Young's double-slit experiment

CIE A-Level Physics· 9702 Unit 8, Topic 4· 10 min read

1. Experimental Setup and Core Requirements★★☆☆☆⏱ 3 min

Young's double-slit experiment provided definitive evidence that light behaves as a wave, by producing a stable interference pattern that cannot be explained by particle theory.

📘 Definition

Coherent Sources

Two wave sources that maintain a constant phase difference and have the same frequency. Coherence is required to produce a visible, stable interference pattern.

Example:

A laser passed through a single slit produces coherent light incident on the double slits.

Monochromatic light source (produces light of a single wavelength)
Narrow single slit to create a coherent point source
Two identical, parallel, narrow double slits
Screen placed a large fixed distance from the double slits

Exam tip:

Marking schemes award a separate mark for mentioning the single slit requirement.

2. Fringe Spacing Formula and Derivation★★★☆☆⏱ 5 min

🔬 Derivation

Goal:

Derive the fringe spacing formula for Young's double-slit experiment

Starting from:

Constructive interference occurs when path difference = $nλ$ , for integer $n$ . For small angles, $sin θ \approx tan θ$ .

1
Let $a$ = slit separation, $D$ = distance to screen, $Δ x$ = spacing between adjacent bright fringes. For adjacent fringes, path difference increases by $λ$ .
2
From the path difference condition for constructive interference:
3
$a \sin\theta = \lambda$
4
For small angles, $sin θ \approx tan θ = \frac{Δ x}{D}$
5
Substitute and rearrange:
6
$a \frac{\Delta x}{D} = \lambda \implies \Delta x = \frac{\lambda D}{a}$

Result:

The fringe spacing formula is $Δ x = \frac{λ D}{a}$ , where all quantities are measured in the same length unit.

📘 Definition

Fringe Spacing

\Delta x

The distance between two consecutive bright fringes (or two consecutive dark fringes) in the interference pattern.

📐 Worked Example

A Young's double-slit experiment has slit separation $0.5$ mm, screen distance $2.5$ m, and light of wavelength $600$ nm. Calculate the fringe spacing in mm.

1
Convert all quantities to SI units (metres):
2
$a = 0.5 \times 10^{-3} \text{ m}, \quad D = 2.5 \text{ m}, \quad \lambda = 600 \times 10^{-9} \text{ m}$
3
Substitute into the fringe spacing formula:
4
$\Delta x = \frac{(600 \times 10^{-9})(2.5)}{0.5 \times 10^{-3}} = 3.0 \times 10^{-3} \text{ m}$
5
Convert to millimetres as requested: $3.0 \times 1 0^{- 3}$ m = $3.0$ mm.

Exam tip:

Unit conversion is the most commonly tested error here. Always check all units match before calculating.

3. Effects of Changing Experimental Parameters★★★☆☆⏱ 4 min

Exam questions very commonly ask you to predict how the interference pattern changes when one experimental parameter is altered. All changes follow directly from the formula $Δ x = \frac{λ D}{a}$ .

Changed Parameter	Effect on $Δ x$	Effect on Pattern
Increase slit separation $a$	$Δ x$ decreases	Fringes move closer together
Decrease screen distance $D$	$Δ x$ decreases	Fringes move closer together
Increase wavelength $λ$	$Δ x$ increases	Fringes move further apart
White light instead of monochromatic	$Δ x$ varies with $λ$	Central white fringe, coloured fringes either side

📐 Worked Example

Original fringe spacing is $2.0$ mm. If slit separation is doubled and screen distance is halved, what is the new fringe spacing?

1
Original: $Δ x_{1} = \frac{λ D _{1}}{a _{1}} = 2.0$ mm
2
New parameters: $a_{2} = 2 a_{1}$ , $D_{2} = \frac{D _{1}}{2}$
3
$\Delta x_2 = \frac{\lambda D_2}{a_2} = \frac{\lambda (D_1/2)}{2a_1} = \frac{1}{4} \Delta x_1$
4
New fringe spacing = $\frac{2.0}{4} = 0.5$ mm

4. Key Pattern Features★★☆☆☆⏱ 3 min

Double-slit interference patterns have consistent features you need to recall for exams:

The central fringe is always bright (zero path difference = constructive interference)
All bright fringes are approximately equally spaced for small angles
Bright fringe intensity gradually decreases as you move away from the central fringe
Dark fringes have near-zero intensity from complete destructive interference

✓ Quick check

Check your understanding:

Which change increases fringe spacing in Young's experiment?
- Increase slit separation $a$
- Decrease screen distance $D$
- Increase wavelength $λ$
- Use white light instead of monochromatic
Reveal answer
2 —
Correct. From $Δ x = \frac{λ D}{a}$ , only increasing wavelength increases fringe spacing. All other options decrease or do not uniformly increase spacing.

5. Common Pitfalls

Wrong move:

Forgetting to convert all quantities to the same unit before calculation

Why:

Wavelength is usually given in nanometres and slit separation in millimetres, so mixing units gives an incorrect order of magnitude

Correct move:

Convert all lengths to metres (or all to millimetres) before substituting into the formula

Wrong move:

Dividing the total distance across $n$ fringes by $n$ to get $Δ x$

Why:

There are $n - 1$ gaps between $n$ fringes, so dividing by $n$ underestimates fringe spacing

Correct move:

Divide the total distance across $n$ fringes by $(n - 1)$ to get the fringe spacing

Wrong move:

Claiming the central fringe is dark in a double-slit pattern

Why:

Zero path difference at the centre gives constructive interference, so it must be bright

Correct move:

Always state that the central fringe is bright for double-slit interference

Wrong move:

Omitting the single slit when describing the setup for a laser source

Why:

Even lasers require the single slit to expand the beam and uniformly illuminate the double slits

Correct move:

Always include the single slit in your description, regardless of the light source

Wrong move:

Confusing fringe spacing with the distance from the centre to the $n$ th fringe

Why:

$Δ x$ is the gap between adjacent fringes, not the position of a single fringe

Correct move:

Remember that the position of the $n$ th fringe is $x_{n} = n Δ x$ , so $Δ x = \frac{x _{n}}{n}$

6. Quick Reference Cheatsheet

Quantity	Symbol	Key Rule
Fringe spacing	$Δ x$	$Δ x = \frac{λ D}{a}$
Slit separation	$a$	Distance between two double slits
Screen distance	$D$	Distance from slits to screen
Increase $a$		$Δ x$ decreases, fringes closer
Increase $D$		$Δ x$ increases, fringes further apart
Increase $λ$		$Δ x$ increases, fringes further apart
Central fringe		Always bright for double-slit

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.

2022 · 2
Calculate fringe spacing
2023 · 1
Effect of changing slit separation
2021 · 2
Describe full experimental setup

What's Next

Young's double-slit experiment is the foundation of wave interference and superposition topics in CIE A-level Physics. The fringe spacing formula and coherence requirements you learned here directly apply to diffraction gratings, a common follow-on topic that is heavily tested in both multiple choice and written papers. These principles also underpin experimental techniques for measuring the wavelength of light, and explain other superposition effects like thin film interference. Understanding double-slit interference also helps you distinguish between interference and diffraction effects, which is a common exam theme.