Part 4 · Chapter 36

Diffraction

Interference taught us that two waves can add to brightness or cancel to darkness. Diffraction is the same wave physics let loose at a single edge or opening: light bends around obstacles and flares through narrow slits, painting bands of light and shadow exactly where the tidy rays of geometrical optics had promised a clean boundary. This one idea explains why even a flawless lens cannot focus to a true point, why two distant stars blur into a single smudge, how the pits on a CD throw rainbows across the room, and how the regular spacing of atoms in a crystal can be read straight off the pattern that X-rays scatter onto film. We start with a single slit, build the full mathematics of its intensity, and end by turning gratings and crystals into the most precise rulers in all of physics.

Fundamentals of Physics Prof. Mithun Mondal Reading time ≈ 85 min
i What you'll learn
  • Diffraction is the flaring of a wave at an edge or aperture — pure interference of the Huygens wavelets across one opening — and the surprising Fresnel bright spot at the center of a circular shadow is its decisive proof.
  • Single-slit diffraction places its dark fringes at \(a\sin\theta = m\lambda\) with \(m = 1,2,3,\dots\) — note \(m=0\) is the bright center, not a minimum — and the narrower the slit, the wider the central blaze.
  • The full pattern follows from phasors as \(I(\theta) = I_m\!\left(\dfrac{\sin\alpha}{\alpha}\right)^{2}\) with \(\alpha = \dfrac{\pi a}{\lambda}\sin\theta\).
  • A circular aperture blurs a point into an Airy disk whose first dark ring sits at \(\sin\theta = 1.22\dfrac{\lambda}{d}\), so two sources are barely resolved at Rayleigh's criterion \(\theta_R = 1.22\dfrac{\lambda}{d}\).
  • A real double slit shows interference fringes \(\cos^{2}\beta\) riding under a single-slit diffraction envelope: \(I(\theta) = I_m(\cos^{2}\beta)\!\left(\dfrac{\sin\alpha}{\alpha}\right)^{2}\).
  • A diffraction grating sharpens the maxima to thin lines at \(d\sin\theta = m\lambda\), with dispersion \(D = \dfrac{m}{d\cos\theta}\) and resolving power \(R = Nm\); X-ray diffraction reads crystal structure through Bragg's law \(2d\sin\theta = m\lambda\).
Section 36-1

What Is Physics?

One persistent goal of physics is to understand light well enough to control it, and the wave property that gives us the most control is diffraction — the spreading of a wave when part of it is blocked or squeezed through an opening. It is far more than a curiosity. Diffraction sets a hard ceiling on the sharpness of every microscope, telescope, and camera ever built, because no lens, however perfect, can focus light to a point smaller than its own diffraction blur. Run the same physics in reverse and it becomes a measuring tool of astonishing precision: the iridescent flash of a CD or DVD is diffraction off its rows of pits, the colors of a beetle's shell or a peacock's feather are structural, and — most consequentially — firing X-rays at a crystal and reading the diffraction pattern is how we learned the atomic arrangement of salt, metals, and the double helix of DNA. To see how a single opening can carry this much information, we return once more to light as a wave.

Section 36-2

Diffraction and the Wave Theory of Light

Send light of wavelength \(\lambda\) through a long narrow slit of width \(a\) onto a distant screen and you do not get a sharp bright stripe the shape of the slit. Instead you find a broad, bright central band flanked by progressively fainter secondary bands, separated by dark fringes — a diffraction pattern. Geometrical optics, with its straight-line rays, cannot explain a single feature of it. The wave theory can, and the mechanism is exactly the one from the last chapter: by Huygens' principle every point across the open slit is a source of secondary wavelets, those wavelets travel slightly different distances to any point on the screen, and there they interfere. Diffraction is interference — interference of a wave with itself, summed over a continuous opening.

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The Fresnel bright spot — diffraction's smoking gun
When light passes a small round disk, wavelets diffracting around its entire rim travel equal distances to the center of the shadow, arrive in phase, and produce a bright spot exactly where geometry predicts deepest darkness.

In 1819 Poisson, defending the ray picture, derived this absurd-sounding spot as a reductio ad absurdum of Fresnel's wave theory — and then Arago went and saw it. Far from killing the wave theory, the bright spot at the heart of a circular shadow became its most dramatic confirmation. It survives today under the names Fresnel bright spot and Poisson spot, a reminder that when ray optics and wave optics disagree, light sides with the waves.

Section 36-3

Diffraction by a Single Slit: Locating the Minima

To find where the screen goes dark, use a clever pairing trick. Imagine the slit of width \(a\) divided into a great many thin strips, each a Huygens source. Pair the top strip with the strip just below the middle, the next-to-top with the next-below-middle, and so on, so the two members of every pair are separated by \(a/2\). With the screen far away the rays head off essentially parallel at angle \(\theta\), and the two rays in a pair differ in path by \((a/2)\sin\theta\). When that half-slit path difference equals half a wavelength, every pair cancels and the whole slit produces darkness.

Single-slit minima (dark fringes)
\[ a\sin\theta = m\lambda, \qquad m = 1, 2, 3, \dots \]
Setting \((a/2)\sin\theta = \lambda/2\) gives the first minimum \(a\sin\theta = \lambda\); splitting the slit into four, six, … strips gives the rest. Beware the trap: this is the formula for dark fringes, the mirror image of the double-slit rule, and \(m=0\) is excluded because at \(\theta = 0\) all wavelets arrive in phase — the bright central maximum.
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Narrow slits spread the light
The central maximum runs from the first minimum on one side to the first on the other, so its angular half-width is set by \(\sin\theta = \lambda/a\) — shrink \(a\) and the blaze fans out.

This is the heart of diffraction's reputation. When the slit is wide compared to the wavelength (\(a \gg \lambda\)) the minima crowd toward \(\theta = 0\) and the bright patch is a near-perfect image of the slit — geometrical optics. Squeeze the slit toward \(a \approx \lambda\) and the light floods the whole screen. You cannot pin a wave down without making it spread; that tension reappears, deepened, in quantum mechanics.

Section 36-4

Intensity in Single-Slit Diffraction

The minima tell us where the screen is dark; to know its brightness everywhere we again add the wavelets as phasors. The slit is sliced into \(N\) tiny strips whose phasors each turn by the same small angle relative to the next, so the chain of phasors bends into the arc of a circle. At \(\theta = 0\) the phasors line up straight and the resultant is maximal; off-axis the arc curls up, the chord shrinks, and where it closes into a full circle the resultant is zero — a minimum. Carrying the geometry through gives a compact intensity law.

Single-slit intensity
\[ I(\theta) = I_m\left(\frac{\sin\alpha}{\alpha}\right)^{2}, \qquad \alpha = \frac{\pi a}{\lambda}\sin\theta \]
Here \(I_m\) is the central (\(\theta = 0\)) intensity and \(\alpha\) is half the phase spread across the slit. As \(\alpha \to 0\), \(\sin\alpha/\alpha \to 1\), recovering the central peak. The dark fringes return whenever \(\alpha = m\pi\), i.e. \(a\sin\theta = m\lambda\), in perfect agreement with the pairing argument.
The secondary maxima are feeble. The bright bands between the dark fringes do not sit exactly halfway; they fall a little short of \(\alpha = \tfrac32\pi, \tfrac52\pi, \dots\) and they fade fast. The first secondary maximum carries only about \(4.7\%\) of the central intensity, the next about \(1.7\%\). Nearly all the diffracted energy lives in the central blaze — which is precisely why that central peak alone sets the resolution limit of an instrument.
Section 36-5

Diffraction by a Circular Aperture

Lenses, mirrors, telescopes, and your own pupil are round, not slit-shaped, so their diffraction pattern is a bright central disk — the Airy disk — surrounded by faint rings. The mathematics is harder (the aperture is two-dimensional), but the result differs from the slit only by a numerical factor: a \(1.22\) sneaks in where the circular geometry is summed.

First dark ring of a circular aperture (diameter d)
\[ \sin\theta = 1.22\,\frac{\lambda}{d} \]
The point image of a distant star is therefore never a point but a disk of angular radius \(\theta\). Two stars closer together than this will overlap into a single blur no matter how good the optics, because the limit is set by the wave nature of light, not by lens quality.
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Rayleigh's criterion for resolution
Two point sources are just resolvable when the central maximum of one falls on the first minimum of the other, i.e. when their angular separation reaches \(\theta_R = 1.22\dfrac{\lambda}{d}\).

The lesson is built into the formula. To see finer detail (smaller \(\theta_R\)) you make the aperture \(d\) bigger — hence giant telescope mirrors — or use a shorter wavelength \(\lambda\) — hence the electron microscope, where the tiny de Broglie wavelength of fast electrons resolves individual atoms, far beyond anything visible light can reach.

Section 36-6

Diffraction by a Double Slit

In Chapter 35 we pretended the two slits were infinitely narrow, so the fringes had uniform brightness across the screen. Real slits have width, and each one diffracts. The honest pattern is therefore two patterns at once: the fine interference fringes set by the slit separation \(d\), multiplied by the broad diffraction envelope set by the slit width \(a\). The closely spaced bright fringes you expect are still there — but their heights are scaled down by the single-slit curve.

Double-slit intensity with finite slit width
\[ I(\theta) = I_m\,(\cos^{2}\beta)\left(\frac{\sin\alpha}{\alpha}\right)^{2}, \qquad \beta = \frac{\pi d}{\lambda}\sin\theta, \quad \alpha = \frac{\pi a}{\lambda}\sin\theta \]
The \(\cos^{2}\beta\) factor is the two-slit interference of Chapter 35 (maxima at \(d\sin\theta = m\lambda\)); the \((\sin\alpha/\alpha)^2\) factor is this chapter's single-slit diffraction. Set \(a \to 0\) and the envelope flattens to \(1\), returning the ideal double-slit fringes.
Missing orders. Wherever a diffraction minimum lands exactly on an interference maximum, that bright fringe is wiped out — a "missing order." It happens when \(d/a\) is a whole number: if \(d = 5a\), for instance, the \(m = 5, 10, 15, \dots\) interference fringes vanish under the first, second, third diffraction zeros. Counting the fringes inside the central diffraction peak is a quick way to read off the ratio \(d/a\) straight from a photograph of the pattern.
Section 36-7

Diffraction Gratings

Now replace two slits with thousands, or tens of thousands, ruled at a uniform spacing \(d\) (the grating's line spacing, often quoted as lines per millimeter). The condition for a bright line is unchanged from the double slit — the waves from adjacent rulings must differ by a whole number of wavelengths — but having so many slits transforms the result. The maxima do not merely brighten; they sharpen into needle-thin lines with almost total darkness between them.

Grating maxima (lines)
\[ d\sin\theta = m\lambda, \qquad m = 0, 1, 2, \dots \]
The integer \(m\) is the order. Because the line position depends on \(\lambda\), white light is fanned out into a spectrum at every order — this is how a grating spectroscope sorts light by wavelength, far more cleanly than a prism.
Half-width of a grating line
\[ \Delta\theta_{\text{hw}} = \frac{\lambda}{N d\cos\theta} \]
Here \(N\) is the number of rulings illuminated. The more lines that share in the interference, the narrower each maximum becomes — sharpness scales as \(1/N\). A grating with \(N\) in the tens of thousands produces lines so fine that wavelengths differing by a fraction of a nanometer stand cleanly apart.
Section 36-8

Gratings: Dispersion and Resolving Power

To judge a grating as a spectroscopic tool we ask two separate questions. First, how far apart in angle does it throw two nearby wavelengths? That is its dispersion. Second, how close in wavelength can two lines be and still be seen as two? That is its resolving power. They are not the same property — a grating can spread the colors widely yet still smear two close lines together, or vice versa — and a good instrument needs both.

Dispersion
\[ D = \frac{\Delta\theta}{\Delta\lambda} = \frac{m}{d\cos\theta} \]
Dispersion is large for a finely ruled grating (small \(d\)) and for higher orders (large \(m\)). It measures the angular separation produced per unit wavelength difference — how far the colors fan apart.
Resolving power
\[ R = \frac{\lambda_{\text{avg}}}{\Delta\lambda} = N m \]
Resolving power depends only on the total number of rulings \(N\) and the order \(m\) — not on the spacing. To split two very close spectral lines you need many lines illuminated, which is why research gratings are ruled with tens of thousands of grooves across a few centimeters.
The sodium doublet test. The classic benchmark is the pair of yellow sodium lines at \(589.00\) and \(589.59\,\mathrm{nm}\), separated by just \(0.59\,\mathrm{nm}\). To resolve them in first order needs \(R = \lambda/\Delta\lambda \approx 589/0.59 \approx 1000\) rulings — easily met by any decent grating, which is why this doublet is the standard demonstration that a grating outperforms a prism.
Section 36-9

X-Ray Diffraction

To diffract a wave you need a grating whose spacing is comparable to the wavelength. Visible light has \(\lambda \approx 500\,\mathrm{nm}\); X-rays have \(\lambda \approx 0.1\,\mathrm{nm}\), about the spacing of atoms in a solid. So a crystal is a ready-made, three-dimensional diffraction grating for X-rays, with its own atoms as the rulings. Treat the parallel planes of atoms as faint mirrors: an X-ray beam reflects weakly off each plane, and the reflections from successive planes interfere. They reinforce only when the extra path between adjacent planes is a whole number of wavelengths.

Bragg's law
\[ 2d\sin\theta = m\lambda, \qquad m = 1, 2, 3, \dots \]
Here \(d\) is the spacing of the atomic planes and \(\theta\) is measured from the plane's surface (the glancing angle), not from the normal — a frequent slip. Knowing \(\lambda\) and measuring the angles of the bright reflections lets you solve for the interplanar spacings, and from a full set of them, the entire atomic architecture of the crystal.
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Diffraction as the microscope of the atomic world
Because the pattern depends on the geometry of the lattice, reading X-ray reflections backward reconstructs where the atoms sit.

This is among the most powerful techniques in all of science. Bragg's law (a 1915 Nobel Prize for father and son) gave us the structure of metals, minerals, and salts; it underlies the determination of DNA's double helix and the shapes of countless proteins. A wave property that began as the annoying blur at the edge of a shadow turned out to be the key that unlocked the architecture of matter itself.

Worked Examples

Putting It to Work

1 Width of the central diffraction peak

Problem. A helium–neon laser (\(\lambda = 633\,\mathrm{nm}\)) shines through a single slit of width \(a = 0.20\,\mathrm{mm}\) onto a screen \(D = 2.0\,\mathrm{m}\) away. How wide is the bright central band?

Solution. The central peak spans from the first minimum on one side to the first on the other. Locate the first minimum with \(a\sin\theta = \lambda\) (i.e. \(m=1\)).

sin θ = λ/a, then y = D tan θ
\[ \sin\theta = \frac{\lambda}{a} = \frac{633\times10^{-9}}{0.20\times10^{-3}} = 3.17\times10^{-3}, \qquad y_1 = D\tan\theta \approx (2.0)(3.17\times10^{-3}) \approx 6.3\,\mathrm{mm} \]

The central band runs from \(-y_1\) to \(+y_1\), so its full width is \(2y_1 \approx 13\,\mathrm{mm}\). Halve the slit width and this blaze doubles — narrower slits spread the light more, the signature behaviour of diffraction.

2 Relative intensity off-axis

Problem. For a single slit, find the intensity (as a fraction of the central \(I_m\)) at the angle where the path difference across the slit is \(a\sin\theta = 1.25\lambda\).

Solution. Convert to the phase variable \(\alpha = (\pi a/\lambda)\sin\theta = \pi(1.25) = 3.93\,\mathrm{rad}\), then apply the intensity law.

I/I_m = (sin α / α)²
\[ \frac{I}{I_m} = \left(\frac{\sin\alpha}{\alpha}\right)^{2} = \left(\frac{\sin 3.93}{3.93}\right)^{2} = \left(\frac{-0.707}{3.93}\right)^{2} \approx 0.032 \]

So that point holds only about \(3.2\%\) of the central brightness — it sits near the first secondary maximum, confirming how quickly the side bands fade. (Use the angle in radians inside \(\sin\alpha\); mixing degrees here is the most common error.)

3 Resolving power of a telescope

Problem. The Hubble Space Telescope has a primary mirror of diameter \(d = 2.4\,\mathrm{m}\). Working at \(\lambda = 550\,\mathrm{nm}\), what is the smallest detail it can resolve on the Moon, a distance \(L = 3.8\times10^{8}\,\mathrm{m}\) away?

Solution. Find the limiting angle with Rayleigh's criterion, then convert the angle to a distance at the Moon.

θ_R = 1.22 λ/d, then x = θ_R L
\[ \theta_R = 1.22\frac{\lambda}{d} = 1.22\frac{550\times10^{-9}}{2.4} \approx 2.8\times10^{-7}\,\mathrm{rad}, \qquad x = \theta_R L \approx (2.8\times10^{-7})(3.8\times10^{8}) \approx 1.1\times10^{2}\,\mathrm{m} \]

Hubble could just separate two features about \(100\,\mathrm{m}\) apart on the lunar surface — a hard limit imposed by diffraction, not by the polish of the mirror. A bigger mirror, or a shorter wavelength, is the only way to do better.

4 A grating spectroscope

Problem. A grating ruled with \(1000\,\text{lines/mm}\) is illuminated over a width that exposes \(N = 10\,000\) rulings by light of \(\lambda = 600\,\mathrm{nm}\). Find (a) the angle of the first-order line and (b) the smallest wavelength difference it can resolve in first order.

Solution. The line spacing is \(d = 1/(1000\,\text{mm}^{-1}) = 1.00\times10^{-6}\,\mathrm{m}\). Use the grating equation for the angle and \(R = Nm\) for the resolution.

d sin θ = mλ ; Δλ = λ/(Nm)
\[ \sin\theta = \frac{m\lambda}{d} = \frac{(1)(600\times10^{-9})}{1.00\times10^{-6}} = 0.600 \;\Rightarrow\; \theta \approx 36.9^\circ \]
Resolving power R = Nm
\[ R = Nm = (10\,000)(1) = 10\,000, \qquad \Delta\lambda = \frac{\lambda}{R} = \frac{600\,\mathrm{nm}}{10\,000} = 0.060\,\mathrm{nm} \]

The first-order line sits at \(36.9^\circ\), and two wavelengths as close as \(0.06\,\mathrm{nm}\) show up as separate lines — comfortably enough to split the sodium doublet ten times over.

Review

Chapter Summary

✓ What diffraction is

The flaring of a wave at an edge or aperture — self-interference of Huygens wavelets. The Fresnel bright spot at a circular shadow's center proved the wave theory.

✓ Single-slit minima

\(a\sin\theta = m\lambda\), \(m = 1,2,3,\dots\) (dark fringes; \(m=0\) is the bright center). Narrower slits widen the central peak.

✓ Single-slit intensity

\(I = I_m\!\left(\dfrac{\sin\alpha}{\alpha}\right)^{2}\), \(\alpha = \dfrac{\pi a}{\lambda}\sin\theta\). Side maxima are weak (first ≈ 4.7% of center).

✓ Circular aperture

Airy disk; first dark ring at \(\sin\theta = 1.22\dfrac{\lambda}{d}\).

✓ Rayleigh's criterion

Just resolved when \(\theta_R = 1.22\dfrac{\lambda}{d}\). Bigger \(d\) or smaller \(\lambda\) sees finer detail.

✓ Double-slit envelope

\(I = I_m(\cos^{2}\beta)\!\left(\dfrac{\sin\alpha}{\alpha}\right)^{2}\); fringes ride under a diffraction envelope, giving missing orders when \(d/a\) is an integer.

✓ Gratings

Sharp lines at \(d\sin\theta = m\lambda\); line half-width \(\Delta\theta = \dfrac{\lambda}{Nd\cos\theta}\).

✓ Dispersion & power

Dispersion \(D = \dfrac{m}{d\cos\theta}\); resolving power \(R = Nm\).

✓ X-ray diffraction

Bragg's law \(2d\sin\theta = m\lambda\) (\(\theta\) from the plane) reads crystal structure.

Practice

Problems

Unless stated otherwise, take air to have \(n = 1.00\) and assume the screen is far from the slit (Fraunhofer diffraction), so the rays to any point are effectively parallel. Keep angles consistent: use radians inside \(\sin\alpha\), and remember that single-slit minima obey \(a\sin\theta = m\lambda\) with \(m \ge 1\), while grating maxima obey \(d\sin\theta = m\lambda\) with \(m \ge 0\). For Bragg reflections, measure \(\theta\) from the atomic plane, not its normal.

  1. Monochromatic light of wavelength 550 nm passes through a single slit of width \(0.025\,\mathrm{mm}\). (a) At what angle does the first diffraction minimum appear? (b) How wide is the central maximum on a screen 1.50 m away?
  2. A single slit is illuminated by light containing two wavelengths, \(\lambda_1\) and \(\lambda_2\). The first minimum of \(\lambda_1\) coincides with the second minimum of \(\lambda_2\). If \(\lambda_2 = 420\,\mathrm{nm}\), find \(\lambda_1\).
  3. Light of wavelength 633 nm falls on a slit of width \(a = 6.0\,\mu\mathrm{m}\). How many diffraction minima appear on each side of the central maximum (i.e., the largest \(m\) with a physical solution)?
  4. For a single slit, show by direct substitution that the intensity at the first diffraction minimum is zero, and compute \(I/I_m\) at the point where \(\alpha = 2.0\,\mathrm{rad}\).
  5. A single slit forms a diffraction pattern. At what angle (in terms of \(\lambda/a\)) is the intensity reduced to half its central value? (Hint: solve \((\sin\alpha/\alpha)^2 = 0.5\) numerically; the root is near \(\alpha \approx 1.39\,\mathrm{rad}\).)
  6. The pupil of a human eye has a diameter of about \(4.0\,\mathrm{mm}\) in daylight. Taking \(\lambda = 550\,\mathrm{nm}\), estimate the smallest angular separation the eye can resolve, and the corresponding separation of two points at the near distance of 25 cm.
  7. Two stars are separated by an angle of \(1.0\times10^{-5}\,\mathrm{rad}\). What minimum telescope-mirror diameter is needed to resolve them at \(\lambda = 500\,\mathrm{nm}\)?
  8. A car's headlights are 1.4 m apart. At \(\lambda = 550\,\mathrm{nm}\) and a pupil diameter of 5.0 mm, at what maximum distance could a human eye, limited only by diffraction, resolve them as two lights?
  9. A double slit has slit separation \(d = 0.20\,\mathrm{mm}\) and slit width \(a = 0.040\,\mathrm{mm}\). How many bright interference fringes lie within the central diffraction peak?
  10. In a two-slit pattern, the fourth-order interference maximum is missing. What is the ratio \(d/a\) of slit separation to slit width?
  11. A diffraction grating has \(N = 4000\) rulings spread over a width of \(2.0\,\mathrm{cm}\). Find the spacing \(d\), and the angle of the second-order line for \(\lambda = 589\,\mathrm{nm}\).
  12. A grating with 600 lines/mm is illuminated by white light (400–700 nm). Find the angular width of the first-order visible spectrum.
  13. Show that for the grating of Problem 12, the first-order and second-order spectra overlap. Over what wavelength range does the overlap occur?
  14. What is the smallest number of rulings a grating must have to resolve the sodium doublet (\(589.00\) and \(589.59\,\mathrm{nm}\)) in (a) first order and (b) second order?
  15. X-rays of wavelength \(0.165\,\mathrm{nm}\) produce a first-order Bragg reflection from a set of crystal planes at a glancing angle of \(\theta = 22.0^\circ\). (a) Find the interplanar spacing \(d\). (b) At what angle would the second-order reflection appear, if it exists?
Tip: three habits carry this whole chapter. First, keep the two "\(\sin\theta\)" formulas straight by remembering what they describe: \(a\sin\theta = m\lambda\) with \(m\ge 1\) locates the dark fringes of one slit (here \(a\) is the slit width), while \(d\sin\theta = m\lambda\) with \(m\ge 0\) locates the bright lines of a grating (here \(d\) is the slit separation) — the same algebra, opposite meanings. Second, for any intensity question convert to the phase variables first, \(\alpha = (\pi a/\lambda)\sin\theta\) and \(\beta = (\pi d/\lambda)\sin\theta\), work the sines and cosines in radians, and only then square — and recall a real double slit is the product \(\cos^{2}\beta\) (fine fringes) times \((\sin\alpha/\alpha)^2\) (broad envelope). Third, separate a grating's two virtues: dispersion \(D = m/(d\cos\theta)\) tells you how far apart the colors fan, but resolving power \(R = Nm\) — which depends only on how many rulings are lit — tells you whether two close lines stay distinct, and for Bragg work always read \(\theta\) off the plane, never its normal.