The fringe spacing equation: $w = lambda D / s$

Interference & Diffraction — AQA A-Level Physics

w = \frac{\lambda D}{s}

$w$: fringe spacing (m)
$\lambda$: wavelengthThe minimum distance between two points on a wave that are in phase (e.g. crest to crest). Measured in metres (m). of light (m)
$D$: distance from slits to screen (m)
$s$: slit separation (m)

D is typically several metres (the largest dimension).
s is the separation between the two slits (typically in mm -- the smallest dimension).
w is the distance between adjacent bright fringes on the screen (typically in mm or cm).
Fringe spacing increases if: wavelengthThe minimum distance between two points on a wave that are in phase (e.g. crest to crest). Measured in metres (m). increases, slit-to-screen distance D increases, or slit separation s decreases.

Worked Example

A 750 THz laser produces fringes with a separation between P and Q (9 fringes apart) of 15 mm on a screen 4.5 m away. Calculate the slit separation.

Show Solution

Calculate the wavelength

$$\lambda = \frac{c}{f} = \frac{3 \times 10^{8}}{750 \times 10^{12}} = 4 \times 10^{-7} \text{ m} = 400 \text{ nm}$$

Calculate the fringe spacing

P and Q are separated by 9 fringes: $w = \frac{15 \times 10^{-3}}{9} = 1.67 \times 10^{-3}$ m

Rearrange and substitute

$$s = \frac{\lambda D}{w} = \frac{4 \times 10^{-7} \times 4.5}{1.67 \times 10^{-3}} = 1.1 \times 10^{-3} \text{ m} = 1.1 \text{ mm}$$

Answer

$s = 1.1$ mm

Common Mistake MEDIUM

Students often: Mixing up w, s and D because they are all distances.
Instead: Label the diagram: D = slit-to-screen (largest), s = slit separation (smallest), $w = fringe spacing (on$ the screen). Always check units.