1 of λ does when m(λ + Δλ) = (m + 1)λ, so the clean span in order m, the free spectral range, is ΔλFSR = λ/m. In first order at 400 nm that is 400 nm: nothing visible overlaps, because second-order 400 nm arrives with first-order 800 nm, in the infrared. In second order the span at 400 nm is 200 nm, and third-order 400 nm coincides with second-order 600 nm, violet on orange, so any spectrometer working above first order carries an order-sorting filter or a cross-dispersing prism. The trade is exact: the number of resolvable elements in one order is ΔλFSR/Δλₘᵢₙ = (λ/m)/(λ/mN) = N, whatever m is. A higher order shares the same N elements over a narrower span, finer but shorter; echelle gratings push this to m ≈ 50–100 with a second disperser to unstack the orders.
6Why a real grating resolves less than mN
mN assumes every groove is lit, identical, and exactly in place. Illumination first: R = W sin θ/λ counts the width the beam covers, so a 12 mm beam on a 30 mm grating gets 12 mm worth of grooves and the ruled width is irrelevant. Groove position errors next: an error ε in one groove shifts its phasor by 2π m ε/d in order m, so random errors of rms σ leave only a fraction e(−(2π m σ/d)²) of the peak intensity — 94% for σ = d/50 in second order — and scatter the rest into a floor between orders that hides faint lines beside bright ones. Periodic errors from the ruling engine are worse: a sinusoidal wobble of amplitude ε produces ghost lines beside every true one with relative intensity (π m ε/d)², about 10⁻³ for ε = d/100 in first order, which a spectroscopist can mistake for a real line. Finally the model is scalar and far-field: it predicts line shape and position but not how much light each order gets, which the groove profile (the blaze) decides and, once grooves approach λ, needs vector theory. mN is a design ceiling; the entrance slit usually sets what you actually get.
Key equations
Read the units, assumptions and limits before substituting.
N-slit intensity
I(θ) = I₀ sin²(Nδ/2) / sin²(δ/2), with δ = 2π d sin θ/λ
The whole pattern in one line: principal maxima of N² I₀ where δ = 2πm, with N − 1 zeros and N − 2 weak secondary maxima between them.
I₀ one slit's intensity at that angle (W m⁻²); δ phase step between neighbours (rad); d spacing (m); λ wavelength (m)Grating equation, normal incidence
d sin θₘ = mλ
Fixes where the orders fall. It contains no N, because adding slits never moves a principal maximum.
m = 0, ±1, ±2 … the order; θₘ the direction of the mth principal maximum; |m| cannot exceed d/λAngular half-width of a principal maximum
Δθ = λ / (N d cos θₘ) = λ / (W cos θₘ)
The line narrows as 1/N while its peak grows as N², so the power per order rises as N, exactly as energy conservation demands.
Δθ in rad, from the peak to its first zero; W = Nd the illuminated width (m), not the ruled widthAngular dispersion
dθ/dλ = m / (d cos θₘ) = tan θₘ / λ
How far apart two wavelengths land. The second form says it depends only on the angle: any grating that puts a line at 45° disperses it the same.
rad m⁻¹ in SI, usually quoted in mrad nm⁻¹; the second form substitutes m/d = sin θₘ/λChromatic resolving power
R = λ/Δλₘᵢₙ = mN = W sin θₘ / λ
Capped at W/λ, the illuminated width in wavelengths, because sin θₘ ≤ 1: a 20 mm grating at 500 nm can never exceed 40 000.
Δλₘᵢₙ the smallest separation resolved by the Rayleigh criterion; R dimensionless; N the slits actually illuminatedFree spectral range
ΔλFSR = λ/m
Each order holds ΔλFSR/Δλₘᵢₙ = N resolvable elements: a higher order buys finer resolution with a proportionally shorter clean span.
the wavelength span in order m before the start of order m + 1 lands on top of it; in nm when λ is in nmInteractive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Keep Δλ at 30 nm and raise N from 6: the λ₁ line narrows as 1/N while its peak never moves, and at N = 19 the grey upright crosses the dashed one as the dip between the two peaks of the sum reaches the Rayleigh depth, about 80% of the peaks. Then return N to 6 and raise the order to 4, which resolves the same pair by spreading the lines instead of narrowing them.
Live results
- ORDER ANGLE θₘ
- 12.7 °
- LINE HALF-WIDTH λ/(Nd cos θ)
- 37.6 mrad
- DISPERSION dθ/dλ
- 0.41 mrad nm⁻¹
- RESOLVABLE Δλₘᵢₙ = λ/(mN)
- 91.7 nm
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Illuminated slits N: 6; Order m: 1; Line separation Δλ: 30 nm
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
First-year optics leaves the picture that a grating's lines are sharp because its slits are narrow, and that the more lines it has ruled the better it resolves. Neither holds. The slit width a enters only through the single-slit envelope, which says how bright each order is and which are missing; narrowing a spreads that envelope and dims everything without sharpening a single line. Sharpness comes from the count N of slits interfering — the half-width is λ/(N d cos θ) whatever a is — and N is the number the beam actually illuminates, so a 50 mm grating behind a 10 mm beam is a 10 mm grating. Nor can the order be raised without limit to make mN as large as you like: m ≤ d/λ, so mN ≤ W/λ, the illuminated width in wavelengths.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
A transmission grating ruled at 300 lines mm⁻¹ is lit at normal incidence by a mercury lamp. Find the angle of the 546.1 nm green line in third order, the angular dispersion there in degrees per nanometre, and the highest order in which this line can appear at all.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 4 steps revealed.
- Spacing: d = 1 mm ÷ 300 = 3.333 μm. The grating equation d sin θ = mλ with m = 3 gives sin θ₃ = 3 × 546.1 nm ÷ 3333 nm = 0.4915, so θ₃ = 29.4°.
- Angular dispersion: D = m/(d cos θ₃) = 3 ÷ (3.333 × 10⁻⁶ m × 0.8709) = 1.033 × 10⁶ rad m⁻¹. Per nanometre that is 1.033 × 10⁻³ rad nm⁻¹, and × 180/π gives 0.0592° nm⁻¹.
- Check with the angle-only form: tan θ₃/λ = 0.5643 ÷ 546.1 nm = 1.033 × 10⁻³ rad nm⁻¹, the same number, as it must be.
- Highest order: sin θ ≤ 1 requires m ≤ d/λ = 3333 ÷ 546.1 = 6.10, so m = 6 is the last order, at θ₆ = arcsin(0.983) = 79.4°, almost grazing the grating.
Answer. θ₃ = 29.4°; D = 1.03 mrad nm⁻¹ = 0.0592° nm⁻¹; the line appears up to sixth order, at 79.4°.
MediumExample 2
A monochromator's plane grating has 1200 grooves mm⁻¹ across a ruled width of 30 mm and is used at normal incidence. The collimator delivers a beam 12 mm wide. Two emission lines at 656.20 nm and 656.23 nm must be separated in first order. Are they? Would switching to second order help? What beam width would do the job, and is it available?
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 4 steps revealed.
- Only the lit grooves count: N = 1200 mm⁻¹ × 12 mm = 14 400. In first order R = mN = 14 400, so Δλₘᵢₙ = 656.2 nm ÷ 14 400 = 0.0456 nm. The lines are 0.030 nm apart — not resolved; they blend into one profile.
- Second order: d = 1 mm ÷ 1200 = 833.3 nm, and d sin θ = 2λ would need sin θ = 1312.4 ÷ 833.3 = 1.57. No such angle exists at normal incidence; the first order itself already sits at sin θ = 0.7874, θ = 51.9°.
- So the only lever is N. The requirement R ≥ λ/Δλ = 656.2 ÷ 0.030 = 2.19 × 10⁴ needs N ≥ 21 900 grooves, a lit width W = 21 900 ÷ 1200 mm⁻¹ = 18.2 mm.
- The ruling is 30 mm wide, so an 18.2 mm beam is available; a 20 mm beam gives R = 24 000 and Δλₘᵢₙ = 0.027 nm, resolving the pair with 10% to spare. Check the cap: W sin θ/λ = 18.2 mm × 0.7874 ÷ 656.2 nm = 2.18 × 10⁴, matching mN = 2.19 × 10⁴ to within the rounding of W, consistent.
Answer. Not resolved at 12 mm (Δλₘᵢₙ = 0.046 nm); second order is geometrically impossible; widen the beam to at least 18.2 mm of the 30 mm ruling.
HardExample 3
A grating 20.0 mm wide with 100.0 lines mm⁻¹ is used at normal incidence to study the region around 500.0 nm. Find the highest usable order and, in that order, the angle, resolving power, smallest resolvable Δλ and free spectral range. Which wavelengths from the neighbouring orders land on top of 500.0 nm, and how wide a band-pass filter keeps the order clean? Finally confirm R = W sin θ/λ and compare it with the ceiling W/λ.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- d = 10.00 μm, so d/λ = 20.00: order 20 would sit at sin θ = 1, grazing along the grating and unusable. The highest usable order is m = 19, at sin θ = 19 × 500.0 ÷ 10 000 = 0.950, θ = 71.8°.
- N = 100.0 mm⁻¹ × 20.0 mm = 2000 lines, so R = mN = 38 000 and Δλₘᵢₙ = 500.0 nm ÷ 38 000 = 0.0132 nm. Free spectral range: ΔλFSR = λ/m = 500.0 ÷ 19 = 26.3 nm.
- Overlaps: order 18 of λ' lands at the same angle when 18λ' = 19 × 500.0 nm, λ' = 527.8 nm; order 20 when 20λ' = 9500 nm, λ' = 475.0 nm. A filter passing 486.8–513.2 nm, one free spectral range centred on the line, admits neither.
- Cap: W sin θ/λ = (20.0 × 10⁻³ m)(0.950) ÷ (500.0 × 10⁻⁹ m) = 38 000, equal to mN as the algebra promised. The ceiling W/λ = 40 000 is reached only at sin θ = 1, the unusable order 20.
- Contrast first order: R = 2000, Δλₘᵢₙ = 0.250 nm, FSR 500 nm, θ = 2.87°. Both orders hold FSR/Δλₘᵢₙ = 2000 = N resolvable elements; order 19 spends them over 26 nm at nineteen times the fineness, and its dispersion, 19/(d cos 71.8°) = 6.09 mrad nm⁻¹, is 61 times first order's.
Answer. m = 19 at θ = 71.8°: R = 38 000, Δλₘᵢₙ = 0.0132 nm, ΔλFSR = 26.3 nm; 527.8 nm (order 18) and 475.0 nm (order 20) overlap, so filter to 487–513 nm; R = W sin θ/λ = 38 000 against the ceiling W/λ = 40 000.
Quick check
Select an answer and explain it before checking.
Review & continue
A diffraction grating is a periodic array of N slits or grooves of spacing d whose far-field interference throws light of wavelength λ into sharp principal maxima at d sin θ = mλ, each of angular half-width λ/(N d cos θ).
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.
Optional supporting lessons
These chapter resources can help with earlier ideas; they are not additional required Physics III completion tasks.