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Diffraction

Key idea: Explain diffraction as spreading of waves, and predict when diffraction is significant using the wavelength-to-aperture size idea (A Level Physics).

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Explain single-aperture diffraction and apply first-minimum and Rayleigh criteria.
  • Relate phase difference to separations in time and position.
  • Explain coherent two-source interference using phase and path difference.

1. Definitions (Must Know)

A. Diffraction

Diffraction is the spreading of waves when they pass through an aperture (gap) or around an obstacle, so waves enter the “shadow” region.

B. Significant diffraction (rule of thumb)

Diffraction is most obvious when the wavelength is comparable to the size of the gap/obstacle: λ ∼ a where a is a characteristic size (gap width or obstacle width).

2. Key Ideas (What Earns Marks)

  • Diffraction is a wave property (sound, water waves, light).
  • Compare the wavelength λ with the gap/obstacle size a:
    • λ ≪ a: little spreading (sharp shadow)
    • λ ∼ a: strong spreading
    • λ ≫ a: very strong spreading, but the transmitted wave may be weak because little energy passes through a tiny opening
  • In exam explanations, use “wavelength comparable to gap size” rather than vague words like “small gap”.
Links to calculation pages

Diffraction calculations are usually done in:

3. Detailed Explanations

A. Why waves spread into the shadow region

You can think of each point across a wavefront as producing secondary wavelets.

When the wavefront is forced through a narrow opening, those wavelets spread out, so the wave travels into regions that would be “geometrically shadowed”.

B. Why the wavelength-to-gap ratio matters

  • If the gap is much larger than λ, the wavefront passes through almost unchanged, so the wave continues mostly straight.
  • If the gap is comparable to λ, the opening strongly disturbs the wavefront, so spreading is obvious.

C. Everyday comparisons (why sound “bends” more than light)

Typical wavelengths:

  • sound (conversation): λ ∼ 0.3–3 m
  • visible light: λ ∼ 5 × 10⁻⁷ m

Doors and corridors have sizes similar to sound wavelengths but enormously larger than light wavelengths, so sound diffracts noticeably while light usually forms sharp shadows.

4. Common Mistakes

  • Saying “diffraction only happens for small gaps” (it happens always; it is only obvious for certain size ratios).
  • Explaining with “waves bend because they want to” (use the size comparison: λ vs a).
  • Mixing diffraction with refraction (refraction needs a change in wave speed between media).

5. Exam Tips

  • If asked “when is diffraction significant?”, write: “when λ is of the same order as the gap/obstacle size”.
  • Use concrete examples: “sound around a doorway” vs “light around a doorway”.
  • If a question mentions “sharp shadow”, it implies λ ≪ a.

6. Worked Examples

Modelled example 1

Predict which wave diffracts more (qualitative)

Core

Problem

A door is 0.80 m wide. Compare diffraction of sound with λ = 0.40 m and light with λ = 500 nm.
Study the worked solution

  1. Compare sound wavelength with the door

    Method

    Sound diffracts significantly.

    Reason

    The sound wavelength is the same order as the 0.80 m opening.

    Working

    λ/a = 0.40/0.80 = 0.50

  2. Compare light wavelength with the door

    Method

    Visible light has negligible spreading at this scale.

    Reason

    500 nm = 5.0 × 10⁻⁷ m is tiny compared with the doorway.

    Working

    λ/a ≈ 6.3 × 10⁻⁷≪1

  3. State the observable consequence

    Method

    Sound reaches around the doorway while light forms a much sharper shadow.

    Reason

    The wavelength-to-opening ratio, not the absolute label “sound” or “light”, controls how obvious diffraction is.

    Working

    0.50≫6.3 × 10⁻⁷, so sound spreads much more.

Guided practice 2

“Same gap, different wavelength”

About 3 min

Problem

Water waves of wavelength 4.0 cm pass through a gap of width 20 cm. Will the diffraction be strong?

Try this before viewing the solution

Diffraction strength

Hints

Hint 1: form a size ratio
Calculate λ/a = 4.0/20 and compare it with one.
View solution step by step

  1. Compare the scales

    Method

    λ/a = 0.20.

    Reason

    The opening is substantially wider than one wavelength.

    Working

    4.0 cm/20 cm = 0.20

  2. Predict the pattern

    Method

    The diffraction is weak rather than strong.

    Reason

    When λ≪ a, most of the transmitted wavefront continues nearly straight.

    Working

    λ/a = 0.20 ⇒ weak spreading

Common misconception 3

Compare diffraction for two gaps (same wavelength)

Find and correct the mistake

Learner claim

Water waves of wavelength 5.0 cm pass through gaps of width 5.0 cm and 25 cm. A learner says the wider gap causes more spreading because more wave enters it. Diagnose the claim.

Try this before viewing the solution

Gap with stronger spreading

View solution step by step

  1. Test the equal-width case

    Method

    The 5.0 cm gap gives strong diffraction.

    Reason

    Its width is equal to the wavelength, so λ/a = 1.

    Working

    5.0/5.0 = 1

  2. Test the wider gap

    Method

    The 25 cm gap gives weaker diffraction.

    Reason

    Here λ/a = 0.20, so the wavefront is less disturbed relative to its wavelength.

    Working

    5.0/25 = 0.20

  3. Diagnose the reasoning

    Method

    The learner has confused transmitted amount with angular spreading.

    Reason

    A wider gap may transmit more energy, but the diffraction angle is governed by wavelength relative to aperture size.

    Working

    transmitted energy ≠ diffraction strength

Examiner practice 4

Will microwaves diffract through a doorway?

2 marks

Examination question

Microwaves of wavelength λ = 3.0 cm are incident on an opening of width a = 1.0 cm. State and explain whether diffraction will be significant. [2 marks]

Try this before viewing the solution

Will diffraction be significant?

View solution step by step

  1. Compare wavelength and aperture

    1 mark

    Method

    The wavelength is comparable to, and larger than, the opening width.

    Reason

    λ/a = 3.0 is not much less than one.

    Working

    3.0 cm/1.0 cm = 3.0

  2. State the prediction

    1 mark

    Method

    Diffraction will be significant and the transmitted wave will spread strongly.

    Reason

    Strong spreading occurs when the aperture is of the same order as the wavelength or smaller.

    Working

    λ/a = 3.0 ⇒ significant diffraction

Challenge 5

Using speed and frequency to estimate diffraction

Minimal support

Independent transfer

A sound wave of frequency f = 680 Hz travels at v = 340 m s⁻¹. Find its wavelength and decide whether it will diffract noticeably around a 1 m wide doorway.

Try this before viewing the solution

Unit: m
Diffraction prediction

Hints

Hint 1: find the missing wavelength
Rearrange the wave equation to λ = v/f.
View solution step by step

  1. Calculate wavelength

    Method

    λ = 0.50 m.

    Reason

    Wave speed equals frequency multiplied by wavelength.

    Working

    λ = v/f = 340/680 = 0.50 m

  2. Compare with the doorway

    Method

    The ratio is λ/a = 0.50.

    Reason

    The wavelength is the same order as the 1 m opening.

    Working

    0.50 m/1.0 m = 0.50

  3. Predict the spreading

    Method

    The sound will diffract noticeably around the doorway.

    Reason

    A wavelength comparable with the opening produces appreciable spreading.

    Working

    λ∼ a ⇒ noticeable diffraction

7. Mind Stretchers

Mind stretcher 1: Why does a very tiny hole give a weak pattern?Extension

Light passes through a pinhole whose diameter is far smaller than the wavelength. Diffraction is very strong, but the screen looks dim.

Explain why.

Show Answer

The pinhole allows only a very small amount of energy through (tiny area), so the transmitted wave is weak even though it spreads out strongly.

Mind stretcher 2: Improving “sharpness” with wavelength or apertureExtension

You want a wave to travel through an opening with as little spreading as possible. Should you use a shorter wavelength or a longer wavelength (assuming the opening size is fixed)? Explain.

Show Answer

Use a shorter wavelength. If λ ≪ a, diffraction is small and the wave spreads less after the opening.

Mind stretcher 3: Optional (Enrichment)Extension

For a single slit of width b, the first minimum occurs at: b sin θ = λ

This turns the qualitative diffraction idea into a quantitative angle/position calculation.

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027