
Optional unit · institution-dependent · Interactive lesson
Multiple Slits & Diffraction Gratings
Two slits give broad fringes you can see. A few thousand give narrow lines you can measure, which is how a ruled strip of glass tells sodium's two yellow lines apart. This is the arithmetic that turns a slit count into resolving power.
UPII-14.03 · Lesson 3 of 8 in this unit. Position is not a completion score.Optional unit · institution-dependent. Your university may teach this optics or introductory modern-physics material separately. Follow your institution’s required sequence. Optional work is not a required completion gate.
Build the model
More slits do not move the orders. They sharpen them, and that is the instrument.
The key idea
A diffraction grating is a large number of identical, equally spaced slits, and it sends sharp principal maxima in the directions where the path difference between neighbouring slits is a whole number of wavelengths.
Example. A grating ruled with 600 lines per millimetre has d = 1.667 µm. In 633 nm light the first order sits at sin θ₁ = 633 nm / 1667 nm = 0.3798, so θ₁ = 22.3°; since d/λ = 2.63, only two orders exist on each side.
The complete physical model and assumptions
A grating is the double slit taken to its limit: N identical slits, equally spaced by d, all fed by one wavefront, so the wave from each slit arrives at a distant screen exactly one phase step φ = 2π d sin θ / λ behind its neighbour. Adding N phasors of equal length that each turn by the same angle gives a chain whose links lie on a circle, and its chord gives I = I₁ [sin(Nφ/2) / sin(φ/2)]². Every property of the instrument is in that ratio. Where φ is a whole multiple of 2π the chain straightens, all N slits add in step, and the intensity is N² times one slit's: these are the principal maxima, at d sin θ = mλ, whose positions depend on d, λ and m but never on N. Between them the chain closes on itself N − 1 times, so N − 1 zeros are packed into each gap and every maximum narrows as 1/N. That is the point of ruling thousands of lines: extra slits do not move a spectral line, they make it thin enough to measure against its neighbour. The model assumes slits narrow enough and light coherent enough that all N contributions really do add, and it breaks down when the slit width's own envelope kills an order, when the coherence length is shorter than the path spread N d sin θ, or when d < λ and no order but the straight-through one exists.
1Add N phasors with one phase step between them
Each slit sends a wave towards the same distant point, and because the slits are equally spaced by d, each wave lags its neighbour by the same path difference d sin θ. In phase that is φ = 2π d sin θ / λ. The resultant is a sum of N phasors of equal length A₁, each turned by the same angle φ from the last — a chain of equal links bending through a constant angle, so the links lie on a circle and the resultant is its chord. The geometry gives A = A₁ sin(Nφ/2) / sin(φ/2), and squaring gives I = I₁ [sin(Nφ/2) / sin(φ/2)]². Test it at N = 2, where it collapses to 4 I₁ cos²(φ/2), the double-slit result you already have. Everything a grating does to light is a property of this one ratio, so read the sections that follow as the places where that ratio is large, zero, or moving.
2Where all N phasors line up: the grating equation
When φ is a whole multiple of 2π the chain straightens out: every slit contributes in step, the amplitude is N A₁ and the intensity is N² I₁. These are the principal maxima, and φ = 2πm rearranges to d sin θ = mλ, with the integer m called the order. Two features of this equation matter. First, N is absent, so adding slits cannot shift a line by any angle at all — only d, λ and m place it. Second, sin θ ≤ 1 caps the order at m ≤ d/λ. That is the cap the 600 lines per millimetre grating above runs into at 633 nm: it stops at m = 2, and that second order lies at sin θ = 0.760, θ = 49.4°. If d were smaller than λ, nothing but the straight-through order would survive, which is why gratings for visible light are ruled at hundreds of lines per millimetre rather than tens of thousands.
3The N − 1 zeros between orders make each line narrow
Between two principal maxima the numerator sin(Nφ/2) vanishes N − 1 times while the denominator does not, giving N − 1 exact zeros and, between them, N − 2 weak secondary maxima that fade as N grows. The zeros sit at d sin θ = (p/N)λ for every integer p that N does not divide, so the first zero past order m lies only λ/(N d) further along in sin θ. In angle, the half-width of a principal maximum is Δθ = λ / (N d cos θₘ), shrinking as 1/N. Put numbers on it: 600 lines per millimetre, a beam covering 5.00 mm so that N = 3000 lines, first order at 589 nm. Then θ = 20.7° and Δθ = 1.26 × 10⁻⁴ rad, about 26 arcseconds. The energy bookkeeping agrees: each peak is N² times brighter but 1/N as wide, so the light in one order grows as N, exactly as N slits passing N times the power must.
4Dispersion: differentiate the grating equation
How far apart do two nearby wavelengths land? Hold the order fixed and differentiate d sin θ = mλ, which gives d cos θ dθ = m dλ, so the angular dispersion is D = dθ/dλ = m / (d cos θₘ), in radians per metre of wavelength. Two wavelengths a gap Δλ apart arrive Δθ ≈ D Δλ apart. For the 600 lines per millimetre grating in first order at 589 nm, D = 1 / (1.667 × 10⁻⁶ m × 0.9355) = 6.41 × 10⁵ rad m⁻¹, so the two sodium D lines, 0.59 nm apart, are pushed 3.8 × 10⁻⁴ rad or 0.022° apart. Notice what D does and does not contain: it grows with the order m, with fine ruling 1/d, and with cos θ shrinking far out in the pattern — but N is nowhere in it. Illuminating more lines spreads the spectrum not at all.
5Resolving power compares the spread with the width
A separation means nothing until it is compared with the width of what is being separated. The Rayleigh criterion calls two wavelengths just resolved when the principal maximum of one falls on the first zero of the other: D Δλ = λ / (N d cos θₘ). Substitute D = m / (d cos θₘ) and the geometry cancels completely, leaving R = λ/Δλ = m N. The 3000-line patch above has R = 3000 in first order, while splitting the sodium doublet needs only R = (589.0 nm)/(0.59 nm) = 998, so those lines stand 3.0 half-widths apart and separate comfortably. The N here is the number of lines the beam actually covers, not the number ruled on the plate: mask the beam down to 1.0 mm and N falls to 600, R falls to 600, and the doublet closes into one line while every angle in the spectrum stays exactly where it was.
6Missing orders: the envelope multiplies the comb
Real slits have a width a, and one slit of width a sends nothing in the directions where a sin θ = pλ. The pattern of the grating is the product of that envelope with the N-slit factor: I = I₀ [sin β / β]² [sin(Nφ/2) / sin(φ/2)]², where β = π a sin θ / λ. A principal maximum survives only where the envelope is non-zero, and the two conditions collide when mλ/d = pλ/a, that is m = p (d/a). So an order dies only when some whole p makes p (d/a) a whole number, which needs d/a to be a ratio of whole numbers: rule d = 4a and orders 4, 8 and 12 are absent however many slits you add, because zero times N² is still zero, while d = 2.5a costs you orders 5, 10 and 15 and nothing below them. The envelope also ranks the brightnesses. With d/a = 5 it leaves order 1 at 88% of the central value and order 6 at 2.4%, which is why high orders disperse well and look faint.
Key equations
Read the units, assumptions and limits before substituting.
Grating equation
d sin θₘ = m λ
Fixes where every principal maximum lands. The number of slits does not appear, so it cannot move a line.
d = slit spacing (m), θₘ = angle from the normal, λ = wavelength (m), m = 0, ±1, ±2, …Intensity of N equal slits
I = I₁ [sin(Nφ/2) / sin(φ/2)]²
One ratio holds the whole pattern: peaks of N² I₁ with N − 1 zeros between each pair.
φ = 2π d sin θ / λ is the phase step between neighbours (rad); I₁ = intensity of one slit alone (W m⁻²)Half-width of a principal maximum
Δθ = λ / (N d cos θₘ)
Double the illuminated width and every line halves in width while staying exactly where it was.
Δθ in rad, peak to the neighbouring zero; N = the lines the beam actually covers, not the lines ruledAngular dispersion
D = dθ/dλ = m / (d cos θₘ)
Converts a wavelength gap into an angle gap, Δθ ≈ D Δλ. Higher order or finer ruling spreads more.
D in rad m⁻¹, radians of angle per metre of wavelength; θₘ is the angle of that same orderChromatic resolving power
R = λ / Δλₘᵢₙ = m N
Says whether two close lines split at all. Mask the beam down and R falls with N.
Δλₘᵢₙ = smallest wavelength gap separable at λ (m); m = order; N = illuminated linesMissing orders
m = p (d/a), p = 1, 2, 3, …
Any m this makes a whole number lands on an envelope zero and vanishes, so nothing goes missing unless d/a is a ratio of whole numbers.
a = width of one slit (m); the single-slit envelope has zeros at a sin θ = p λInteractive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Drag the slit count from 2 to 12: no maximum moves, each one sharpens, and N − 2 faint secondary peaks fill every gap. Then take d/a from 3 down to 2 and watch the missing order jump from the third to the second.
Live results
- HIGHEST ORDER
- 3
- PEAK HALF-WIDTH
- 0.200 × λ/d in sin θ
- PEAK INTENSITY
- 25 × one slit
- ENVELOPE ZERO AT m =
- 3.0
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. Slits illuminated N: 5; Spacing / slit width d/a: 3.0; Spacing / wavelength d/λ: 3.5
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
Two different counts are both called N, and swapping them is the standard error here. The lines per millimetre set the spacing d, and d sets every angle in the pattern. The number of lines the beam actually covers sets the width of each line and the resolving power R = mN. Double the lines per millimetre and d halves, so the orders swing wider apart. Double the illuminated width instead and not one angle changes: each line simply becomes half as wide. Neither is more resolution in the other's sense. A coarse grating flooded over 50 mm outresolves a fine one lit over 1 mm, and masking a grating down costs resolving power while leaving the spectrum exactly where it was.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
A transmission grating ruled with 500 lines per millimetre is illuminated at normal incidence by a 532 nm green laser. Find the slit spacing, the angle of the first-order maximum, and the highest order that leaves the grating.
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 first: d is the reciprocal of the line density, d = 1 mm / 500 = 2.000 × 10⁻⁶ m. Lines per millimetre is not a spacing, and the grating equation wants the spacing.
- First order from d sin θ = mλ: sin θ₁ = (1)(532 × 10⁻⁹ m) / (2.000 × 10⁻⁶ m) = 0.2660, so θ₁ = 15.43°.
- The highest order needs sin θ ≤ 1, so m ≤ d/λ = (2.000 × 10⁻⁶ m) / (532 × 10⁻⁹ m) = 3.76, and m must be a whole number: m = 3.
- Check the edge: sin θ₃ = 3 × 0.2660 = 0.7980, θ₃ = 52.94°, while m = 4 would demand sin θ = 1.064, which no direction can deliver.
Answer. d = 2.000 µm, θ₁ = 15.43°, and orders up to m = 3 (θ₃ = 52.9°) appear on each side of the straight-through beam.
MediumExample 2
A grating of 1200 lines per millimetre is used in first order with the sodium doublet, 589.00 nm and 589.59 nm, and a beam 2.00 mm wide covers it. Find the angle of each line, their angular separation, and whether the doublet is resolved.
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 = 1 mm / 1200 = 8.3333 × 10⁻⁷ m. First order gives sin θ = λ/d: 0.706800 for 589.00 nm and 0.707508 for 589.59 nm.
- So θ = 44.975° and 45.033°, a separation of 0.0574°, which is 1.00 × 10⁻³ rad.
- Confirm with dispersion: D = m/(d cos θ) = 1 / (8.3333 × 10⁻⁷ m × 0.7074) = 1.696 × 10⁶ rad m⁻¹, so Δθ ≈ D Δλ = (1.696 × 10⁶ rad m⁻¹)(0.59 × 10⁻⁹ m) = 1.00 × 10⁻³ rad, the same to three figures.
- Now the width. The beam covers N = 1200 mm⁻¹ × 2.00 mm = 2400 lines, so R = mN = 2400, against the R = λ/Δλ = (589.0 nm)/(0.59 nm) = 998 this doublet demands.
- Equivalently, the half-width is λ/(N d cos θ) = 4.16 × 10⁻⁴ rad, so the peaks sit 2.4 half-widths apart, more than the one half-width Rayleigh asks for.
Answer. θ = 44.975° and 45.033°, separated by 1.00 × 10⁻³ rad (0.0574°). With R = 2400 available against 998 needed, the doublet is cleanly resolved.
HardExample 3
A grating has 250 lines per millimetre and each slit is 0.80 µm wide. A beam 4.00 mm across, of 650 nm light, hits it at normal incidence. Which orders appear and which is missing? In the highest order that does appear, what is the smallest resolvable wavelength difference, and how bright is that order compared with the centre of the envelope?
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.
- Geometry: d = 1 mm / 250 = 4.00 × 10⁻⁶ m and a = 0.80 × 10⁻⁶ m, so d/a = 5.00, a whole number, which means missing orders will exist.
- Cutoff: m ≤ d/λ = (4.00 × 10⁻⁶ m) / (650 × 10⁻⁹ m) = 6.15, so orders 1 to 6 are geometrically available, the sixth at sin θ = 0.9750, θ = 77.2°.
- Envelope zeros fall at m = p (d/a) = 5, 10, …, so order 5 lands on a zero and is missing: orders 1, 2, 3, 4 and 6 appear.
- Resolving power: the beam covers N = 250 mm⁻¹ × 4.00 mm = 1000 lines, so in the highest surviving order R = mN = 6 × 1000 = 6000 and Δλ = λ/R = (650 nm)/6000 = 0.108 nm.
- Brightness: β = π a sin θ / λ = π m (a/d) = 1.2π, that is 216°, so [sin β / β]² = (−0.5878/3.7699)² = 0.024.
Answer. Orders 1, 2, 3, 4 and 6 appear; order 5 is missing. In sixth order R = 6000, so Δλ = 0.108 nm at 650 nm, from a peak carrying only 2.4% of the envelope's central brightness.
Quick check
Select an answer and explain it before checking.
Review & continue
A diffraction grating is a large number of identical, equally spaced slits, and it sends sharp principal maxima in the directions where the path difference between neighbouring slits is a whole number of wavelengths.
- 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 II completion tasks.