17.2Interference and Diffraction
Diffraction — American System Physics Revision Notes & Topical Questions
10 min read 7 parts Lesson 4 of 5 1 quiz
Describe diffraction of waves at a gap and at an edge.Single slit, first dark minimaw sin θ = λ (for small angles θ ≈ λ/w, θ in radians).Single-slit dark fringesw sin θ = mλ, m = 1, 2, 3 …Width of central maximum on a screen at distance L: 2x1 = 2λL/w (m).Rayleigh criterion, circ…
- Wave equation still applies
- Huygens' principle explains diffraction
- For the slit above, a screen is 2.0 m away. How wide is the central maximum?
- Aperture (gap width, w)
- Single-slit diffraction pattern
- Diffraction condition for large effect
Section 1 of 7Diffraction
Section 2 of 7Process & sequence
Plane waves passing through a gapwrite these steps in this order; each step is a marking point.
Section 3 of 7Compare
| Gap width compared with λ | Amount of spreading | Example |
|---|---|---|
| Much larger than λ | Very little; mainly straight-through beam with slight edge bending | Light through a doorway |
| Similar to λ | Large; wavefronts become nearly semicircular | Sound through a doorway (λ ≈ 1 m) |
| Smaller than λ | Wave spreads strongly but little energy gets through | Microwaves and a mesh screen |
Longer wavelength with the same gap gives more spreading: that is why you can hear round corners but not see round them.
| Effect | Cause | Changes |
|---|---|---|
| Diffraction | Wave passes gap/edge | Direction spreads; λ, v, f unchanged |
| Refraction | Wave enters new medium | Speed and λ change; f unchanged |
| Interference | Two coherent waves overlap | Amplitude pattern (maxima/minima) |
Section 4 of 7Worked examples
- sin θ = λ/w = 600 × 10−9 / 0.12 × 10−3 = 5.0 × 10−3, so θ = 0.29° (5.0 × 10−3 rad).
- Distance to first minimum x1 = λL/w = 600 × 10−9 × 2.0 / 0.12 × 10−3 = 1.0 × 10−2 m.
- Width = 2x1 = 2.0 × 10−2 m (2.0 cm).
- θ = 1.22λ/D = 1.22 × 550 × 10−9 / 0.10 = 6.7 × 10−6 rad.
Section 5 of 7Key terms
Section 6 of 7Command terms
Section 7 of 7Exam technique
Keywords the mark scheme looks for in this lesson
How to get full marks (examiner guidance)
Words to avoid
Instant feedbackCheck yourself
Q1. Diffraction is greatest when the gap width is
- A. much larger than λ
- B. about equal to λ
- C. zero
- D. equal to the amplitude
Show answer
B. Maximum spreading occurs for gaps comparable with the wavelength.
Q2. After diffraction through a gap, the wavelength
Show answer
C. Diffraction does not change λ, f or v.
Q3. We can hear round a corner but not see round it because
Show answer
B. Sound λ ≈ door width so it diffracts strongly; light λ is tiny.
Q4. A slit is made narrower. The central maximum becomes
- A. narrower and brighter
- B. wider
- C. unchanged
- D. coloured
Show answer
B. sin θ = λ/w, so smaller w gives larger θ and a wider central maximum.
Q5. λ = 500 nm, w = 0.050 mm. sin θ for the first minimum is
Show answer
A. sin θ = 5.0×10⁻⁷/5.0×10⁻⁵ = 1.0×10⁻².
Q6. Which improves the resolution of a telescope?
Show answer
C. θ = 1.22λ/D; larger D gives a smaller resolvable angle.
Marked practiceQuizzes for this lesson
Last updated:
