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Young's double-slit experiment

Key idea: Two coherent slits produce bright and dark fringes as their waves reinforce or cancel. Near the centre, fringe spacing is λD/a, so measuring it gives the wavelength.

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  • GCE A-Level H2 Physics 2027
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Light passing through two narrow slits produces a row of bright and dark fringes on a screen. Their spacing lets you measure the wavelength, while their contrast depends on how well the two waves cancel. This lesson shows where the fringes form, derives the fringe spacing x = λ D/a, and uses it to find a wavelength from measurements on a screen.

Where the bright fringes form

Light from one source passes through two narrow slits a distance a apart, centre to centre. Because both slits are lit by the same wave, they act as two coherent sources in phase. Diffraction spreads the light from each slit, the two waves overlap, and they interfere on a screen a distance D away.

Young double-slit geometryIn-phase illumination reaches two slit centres separated by a. Both slits send waves to the same point P on a screen. The angle theta is measured from the central axis to the line joining the slit midpoint to P. Screen distance D is measured along the axis; y is the distance from the central point O to P on the screen.Slits to a screenSame-phase illuminationS₁S₂aθPOyD
The slits are illuminated in phase. At a distant screen, Δr ≈ a sinθ. For small angles, y ≈ Dθ and adjacent bright fringes have spacing x ≈ λD/a. The angle and slit separation are exaggerated in this schematic.
Measure spacing between adjacent bright fringesFive bright fringes have equally spaced centres. Four dark fringes lie halfway between them. The middle bright fringe is zero order. The marked distance x joins the centres of two neighbouring bright fringes, rather than the edges of one fringe.Fringes on the screenxbrightdarkn = 0Measure between adjacent centres.
In the ideal narrow-slit, small-angle model, bright-fringe centres are equally spaced. Here x is a centre-to-centre distance; it is not the width of a slit or of a bright stripe.

Take a point P on the screen at an angle θ from the central axis. The screen is far away compared with a, so the two paths to P are almost parallel. The extra distance travelled by the wave from the further slit is then

Δ r = a sin θ

A bright fringe forms where this path difference is a whole number of wavelengths, a sin θ = nλ. At the centre of the screen the paths are equal, so the central fringe is bright.

Fringe spacing

Near the centre of the screen, the angles are small, so sin θ ≈ tan θ = y/D, where y is the distance from the central fringe. The nth bright fringe is therefore at

yₙ = (nλ D)/a

Each bright fringe needs exactly one more wavelength of path difference than the one before, and near the centre the path difference grows in proportion to y. So the bright fringes are evenly spaced, and the fringe separation x between neighbouring bright fringes is

x = (λ D)/a or λ = ax/D

The dark fringes lie halfway between, with the same spacing. The relation holds when D is much larger than a and the fringes are near the centre.

The relation explains why the experiment needs slits less than a millimetre apart. The wavelength of light is less than a millionth of a metre, so with a any larger, x would be too small to see.

Check your understanding 1

Using the same light, give two changes that would each double the fringe separation.

Show answer

Double the distance D to the screen, or halve the slit separation a. Both double λ D/a.

Worked example 1

Wavelength from the fringe separation

Problem

Slits 0.30 mm apart produce bright fringes 4.0 mm apart on a screen 2.0 m away. Find the wavelength of the light.

Show full solution

  1. Identify and convert the lengths

    Method

    Tell the slit separation a from the fringe separation x, and convert both to metres.

    Reason

    Both are small lengths in millimetres, but a is measured at the slits and x on the screen.

    Working

    Answer

    a = 3.0 × 10⁻⁴ m, x = 4.0 × 10⁻³ m, D = 2.0 m

  2. Calculate the wavelength

    Working

    Answer

    λ = ax/D = ((3.0 × 10⁻⁴)(4.0 × 10⁻³))/2.0 = 6.0 × 10⁻⁷ m = 600 nm

Try it yourself 2

Measure across several fringes

Minimal support

Problem

On a screen 1.80 m from slits 0.250 mm apart, the distance from the centre of the first bright fringe to the centre of the ninth is 27.0 mm. Find the wavelength. The screen is then moved to 1.20 m from the slits. What is the distance across nine bright fringes now?

Find the wavelength and the new distance

Show the steps Optional working
Unit: mm
Unit: nm
Show the steps Optional working
Unit: mm
Unit: nm
Unit: mm

Hints

Hint 1

count the gaps

How many spacings lie between the first and ninth fringes?

Hint 2

change the distance

Which quantities change when only the screen moves?

Show solution step by step

  1. Find one spacing

    Method

    Divide the distance by eight.

    Reason

    Nine fringe centres enclose eight spacings.

    Working

    Answer

    x = 27.0/8 = 3.375 mm

  2. Find the wavelength

    Reason

    Use λ = ax/D with every length in metres.

    Working

    Answer

    λ = ((2.50 × 10⁻⁴)(3.375 × 10⁻³))/1.80 = 4.69 × 10⁻⁷ m = 469 nm

  3. Scale to the new distance

    Working

    Answer

    x ∝ D, so the new distance across eight spacings is 27.0 × 1.20/1.80 = 18.0 mm.

What the pattern looks like

With light of one wavelength, the screen shows evenly spaced bright and dark fringes of one colour. With white light, every wavelength has zero path difference at the centre, so the central fringe is white. The other fringes are coloured, because each colour has its own spacing: red fringes, with the longest wavelength, are furthest apart, and the blue edge of each fringe is nearest the centre. Further out, the colours from different fringes overlap and the pattern fades to white.

The fringes also grow fainter away from the centre, because each slit has a finite width and diffracts less light at larger angles.

The two slits must send similar amounts of light. If one slit is narrower, its wave has a smaller amplitude, so the waves no longer cancel completely: the dark fringes become less dark. The fringe separation does not change, because λ, D and a have not changed.

Exam-style question 1

Explain the effect of narrowing one slit

6 marks

Examination question

(a) Explain how bright and dark fringes are formed in a double-slit experiment. [4 marks]

(b) One slit is made narrower while its centre stays in the same place. State and explain the effect on the fringe separation and on the appearance of the dark fringes. [2 marks]

Write your answer before viewing the mark scheme

Show solution step by step

  1. Explain the fringes

    4 marks

    Method

    Link coherent sources to path difference.

    Reason

    Each mark needs a link in the chain from the slits to the screen.

    Working

    Answer

    Both slits are lit by the same source, so they are coherent sources in phase. Light diffracts at each slit, so the two waves overlap and superpose. Where the path difference is a whole number of wavelengths, the waves arrive in phase and interfere constructively: a bright fringe. Where it is a whole number plus half a wavelength, they arrive in antiphase and interfere destructively: a dark fringe.

  2. Narrow one slit

    2 marks

    Working

    Answer

    The fringe separation x = λ D/a is unchanged, because λ, D and a are unchanged. The narrower slit sends a wave of smaller amplitude, so the waves no longer cancel completely and the dark fringes become less dark.

Common mistakes

  • Mixing up the slit separation a with the fringe separation x.
  • Leaving a or x in millimetres while D is in metres.
  • Dividing a distance across N fringes by N instead of N-1.
  • Using x = λ D/a far from the centre, where the angles are no longer small.

Before you move on

Check your understanding 2

In a double-slit experiment, the distance from the first to the sixth bright-fringe centre is 14.0 mm. The slits are 0.400 mm apart and the screen is 2.00 m away. Find the wavelength. A learner divides 14.0 mm by six; how does that affect the calculated wavelength?

Show the steps Optional working
Unit: m
Unit: nm
Show answer

Six fringe centres enclose five spacings, so x = 14.0/5 = 2.80 mm. Hence λ = ax/D = (0.400 × 10⁻³)(2.80 × 10⁻³)/2.00 = 5.60 × 10⁻⁷ m = 560 nm. Dividing by six makes x, and therefore the wavelength, too small by a factor of 5/6, giving about 467 nm.

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