Diffraction

CIE A-Level Physics· Unit 8: Superposition, Topic 5: Diffraction· 15 min read

1. 1. What is Diffraction?★☆☆☆☆⏱ 3 min

📘 Definition

Diffraction

The bending or spreading of waves when they pass through an aperture or around an obstacle. Diffraction occurs for all types of waves.

Example:

Sound diffracting around a doorway, allowing you to hear sound from another room.

Diffraction is most significant when the size of the aperture or obstacle is approximately equal (comparable) to the wavelength of the incident wave.

📐 Worked Example

A gap of width 1 m is cut into a harbour wall. Which of the following will diffract the most: radio waves ( $λ = 100$ m), sound waves ( $λ = 1$ m), visible light ( $λ = 5 \times 1 0^{- 7}$ m)?

1
Recall that diffraction is most significant when the aperture width is approximately equal to the wavelength of the wave.
2
Compare each wavelength to the 1 m gap:
3
$100 \text{ m} \gg 1 \text{ m}, \quad 1 \text{ m} \approx 1 \text{ m}, \quad 5 \times 10^{-7} \text{ m} \ll 1 \text{ m}$
4
Sound waves have a wavelength closest to the aperture width, so they diffract the most.

2. 2. Single Slit Diffraction Formula★★☆☆☆⏱ 5 min

When monochromatic light passes through a narrow single slit and hits a distant screen, it forms a diffraction pattern with a wide, bright central maximum, and weaker, narrower secondary maxima on either side, separated by dark minima (zero intensity).

📘 Definition

nth Minimum

The nth point of zero intensity from the centre of the pattern. The position of the first minimum defines the width of the central maximum.

The angular position of the nth minimum is given by the formula:

a \sin\theta = n \lambda

Where $a$ = slit width, $θ$ = angle from the central line to the nth minimum, $n = 1, 2, 3...$ = order of the minimum, $λ$ = wavelength of the incident light. For small angles (when the screen is far away), $sin θ \approx \frac{y}{D}$ , so the formula approximates to:

a y = n \lambda D

Here, $y$ = distance of the nth minimum from the centre of the pattern, $D$ = distance from the slit to the screen.

📐 Worked Example

A slit of width 0.1 mm is illuminated by red light of wavelength 650 nm. A screen is placed 2.0 m from the slit. Calculate the distance from the central maximum to the first order minimum.

1
Convert all values to SI units:
2
$a = 0.1 \text{ mm} = 1 \times 10^{-4} \text{ m}, \quad \lambda = 650 \text{ nm} = 6.5 \times 10^{-7} \text{ m}, \quad D = 2.0 \text{ m}, \quad n = 1$
3
Rearrange the small angle formula for $y$ :
4
$y = \frac{n \lambda D}{a}$
5
Substitute values:
6
$y = \frac{1 \times 6.5 \times 10^{-7} \times 2.0}{1 \times 10^{-4}} = 1.3 \times 10^{-2} \text{ m} = 13 \text{ mm}$
7
Final answer: The distance is 13 mm (or 0.013 m)

3. 3. Effect of Changing Slit Width and Wavelength★★☆☆☆⏱ 4 min

The width of the central maximum $W$ is twice the distance from the centre to the first minimum ( $n = 1$ ), so from the small angle formula:

W = \frac{2 \lambda D}{a}

If slit width $a$ decreases: $W$ increases, the central maximum becomes wider, all maxima are further apart, and overall intensity decreases because less light passes through.
If slit width $a$ increases: $W$ decreases, the central maximum becomes narrower, all maxima are closer together, and overall intensity increases.
If wavelength $λ$ increases: $W$ increases, so red light produces a wider pattern than blue light for the same slit.
If wavelength $λ$ decreases: $W$ decreases, the central maximum becomes narrower.

📐 Worked Example

Blue light of wavelength 450 nm produces a central maximum of width 10 cm on a screen. What is the new width of the central maximum if the slit width is halved and the wavelength is doubled?

1
We know that $W \propto \frac{λ}{a}$ from the formula $W = \frac{2 λ D}{a}$ .
2
New values: $λ^{'} = 2 λ$ , $a^{'} = \frac{a}{2}$
3
$W' = \frac{2 \lambda' D}{a'} = \frac{2 (2\lambda) D}{a/2} = 4 \times \frac{2 \lambda D}{a} = 4W$
4
Original $W = 10$ cm, so new $W^{'} = 4 \times 10 = 40$ cm

4. 4. Intensity Distribution★★★☆☆⏱ 3 min

The intensity of the maxima decreases rapidly as you move away from the centre. The relative intensities are shown in the table below:

Position of Maximum	Relative Intensity
Central maximum	100%
First secondary maximum	~4%
Second secondary maximum	~1.6%

✓ Quick check

Test your understanding

Which of the following changes will increase the width of the central maximum?
- Decrease slit width
- Decrease wavelength
- Increase slit width
- Increase distance from slit to screen
Reveal answer
[ "Decrease slit width", "Increase distance from slit to screen" ] —
From $W = \frac{2 λ D}{a}$ , width increases when $D$ increases or $a$ decreases. Correct answer is the two options above.

5. Common Pitfalls

Wrong move:

Confusing single slit formula with double slit formula

Why:

Both use $nλ$ , but they describe different phenomena with different variables

Correct move:

Remember: $a sin θ = nλ$ = single slit for minima, $d sin θ = nλ$ = double slit/grating for maxima

Wrong move:

Leaving wavelength in nm or slit width in mm for calculations

Why:

This gives an answer that is wrong by orders of magnitude

Correct move:

Always convert all lengths to SI units (meters) before substituting

Wrong move:

Claiming diffraction does not occur unless aperture size equals wavelength

Why:

Diffraction occurs at all aperture sizes, it is just most significant when comparable to wavelength

Correct move:

Diffraction is significant when aperture size is within one order of magnitude of the wavelength

Wrong move:

Claiming increasing slit width increases diffraction

Why:

Wider apertures produce less diffraction, narrower apertures produce more

Correct move:

Smaller slit width gives a wider diffraction pattern, meaning more diffraction

Wrong move:

Using $a sin θ = nλ$ to find the position of maxima

Why:

The formula only gives positions of minima, not maxima

Correct move:

Maxima are approximately halfway between adjacent minima, calculate them from the minima positions if needed

6. Quick Reference Cheatsheet

Relationship	Formula	Notes
Most significant diffraction	$a \approx λ$	Aperture width comparable to wavelength
Position of nth minimum	$a sin θ = nλ$	$n = 1, 2, 3...$
Small angle approximation	$y = \frac{nλ D}{a}$	$y$ = distance to minimum, $D$ = slit-screen distance
Width of central maximum	$W = \frac{2 λ D}{a}$	Twice distance to first minimum
Decrease slit width $a$	$W \propto \frac{1}{a}$	W increases, intensity decreases
Increase wavelength $λ$	$W \propto λ$	W increases

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · 2
Single slit diffraction calculation
2022 · 1
Effect of changing slit width
2021 · 2
Draw intensity distribution

Going deeper

syllabusCIE 9702 Physics Syllabus

What's Next

Diffraction is a fundamental wave property that is built on in subsequent topics in superposition, most importantly diffraction gratings, which are a very common exam question in CIE A-Level Physics. Understanding single slit diffraction also helps you distinguish between diffraction effects and interference from multiple slits, which is a common source of confusion in multiple choice questions. The principles of diffraction are also used in topics involving wave-particle duality, so mastering this sub-topic will help you with later content in the syllabus.