IB DP Physics: Superposition, double-slit interference, single-slit diffraction and gratings (DP Physics)

This guide covers the second half of sub-topic C.3, Wave phenomena, from the IB Physics guide, first assessment 2025: superposition, double-source interference and Young’s double slit (standard and higher level), then single-slit diffraction, its modulation of the double-slit pattern, and multiple slits and diffraction gratings (additional higher level, HL only). Wavefronts, rays, reflection, refraction and the diffraction of waves around bodies and apertures are taught in a separate unit and are not repeated here.

Use it with the revision notes and the practice questions. Course links: IB DP Physics hub, printable checklist and the free diagnostics.

What this unit covers

Section	What you must be able to do	Level
C.3	Apply superposition to waves and pulses	SL and HL
C.3	Explain why double-source interference needs coherent sources	SL and HL
C.3	Use path difference = nλ (constructive) and (n + ½)λ (destructive)	SL and HL
C.3	Use Young’s double-slit result s = λD/d	SL and HL
C.3	Use single-slit diffraction θ = λ/b and describe its intensity pattern	HL only
C.3	Explain how single-slit diffraction shapes the brightness of double-slit fringes	HL only
C.3	Use nλ = d sin θ for multiple slits and gratings, with white and monochromatic light	HL only

The guide limits patterns to normal incidence, single-slit work to monochromatic light and rectangular slits, and the effect of slit width on single-slit intensity to a qualitative treatment.

Superposition of waves and pulses

When two waves or pulses arrive at the same point, the resultant displacement there is the sum of the individual displacements, with direction (sign) included. This is the principle of superposition. The waves then carry on unchanged after they have passed through each other.

• Two upward pulses overlap to give a larger upward displacement (constructive).
• An upward and a downward pulse of equal size cancel at the instant they coincide (destructive).

Worked example 1. A pulse of height +4.0 cm on a rope meets a pulse of height −3.0 cm. At the instant the centres coincide, the resultant is +4.0 + (−3.0) = +1.0 cm. Once they separate, each pulse has its original shape and size.

For continuous waves, superposition at a point depends on the phase difference between the two waves there. In phase gives a large resultant; in antiphase gives a small or zero one.

Coherent sources

Stable interference needs coherent sources: two sources of the same frequency (so the same wavelength in the same medium) whose phase difference stays constant. If the phase difference changed randomly, maxima and minima would move about faster than any detector could follow and you would see an even average.

Two separate filament lamps are not coherent, because each emits light in short, random bursts. Coherence is achieved by taking the light from one source and splitting it, for example with a laser lighting two slits, so any change in the source affects both beams alike. In the standard demonstration the two slits act as two coherent sources.

Conditions for interference

Take two coherent sources that vibrate in phase. At a point P, the path difference is the difference between the distances from P to each source.

• Constructive interference (maximum): path difference = nλ, for n = 0, 1, 2, …
• Destructive interference (minimum): path difference = (n + ½)λ.

Path difference nλ means the waves arrive in phase, because one is a whole number of wavelengths behind the other. A half-integer number of wavelengths puts them in antiphase.

Worked example 2. Two microwave transmitters, in phase, emit λ = 2.8 cm.

• Point P is 41.3 cm from one and 27.3 cm from the other. Path difference = 14.0 cm. Divide by λ: 14.0 / 2.8 = 5.0, a whole number, so constructive.
• Point Q is 36.4 cm and 26.6 cm away. Path difference = 9.8 cm; 9.8 / 2.8 = 3.5 = (3 + ½), so destructive.

If the resultant at a destructive point is near zero, the energy has not vanished. It appears at the maxima, so the total energy delivered is conserved. Equal-amplitude waves in phase give double the amplitude, and since intensity depends on amplitude squared, the maximum is four times the intensity of a single source.

Young’s double-slit experiment

Monochromatic light passes through a narrow single slit (or comes from a laser), then through two narrow, closely spaced slits of separation d. The overlapping beams give a pattern of equally spaced bright and dark fringes on a screen a distance D away. The central bright fringe lies on the line equidistant from the slits, where the path difference is zero.

The fringe separation s (centre to centre of adjacent bright fringes, or adjacent dark ones) is

s = λD / d

where d is the slit separation and D the distance from slits to screen. The result is a good approximation when D is much greater than d, so the angles are small.

What each variable does:

Change	Effect on s
Longer wavelength (red instead of green)	Fringes spread out
Slits closer together (smaller d)	Fringes spread out
Screen further away (larger D)	Fringes spread out

Worked example 3. A green laser (λ = 532 nm) illuminates slits 0.25 mm apart. The screen is 1.80 m away.

s = λD / d = (532 × 10⁻⁹ × 1.80) / (0.25 × 10⁻³) = 3.83 × 10⁻³ m, so s = 3.83 mm.

Worked example 4 (finding the wavelength). With d = 0.30 mm and D = 2.40 m, the distance across 10 fringe spacings is 4.2 cm. Then s = 0.042 / 10 = 4.2 × 10⁻³ m.

λ = sd / D = (4.2 × 10⁻³ × 0.30 × 10⁻³) / 2.40 = 5.25 × 10⁻⁷ m, so λ = 525 nm.

Measuring across many fringe spacings and dividing is better than measuring one spacing, because the fixed reading uncertainty is shared over many fringes.

The pattern is also evidence for the wave nature of light, since particles travelling in straight lines would not produce alternating bright and dark bands.

Single-slit diffraction (HL only)

Light passing through one narrow slit of width b spreads out and forms a pattern on a distant screen. For a rectangular slit and monochromatic light the pattern has:

• a wide, bright central maximum;
• weaker secondary maxima on both sides, each about half the width of the central one;
• minima between them.

The angle to the first minimum, measured from the central axis, is given by

θ = λ / b

(θ in radians, small angle). The central maximum therefore has angular width 2λ/b, and on a screen at distance D its width is about 2Dλ/b.

Qualitatively: a narrower slit (smaller b) gives a wider pattern and a less intense one, because less light gets through. A wider slit gives a narrower, brighter pattern. Longer wavelengths also spread more.

Worked example 5. Red light λ = 650 nm falls on a slit of width 0.040 mm. The screen is 1.5 m away.

θ = λ / b = (650 × 10⁻⁹) / (0.040 × 10⁻³) = 0.0163 rad.

Width of central maximum = 2Dθ = 2 × 1.5 × 0.01625 = 0.0488 m (about 4.9 cm).

If the slit is made twice as wide, θ halves and the central maximum is about 2.4 cm wide, and brighter.

Modulation of the double-slit pattern (HL only)

Real double slits have slits of finite width b. Each slit produces its own single-slit pattern, and the two patterns overlap on the screen. The result is the double-slit fringes (spacing set by d) inside an envelope shaped like the single-slit pattern (width set by b). This is the modulation of the double-slit pattern by the single-slit one: fringe brightness follows the envelope, and fringes are strong near the centre and fade towards the envelope’s minima.

Where a double-slit maximum falls exactly on a single-slit minimum, that fringe is missing.

Worked example 6. Slits with d = 0.30 mm and b = 0.050 mm are lit by monochromatic light.

• The first single-slit minimum is at sin θ ≈ λ/b. A double-slit maximum of order n is at sin θ ≈ nλ/d.
• They coincide when n/d = 1/b, so n = d/b = 0.30 / 0.050 = 6.
• The sixth-order fringe is missing on each side. Fringes from orders 0, ±1, …, ±5 lie inside the central envelope: 11 bright fringes.

Multiple slits and diffraction gratings (HL only)

A diffraction grating has a large number of equally spaced parallel slits. For normal incidence, the maxima occur at angles θ where

nλ = d sin θ

with n = 0, 1, 2, … the order and d the slit spacing (the distance between adjacent slits). If a grating has N lines per millimetre, d = (1/N) mm.

Compared with two slits, a grating with many slits gives narrower, sharper and brighter maxima, with dark regions between. Because the maxima are narrow, wavelengths can be measured accurately. The maximum order is the largest n with n < d/λ, because sin θ cannot exceed 1.

Monochromatic light: a central (n = 0) maximum and symmetrical maxima on either side.

White light: the central maximum is white, since all wavelengths have zero path difference. For each higher order a continuous spectrum appears, with violet (shorter λ) nearer the centre and red (longer λ) further out, because sin θ is proportional to λ. Higher orders are more spread out and can overlap with neighbouring orders.

Worked example 7. A grating has 750 lines per mm. Light of λ = 610 nm is incident normally.

• d = 1/750 mm = 1.33 × 10⁻⁶ m.
• First order: sin θ = λ/d = 610 × 10⁻⁹ / 1.333 × 10⁻⁶ = 0.4575, so θ = 27.2°.
• Maximum order: d/λ = 2.19, so n ≤ 2. Second order: sin θ = 0.915, θ = 66.2°.
• Maxima seen: n = 0, ±1, ±2, which is 5 in total.

Worked example 8 (white light, same grating). Take λ from 400 nm to 700 nm.

• First order spectrum: 17.5° (violet) to 31.7° (red).
• Second order: violet at sin θ = 0.600, so 36.9°. Wavelengths above d/2 = 667 nm have no second order, so the red end is cut off.
• Third order violet: sin θ = 3 × 400 × 10⁻⁹ / 1.333 × 10⁻⁶ = 0.900, so θ = 64.2°, which lies inside the range of the second-order spectrum (from 36.9° upwards). Orders 2 and 3 overlap; orders 1 and 2 do not.

Common errors

• Using path difference = nλ for destructive interference, or forgetting the ½.
• Using d as the slit width in single-slit work or b as the slit separation in double-slit work.
• Not converting units: lines per mm to metres, mm and nm to metres.
• Quoting s = λD/d for the grating spacing of maxima. Grating maxima follow nλ = d sin θ and are not equally spaced in general.
• Saying a narrower slit gives a narrower diffraction pattern.
• Stating the white central maximum is coloured.
• Calling the half-width of the central maximum its width: the angle θ = λ/b is to the first minimum, so the full width is twice as much.

Next steps

Try the practice questions, then use the revision notes for quick recall.

Official syllabus

Physics guide, first assessment 2025, International Baccalaureate Organization, sub-topic C.3 Wave phenomena (superposition, interference and the additional higher level content on diffraction and gratings).