Young's Double Slit and Fringe Spacing Revision Notes | A Level AQA Physics

1. Why Young's Experiment Changed Physics

In 1801, Thomas Young shone light through two narrow slits and saw something that should not have been possible if light were made of particles: a pattern of alternating bright and dark bands on a screen. That pattern — the interference pattern — was direct evidence that light behaves as a wave.

At A Level, you need to go beyond knowing the experiment happened. You need to explain why the pattern forms, what conditions are required, and how to calculate the spacing between fringes precisely. This lesson builds that understanding from the ground up.

2. Coherence and the Need for Two Coherent Sources

Picture two people on a rope, each flicking it up and down at slightly different rates. The pattern they create is constantly shifting — no stable shape ever forms. The same problem arises with light: if two sources emit waves with a constantly changing phase relationship, any interference pattern washes out almost instantly.

In Young's original experiment, a single narrow source slit illuminated two closely spaced slits. Because both slits were fed by the same wavefront, they acted as two coherent secondary sources — the phase difference between them was fixed at zero. Modern versions of the experiment use a laser, which is itself a coherent source, so the single source slit is no longer needed.

The specification requires you to know both approaches: using a single source with double slits (Young's original method) and using two coherent sources directly (such as a laser). Either way, the physics of the resulting pattern is identical.

3. How the Interference Pattern Forms

Once the two coherent sources are in place, each one sends out circular (or spherical) wavefronts. Where these wavefronts overlap, superposition occurs: the displacements of the two waves add together at every point in space.

The key variable is path difference — the difference in distance from each slit to a particular point on the screen. At the central point directly between the two slits, both waves travel exactly the same distance, so the path difference is zero. The waves arrive in phase and interfere constructively, producing a bright fringe.

Move slightly to one side and the path difference grows. When it reaches exactly half a wavelength ($\frac{\lambda}{2}$), the waves arrive exactly out of phase — crest meets trough — and destructive interference produces a dark fringe. Move further still until the path difference reaches one full wavelength ($\lambda$), and constructive interference returns, giving the next bright fringe. This alternating pattern repeats across the screen.

• Path difference = $n\lambda$ (where $n$ = 0, 1, 2,...) → constructive interference → bright fringe
• Path difference = $(n + \frac{1}{2})\lambda$ (where $n$ = 0, 1, 2,...) → destructive interference → dark fringe
• The central bright fringe ($n = 0$) sits at the midpoint between the two slits

4. The Fringe Spacing Equation

Knowing that bright fringes form at whole-wavelength path differences is useful conceptually, but in practice you need to predict exactly where on the screen each fringe will appear. That is where the fringe spacing equation comes in.

Notice what the equation tells you qualitatively. Fringe spacing $w$ increases if you use a longer wavelength (red light gives wider fringes than blue), increase the screen distance $D$, or decrease the slit separation $s$. These relationships are directly testable in exam questions — and in the required practical.

The equation assumes the small-angle approximation: the angle from the central fringe to any bright fringe is small enough that $\sin\theta \approx \tan\theta \approx \theta$. This holds whenever $D$ is much larger than $s$, which is always the case in a standard Young's experiment setup.

5. Worked Calculations Using $w = \frac{\lambda D}{s}$

The equation looks straightforward, but exam questions routinely give wavelength in nanometres and slit separation in millimetres — then award a mark specifically for correct unit conversion. The worked examples below model the full method, including unit handling.

Worked example: Calculating Fringe Spacing

A double-slit experiment uses light of wavelength 590 nm. The slit separation is 0.40 mm and the screen is placed 1.5 m from the slits. Calculate the fringe spacing. [3 marks]

1. $w = \frac{\lambda D}{s}$ — [M1] Selecting the correct equation earns the method mark.
2. $\lambda = 590 \text{ nm} = 590 \times 10^{-9} \text{ m}$ $s = 0.40 \text{ mm} = 0.40 \times 10^{-3} \text{ m} = 4.0 \times 10^{-4} \text{ m}$ $D = 1.5 \text{ m}$ — Unit conversion is a common mark-losing step. Always convert nm to m (multiply by $10^{-9}$) and mm to m (multiply by $10^{-3}$) before substituting.
3. $w = \frac{590 \times 10^{-9} \times 1.5}{4.0 \times 10^{-4}}$ — [M1] Correct substitution with converted values.
4. $590 \times 10^{-9} \times 1.5 = 885 \times 10^{-9} = 8.85 \times 10^{-7} \text{ m}^2$ — $590 \times 1.5 = 885$, then carry the $\times 10^{-9}$ factor.
5. $w = \frac{8.85 \times 10^{-7}}{4.0 \times 10^{-4}} = \frac{8.85}{4.0} \times 10^{-7-(-4)} = 2.2125 \times 10^{-3} \text{ m}$ — $8.85 \div 4.0 = 2.21625$; exponent: $-7 - (-4) = -3$. Round to 2 significant figures to match the data given.
6. $w = 2.2 \times 10^{-3} \text{ m} = 2.2 \text{ mm}$ — [A1] Correct answer with units. Expressing in mm is acceptable provided the unit is stated.

Answer: $w = 2.2 \times 10^{-3}$ m (2.2 mm)

Now try rearranging the equation. Examiners frequently ask you to find the wavelength or slit separation from measured fringe data — this is exactly what happens in the required practical.

Worked example: Finding Wavelength from Fringe Data

In a Young's double-slit experiment, the fringe spacing is measured as 1.8 mm. The slit separation is 0.30 mm and the screen is 0.90 m from the slits. Calculate the wavelength of the light. [3 marks]

1. $w = \frac{\lambda D}{s}$ \Rightarrow $\lambda = \frac{ws}{D}$ — [M1] Rearranging correctly for $\lambda$.
2. $w = 1.8 \text{ mm} = 1.8 \times 10^{-3} \text{ m}$ $s = 0.30 \text{ mm} = 0.30 \times 10^{-3} \text{ m} = 3.0 \times 10^{-4} \text{ m}$ $D = 0.90 \text{ m}$ — Both $w$ and $s$ are given in mm — convert both before substituting.
3. $\lambda = \frac{1.8 \times 10^{-3} \times 3.0 \times 10^{-4}}{0.90}$ — [M1] Correct substitution.
4. $1.8 \times 10^{-3} \times 3.0 \times 10^{-4} = 5.4 \times 10^{-7} \text{ m}^2$ — $1.8 \times 3.0 = 5.4$; exponents: $-3 + (-4) = -7$.
5. $\lambda = \frac{5.4 \times 10^{-7}}{0.90} = 6.0 \times 10^{-7} \text{ m}$ — $5.4 \div 0.90 = 6.0$; exponent stays $-7$.
6. $\lambda = 6.0 \times 10^{-7} \text{ m} = 600 \text{ nm}$ — [A1] This wavelength corresponds to orange-red visible light, which is a sensible sanity check.

Answer: $\lambda = 6.0 \times 10^{-7}$ m (600 nm)

6. Required Practical — Measuring the Wavelength of Light

Young's double-slit experiment is one of the AQA required practicals. The aim is to measure fringe spacing and use $w = \frac{\lambda D}{s}$ to determine the wavelength of the light source. Understanding the method, variables, and sources of uncertainty is examinable.

Required Practical Link

AQA required practical 9 also allows the wavelength of light to be measured using a diffraction grating. This lesson focuses on the Young double-slit method because it develops the interference model directly, but students should recognise that the diffraction-grating method is an alternative practical route to the same measurement.

7. White Light Interference Pattern

So far the discussion has used monochromatic light — a single wavelength. White light contains a continuous range of wavelengths from approximately 400 nm (violet) to 700 nm (red). When white light passes through a double slit, each wavelength produces its own set of fringes with its own spacing, and these sets overlap on the screen.

At the centre of the pattern, the path difference is zero for every wavelength simultaneously. All colours interfere constructively at the same point, so the central fringe is white. This is a key distinguishing feature of the white-light pattern.

Interpreting the Pattern

Moving away from the centre, different wavelengths reach their first bright fringe at different positions because $w = \frac{\lambda D}{s}$ gives a larger spacing for longer wavelengths. Violet light (short $\lambda$) has its first bright fringe closest to the centre; red light (long $\lambda$) has its first bright fringe furthest from the centre. The result is a spectrum of colours — violet on the inside, red on the outside — on either side of the central white fringe. These are referred to as the coloured interference fringes of the white-light pattern.

Further from the centre, the coloured fringes from different orders begin to overlap and merge, so the pattern becomes less distinct. Only a few orders of coloured fringes are visible before the pattern washes out into a uniform white blur.

The chart above illustrates why the coloured fringes spread outwards: red light produces the widest fringe spacing and violet the narrowest, so they separate spatially as you move away from the central white fringe.

8. Exam Technique and Common Mistakes

Young's double-slit questions appear across all three AQA A Level Physics papers and can carry anywhere from 1 to 6 marks. The calculation questions are reliable mark-earners if you follow a consistent method; the descriptive and explanatory questions are where most marks are dropped.

For extended-response questions (4–6 marks) about the double-slit experiment, structure your answer as: state the condition (coherence) → explain what happens at the slits (two coherent secondary sources) → explain path difference at a general point → link path difference to constructive/destructive interference → state the result (bright/dark fringes). This chain of reasoning is what distinguishes a Level 3 response from a Level 1 response on the mark scheme.

Young's double-slit experiment sits at the heart of the AQA waves topic: it provides direct experimental evidence for the wave nature of light, links the abstract concept of superposition to a measurable, predictable pattern, and gives you a precise tool — $w = \frac{\lambda D}{s}$ — for determining wavelength in the laboratory. Mastering both the conceptual explanation and the calculation method covers all three specification requirements for this topic.