C.3.9—Young double-slit equation

Relate fringe spacing to the apparatus

For Young’s double-slit interference, fringe separation is $s=\lambda D/d$ , where $\lambda$ is wavelength, $D$ is slit-to-screen distance and $d$ is slit separation.

Rearrange before substituting

Use $\lambda=sd/D$ , $d=\lambda D/s$ , or $D=sd/\lambda$ as needed. Measure the separation between adjacent bright or dark fringe centres; if several fringes are measured, divide the total width by the number of intervals.

Check the trends

Fringes spread farther apart when wavelength or screen distance increases, and become closer when slit separation increases. The small-angle model assumes $D\gg d$ and approximately plane wavefronts normal to the slits.

Common trap

Do not use the total width across several fringes as s without dividing by the number of fringe spacings, and do not confuse slit separation d with screen distance D.

How it is assessed

Questions ask you to calculate wavelength from a measured pattern or identify which colour gives the largest fringe separation. The evidence rewards λ = sd/D and the correct wavelength trend.

Command terms

Calculate / What is

What earns marks

Use s = λD/d, rearrange to the requested variable, convert units, and divide a measured multi-fringe width by its number of intervals. Check that longer wavelength gives larger fringe separation.

Watch for

Using total pattern width as one fringe spacing or reversing d and D in λ = sd/D.

Representative question

[Maximum number: 3]

Calculate, in $\mathrm{nm}, \lambda$ .

$s=0.15/8=0.0188\ \mathrm{m}$ .
Use $\lambda=ds/D$ .
$\lambda=450\ \mathrm{nm}$ .

s=\frac{0.15}{8}=0.0188\ \mathrm{m},\qquad \lambda=\frac{ds}{D}=450\ \mathrm{nm}

C.3.9—Young double-slit equation

Use Young’s Double-Slit Equation

C.3.9 Exam Analysis

Assessment in practice

Question 1