9478 / 2027
Superposition overview

Lesson 6 of 9

Measure double-slit fringes

  • I can use lambda = ax/D for double-slit interference, distinguishing slit and fringe separations.Syllabus

    Syllabus 9478, 11(h). Use lambda = ax/D for double-slit interference, distinguishing slit and fringe separations.

Make a guess. You are not marked.

In a double-slit experiment with red light, the two slits are replaced by a pair that are closer together. Nothing else changes. What happens to the bright fringes on the screen?

  1. They move closer together, because the slits themselves are closer together.
  2. They move further apart, because fringe spacing is inversely proportional to slit separation.
  3. They stay the same, because the wavelength is unchanged.
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They move further apart, because fringe spacing is inversely proportional to slit separation.

Fringe spacing x = lambda D / a. A smaller slit separation a gives a larger spacing, so the fringes spread out across the screen.

A double-slit pattern turns the difference between two light paths into a measurable fringe spacing. Keep slit separation, fringe separation and screen position distinct.

One coherent source illuminates two slits. Their outputs overlap on a screen and form stable bright and dark bands when the relative phase remains fixed. Here the slit outputs are in phase, and the relevant polarisation components are compatible.

Label the geometry

  • a: centre-to-centre slit separation.
  • D: perpendicular distance from the slit plane to the screen.
  • x: separation of neighbouring bright centres.
  • y: position on the screen measured from the central maximum.

For a distant screen and small observation angle θ, the path difference is approximately a sin θ, with sin θ ≈ tan θ ≈ y/D. Thus:

Path difference ≈ ay/D
Bright centres: ay/D ≈ nλ
Fringe separation x = λD/a
λ = ax/D

The approximation requires D large compared with the slit separation and small y/D in the region used. It does not make sine, tangent and angle identical at arbitrary angles.

Worked fringe spacing

Use wavelength 600 nm, slit separation 0.30 mm and screen distance 2.0 m.

x = (600 × 10-9 × 2.0)/(0.30 × 10-3)
= 0.0040 m = 4.0 mm

Separate apparatus distances from the enlarged fringe scale

One coherent source, two slits, one screen

Slit spacing, screen distance and fringe position are distinct lengthsOne coherent source illuminates two slits in a barrier. Their centre separation is a, 0.30 millimetres. The perpendicular slit-plane-to-screen distance D is 2.0 metres. A central axis reaches screen coordinate zero; an observation point P is at a positive screen coordinate y above it. Both slit outputs reach P. The sketch explicitly uses different schematic scales for millimetre slit separation and metre screen distance, so its ray angles are not numerical angle measurements. Fringe separation x is measured on the separately enlarged screen view, not between these two slits.ayD = 2.0 mP0SlitsScreenOnesourceSchematic distances: a = 0.30 mm

Enlarged calibrated screen region

Nine bright centres contain eight four-millimetre intervalsA separately enlarged screen strip uses seven and a half drawing units per millimetre. Nine bright centres are at screen coordinates negative sixteen, negative twelve, negative eight, negative four, zero, positive four, positive eight, positive twelve and positive sixteen millimetres. Their span is 32 millimetres and contains eight equal four-millimetre intervals. The central-region model neglects the single-slit envelope and uses equal peak brightness. Grey-green brightness follows a normalised cosine-squared intensity profile, not displacement. Screen coordinate positive ten millimetres lies at the dark centre between the positive eight and twelve bright centres. One four-millimetre interval is separately bracketed.-16-80816Equal-height central-region fringesy = 10 mm: dark32 mm: eight equal intervalsOne interval: x = 4.0 mmScreen coordinate y / mm

The centre-to-centre span is divided by eight intervals, not nine bright centres. Fringe width and slit separation are measured in different places.

The apparatus is schematic in scale. The separate enlarged screen view has a calibrated position axis and nine consecutive bright centres. The illustrated central-region pattern neglects variation from the single-slit envelope.

Nine consecutive bright centres from -16 mm to +16 mm span 32 mm. They contain eight intervals:

x = 32/8 = 4.0 mm
λ = (0.30 × 10-3)(4.0 × 10-3)/2.0
= 6.0 × 10-7 m = 600 nm

Try it step by step

Work it out in steps: Slit separation a in m, then Fringe spacing x = lambda D / a in mm.

  1. Problem 1. Light of wavelength 600 nm falls on two slits 0.50 mm apart. The screen is 2.0 m away. Find the fringe spacing.
  2. Problem 2. Light of wavelength 450 nm falls on two slits 0.25 mm apart. The screen is 1.5 m away. Find the fringe spacing.
  3. Problem 3. Light of wavelength 650 nm falls on two slits 0.40 mm apart. The screen is 3.0 m away. Find the fringe spacing.
Show the answer

Problem 1.

Slit separation a in m: a = 0.50 mm = 5.0 x 10-4 m

Fringe spacing x = lambda D / a in mm: x = 6.0 x 10-7 x 2.0 / (5.0 x 10-4) = 2.4 x 10-3 m = 2.4 mm

Problem 2.

Slit separation a in m: a = 0.25 mm = 2.5 x 10-4 m

Fringe spacing x = lambda D / a in mm: x = 4.5 x 10-7 x 1.5 / (2.5 x 10-4) = 2.7 x 10-3 m = 2.7 mm

Problem 3.

Slit separation a in m: a = 0.40 mm = 4.0 x 10-4 m

Fringe spacing x = lambda D / a in mm: x = 6.5 x 10-7 x 3.0 / (4.0 x 10-4) = 4.9 x 10-3 m = 4.9 mm (2 sf)

Check your understanding

Nine consecutive bright centres span 32 mm in a double-slit pattern. Slit separation is 0.30 mm and perpendicular screen distance is 2.0 m. Under the small-angle approximation, what is the wavelength?

  1. 600 nm.
  2. About 533 nm.
  3. 4800 nm.
Show the answer

600 nm.

The eight intervals give x = 4.0 x 10-3 m. Then lambda = ax/D = (0.30 x 10-3)(4.0 x 10-3)/2.0 = 6.0 x 10-7 m.

Classify a screen position

At y = 10 mm = 0.010 m in the same model:

Path difference ≈ (0.30 × 10-3)(0.010)/2.0
= 1.50 × 10-6 m = 2.5λ

This is a minimum for the stated in-phase sources. The angle is approximately y/D = 0.005 rad, consistent with the small-angle approximation. The result uses the distant-screen model, not an exact finite-distance construction.

Make a wavelength measurement

Record the known slit separation with its units. Measure perpendicular D, keep the screen normal to the central axis and identify the central maximum. Locate several corresponding bright centres, measure their total span and divide by the number of intervals.

Use a band's centre rather than an arbitrary bright edge. A supplied image needs a calibrated spatial scale. Background light and broad fringes limit centre-location precision; a larger image alone does not improve the original measurement resolution. Repeat position readings while retaining the original values and their units.

Practise an exam question

Write your answer first, then tick the mark points it makes.

Exam-style question

Light of wavelength 633 nm passes through two narrow slits 0.50 mm apart. Fringes are seen on a screen 2.0 m away. Calculate the fringe separation. [3]

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