Single Slit Diffraction Revision Notes | A Level AQA Physics

1. Why a Single Slit Produces a Pattern, Not a Spot

Shine a laser through a narrow slit and you might expect a single bright line on the screen behind it. What you actually see is something far more striking: a broad central band of light flanked by a series of dimmer fringes, alternating with dark gaps. This is single slit diffraction, and it reveals something fundamental about the wave nature of light.

The key idea is that diffraction occurs whenever a wave passes through a gap whose width is comparable to the wavelength of the wave. When light passes through a narrow slit, every point across the width of the slit acts as a new source of secondary wavelets — this is Huygens' principle. These wavelets spread out and overlap, and where they meet they interfere, producing the characteristic pattern of bright and dark regions on the screen.

At the centre of the pattern, wavelets from all parts of the slit arrive roughly in phase, so they interfere constructively to produce the bright central maximum. Moving away from the centre, path differences between wavelets from different parts of the slit lead to destructive interference, creating the first dark fringe (minimum) on either side. Further out, partial constructive interference produces the secondary maxima, which are noticeably dimmer than the central maximum.

2. Appearance of the Pattern: Monochromatic Light

Picture the screen in front of you. With monochromatic light — say, a red laser at 650 nm — the single slit diffraction pattern has a very specific appearance that you need to be able to describe precisely in an exam.

• A broad, bright central maximum at the centre of the pattern
• The central maximum is approximately twice as wide as each of the secondary maxima on either side
• A series of narrower, dimmer secondary maxima (subsidiary maxima) on both sides of the central maximum
• Dark fringes (minima) separating each maximum, where destructive interference occurs
• The intensity of the secondary maxima decreases progressively with distance from the centre
• The pattern is symmetrical about the centre

The most important feature to fix in your mind is the intensity distribution: the central maximum dominates. It is both the brightest and the widest feature of the pattern. The secondary maxima are significantly dimmer — approximately 4.7% of the intensity of the central maximum for the first secondary maximum — and each successive one is dimmer still.

The chart above captures the key qualitative point: the central maximum is overwhelmingly brighter than the secondary maxima, and the minima have zero intensity. In an exam, if you are asked to describe the pattern, you must mention the relative brightness as well as the relative widths — both are credit-worthy points.

3. Appearance of the Pattern: White Light

White light contains a continuous range of wavelengths, from violet (around 400 nm) to red (around 700 nm). Because the width of the diffraction pattern depends on wavelength, each wavelength diffracts by a different amount — and this produces a distinctly different pattern from the monochromatic case.

The central maximum with white light appears white (or very nearly so) at its centre. This is because all wavelengths overlap constructively at the very centre of the pattern, so all colours combine to give white light. Moving away from the centre, the colours begin to separate: violet light, having the shortest wavelength, diffracts the least and so its minima appear closest to the centre, while red light, with the longest wavelength, diffracts the most and its minima appear furthest from the centre.

The result is that the edges of the central maximum show a spread of colour — a spectrum — with violet on the inside (closer to the centre) and red on the outside. The secondary maxima also show this spectral colouring, but they overlap with adjacent maxima from other wavelengths, making them appear as smeared, less distinct coloured fringes rather than clean bands.

4. How Slit Width Affects the Central Maximum

Now that you can describe the pattern, the next step is to understand what controls its size — specifically, the width of the central maximum. This is where the qualitative relationship between slit width and diffraction becomes essential.

The relationship is inverse: a narrower slit produces a wider central maximum, and a wider slit produces a narrower central maximum. To see why, think about what happens at the extremes. If the slit is very wide compared to the wavelength, the light passes through almost unimpeded and you get something close to a geometric shadow — a narrow bright line with very little spreading. If the slit is very narrow (approaching the wavelength in size), the wave spreads out dramatically in all directions, producing a very broad central maximum.

The physical reasoning is this: a narrower slit means the path difference between wavelets from opposite edges of the slit is smaller for a given angle. Destructive interference (which creates the first minimum) requires the path difference between opposite-edge wavelets to equal one full wavelength (λ); this condition is therefore met only at a larger angle from the centre, pushing the first dark fringe further out and making the central maximum wider.

5. How Wavelength Affects the Central Maximum

The slit width is not the only variable at play. The wavelength of the incident light also determines how much the wave diffracts, and therefore how wide the central maximum appears.

The relationship here is direct: a longer wavelength produces a wider central maximum. Red light (wavelength around 700 nm) diffracts more than violet light (wavelength around 400 nm) through the same slit. This is exactly why the edges of the central maximum in white light show red on the outside and violet on the inside — the red component has spread furthest from the centre.

The underlying reason is the same as before. For a fixed slit width, a longer wavelength means the path difference between wavelets from opposite edges of the slit reaches one full wavelength (the condition for destructive interference across the slit, path difference = λ) at a larger angle. The first minimum therefore occurs at a larger angle, and the central maximum is wider.

Both relationships can be summarised neatly: the angular half-width of the central maximum is proportional to $\lambda / a$, where $\lambda$ is the wavelength and $a$ is the slit width. You are not required to use this formula quantitatively at A Level, but understanding the ratio helps you reason about any scenario quickly.

6. Required Practical: Observing Single Slit Diffraction

The qualitative relationships you have been studying can be directly observed in the laboratory. The standard approach uses a laser and a set of single slits of different widths, allowing you to see the effect of changing slit width on the diffraction pattern in real time.

One important practical point: the screen must be far enough from the slit for the pattern to be clearly resolved. At close range, the fringes are too close together to distinguish. Increasing the slit-to-screen distance D spreads the pattern out, making measurements more reliable — though it does not change the angular positions of the fringes, only their physical separation on the screen.

7. Exam Technique: Getting Full Marks on Diffraction Questions

Single slit diffraction questions tend to fall into two categories: describe the pattern, or explain how it changes. Each requires a different approach, and examiners are very specific about what earns marks.

For 'describe' questions, you need to state observable features in sequence — no 'because' required. For 'explain' questions, you must link the observation to a physical principle using causal language. Practise distinguishing between these two command words before your exam.

Worked example: Answering a 'Describe and Explain' Diffraction Question

A student shines monochromatic red light through a single narrow slit. Describe the diffraction pattern observed on a screen. The student then replaces the slit with a wider one. Explain how the pattern changes. [4 marks]

1. A bright, broad central maximum is observed at the centre of the pattern. On either side are dimmer, narrower secondary maxima separated from the central maximum and from each other by dark fringes (minima). — [B1] for central maximum being brightest and widest. [B1] for secondary maxima being dimmer and narrower, with dark fringes between them. Do not just say 'bright and dark fringes' — you must distinguish the central maximum from the secondary maxima.
2. The central maximum becomes narrower (less wide). — [B1] for correctly identifying that the central maximum narrows. This is the direct observation — state it clearly before explaining it.
3. A wider slit means the path difference between wavelets from opposite edges of the slit reaches the condition for destructive interference at a smaller angle. The first minimum therefore occurs closer to the centre, so the central maximum is narrower. — [B1] for linking wider slit to smaller angle of first minimum, or equivalently stating that less diffraction occurs with a wider slit. Reference to path difference or the ratio of wavelength to slit width strengthens the answer.

Answer: Bright central maximum (widest and brightest), flanked by dimmer secondary maxima with dark fringes between them. Wider slit produces a narrower central maximum because the first minimum occurs at a smaller angle (less diffraction).

Notice in the worked answer above that the description and explanation are kept separate. Mixing them in a 'describe' question will not earn the explanation marks — examiners award marks for the specific command word being addressed.

To consolidate everything, recall the two core spec requirements: you must be able to describe the appearance of the pattern (for both monochromatic and white light) and explain qualitatively how the central maximum width varies with wavelength and slit width. Both of these have now been covered in teaching and applied in exam-style contexts — you are ready to tackle any question the examiner sets on this topic.