# Diffraction

> CIE A-Level Physics · 9702
> Source: https://www.owlsprep.com/study/cie-9702-u8-diffraction/

This module covers diffraction of waves, the structure of single slit diffraction patterns, the formula for minima positions, and how changing slit width and wavelength affects diffraction. It is a foundational topic for diffraction gratings.

**Prerequisites:** [Wave properties and superposition](https://www.owlsprep.com/study/cie-9702-u8-principles-of-superposition/)

## Learning objectives

- Define diffraction and identify when it is most significant
- Describe the intensity distribution of a single slit diffraction pattern
- Recall and use the formula for positions of minima in single slit diffraction
- Explain how changing slit width and wavelength affects the diffraction pattern

## 1. What is Diffraction?

**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.

> **info**
>
> If the aperture is much larger than the wavelength, very little diffraction occurs, and waves travel approximately in straight lines. If the aperture is much smaller than the wavelength, most of the wave is blocked, so very little passes through.

**Worked example:** A gap of width 1 m is cut into a harbour wall. Which of the following will diffract the most: radio waves ($\lambda = 100$ m), sound waves ($\lambda = 1$ m), visible light ($\lambda = 5 \times 10^{-7}$ 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. Single Slit Diffraction Formula

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).

**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, $\theta$ = angle from the central line to the nth minimum, $n = 1,2,3...$ = order of the minimum, $\lambda$ = wavelength of the incident light. For small angles (when the screen is far away), $\sin\theta \approx \frac{y}{D}$, 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)

> **exam_tip**
>
> Always convert all units to SI (meters) before substituting into the formula, this is the most common mistake in calculations.

## 3. Effect of Changing Slit Width and Wavelength

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:

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

1. 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.
2. If slit width $a$ **increases**: $W$ decreases, the central maximum becomes narrower, all maxima are closer together, and overall intensity increases.
3. If wavelength $\lambda$ **increases**: $W$ increases, so red light produces a wider pattern than blue light for the same slit.
4. If wavelength $\lambda$ **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 $W \propto \frac{\lambda}{a}$ from the formula $W = \frac{2 \lambda D}{a}$.
2. New values: $\lambda' = 2\lambda$, $a' = \frac{a}{2}$
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. Intensity Distribution

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% |

**Check your understanding**

Test your understanding

1. 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

   *Why:* From $W = \frac{2 \lambda D}{a}$, width increases when $D$ increases or $a$ decreases. Correct answer is the two options above.

## Common pitfalls

- **Wrong:** Confusing single slit formula with double slit formula
  - Why it fails: Both use $n\lambda$, but they describe different phenomena with different variables
  - Correct: Remember: $a \sin\theta = n\lambda$ = single slit for minima, $d \sin\theta = n\lambda$ = double slit/grating for maxima
- **Wrong:** Leaving wavelength in nm or slit width in mm for calculations
  - Why it fails: This gives an answer that is wrong by orders of magnitude
  - Correct: Always convert all lengths to SI units (meters) before substituting
- **Wrong:** Claiming diffraction does not occur unless aperture size equals wavelength
  - Why it fails: Diffraction occurs at all aperture sizes, it is just most significant when comparable to wavelength
  - Correct: Diffraction is significant when aperture size is within one order of magnitude of the wavelength
- **Wrong:** Claiming increasing slit width increases diffraction
  - Why it fails: Wider apertures produce less diffraction, narrower apertures produce more
  - Correct: Smaller slit width gives a wider diffraction pattern, meaning more diffraction
- **Wrong:** Using $a \sin\theta = n\lambda$ to find the position of maxima
  - Why it fails: The formula only gives positions of minima, not maxima
  - Correct: Maxima are approximately halfway between adjacent minima, calculate them from the minima positions if needed

## Cheatsheet

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

## 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.

- [Diffraction grating](https://www.owlsprep.com/study/cie-9702-u8-diffraction-grating/)
- [Current of electricity](https://www.owlsprep.com/study/cie-9702-u9-overview/)
- [Electric current](https://www.owlsprep.com/study/cie-9702-u9-electric-current/)

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