Describe how wavelength and gap size affects diffraction through a gap

3.1 General Properties of Waves – Diffraction through a Gap

Learning Objective

Explain how the wavelength (λ) and the width of a gap (a) determine the amount of diffraction when a wave passes through a narrow opening, and describe the resulting diffraction pattern.

What is Diffraction?

Diffraction is the bending and spreading of a wave when it encounters an obstacle or passes through an aperture (a gap or slit). The extent of the spreading depends on the relationship between the wavelength (λ) of the wave and the characteristic dimension of the aperture (a).

Key Relationship (first minima)

The directions of the first minima on either side of the central maximum satisfy the exact condition

a sin θ = λ

For small angles (θ ≲ 30°) we may use the small‑angle approximation

θ ≈ λ / a  (θ in radians)

where

• θ = angle between the central direction and the first minimum
• λ = wavelength of the incident wave
• a = width of the gap (aperture)

Qualitative Diffraction Pattern

• Central maximum – the brightest, most intense region directly opposite the gap.
• First minima – points where destructive interference makes the intensity essentially zero (a sin θ = λ).
• Secondary (side) maxima – weaker bright bands between successive minima.

Effect of Wavelength (λ)

• Longer wavelength (large λ) – increases λ/a, therefore θ becomes larger. The wave “samples” a larger fraction of the aperture and spreads more widely.
• Shorter wavelength (small λ) – reduces λ/a, giving a smaller θ and a more collimated beam.

Effect of Gap Size (a)

• Large gap (a ≫ λ) – λ/a is very small → θ ≈ 0°. Diffraction is negligible; the beam stays narrow and the shadow edge is sharp.
• Gap comparable to wavelength (a ≈ λ) – λ/a is of order unity → θ of a few‑tens of degrees. Noticeable spreading with a well‑defined central maximum and visible side fringes.
• Very narrow gap (a ≪ λ) – λ/a becomes large → θ can approach 90°. The wave spreads almost isotropically; the central maximum dominates the pattern.

Order‑of‑magnitude tip: if a is ten times λ, then λ/a ≈ 0.1 rad ≈ 6°, giving only a slight diffraction spread.

Summary Table

Paper 5/6‑style Experimental Demonstration

1. Set up a monochromatic laser (e.g., red, λ ≈ 650 nm) so that its beam strikes a narrow slit of known width a.
2. Place a screen a distance L (≈ 2–3 m) behind the slit.
3. Measure the distance d from the centre of the bright spot to the first dark fringe on the screen.
4. Calculate the diffraction angle using the small‑angle relation θ ≈ tan⁻¹(d/L).
5. Compare the experimental value of θ with the theoretical prediction θ ≈ λ/a (or use the exact condition a sin θ = λ for higher accuracy).

This procedure develops AO2 (handling data) and AO3 (experimental skills) – both required for the IGCSE exam.

Real‑World Examples (all wave types)

• Visible light (λ ≈ 500 nm) through a slit a = 5 µm (a ≈ 10 λ) – only slight diffraction; sharp shadow.
• Radio waves (λ ≈ 1 m) passing through a doorway a ≈ 1 m – pronounced diffraction, allowing the signal to bend around corners.
• Sound (acoustic) waves (λ ≈ 0.3 m) encountering a gap a ≈ 0.3 m – audible bending around walls, useful in auditorium design.
• Water waves (λ = 0.2 m) through a gap a = 0.05 m (a = 0.25 λ) – fan‑shaped spreading visible on the water surface.

Key Points for Exam Answers

• State the exact condition for the first minima: a sin θ = λ.
• For small angles you may use the approximation θ ≈ λ/a and note its validity (θ ≲ 30°).
• Explain that increasing λ or decreasing a increases λ/a, giving a larger diffraction angle.
• Use an order‑of‑magnitude comparison (e.g., a ≈ 10 λ → very little diffraction) to justify predictions.
• Reference the laser‑slit‑screen experiment or a real‑world example to illustrate the concept.
• Remember the principle applies to all wave types (light, sound, water, radio) when the ratio a/λ is considered.