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Young's Double Slit Experiment

Key idea: Use YDSE to link wavelength, slit separation and fringe spacing, and solve x = λD/a under the small-angle approximation (A Level Physics).

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Explain coherent two-source interference using phase and path difference.
  • Analyse Young double-slit interference and its small-angle assumptions.

1. Definitions (Must Know)

Young double-slit geometry and fringe spacingA monochromatic source illuminates two close slits separated by a. Rays reach a distant screen a distance D away, where equally spaced bright and dark fringes form.monochromatic sourceS₁S₂slit separation aθscreen distance Dxbright-to-bright fringe spacing
Scroll diagram horizontally to read all labels.
A single source illuminates both slits so they are coherent. At small angles, the path difference is a sinθ ≈ ay/D; adjacent maxima are separated by x = λD/a.

A. Coherent sources

Two sources are coherent if they have the same frequency and a constant phase difference.

B. Fringe spacing, x (m)

The fringe spacing, x, is the distance between adjacent bright fringes (or adjacent dark fringes) on the screen.

C. Double-slit geometry symbols

  • slit separation: a (m)
  • distance from slits to screen: D (m)
  • wavelength: λ (m)
  • fringe order: n (integer, with the central maximum at n = 0)

D. YDSE formula (small-angle approximation)

For small angles and a distant screen (D ≫ a): x ≈ (λ D)/a

2. Key Ideas (What Earns Marks)

  • State the assumptions for x = λ D/a: small angle and D≫ a.
  • A path difference of nλ gives a bright fringe (for in-phase slits).
  • Use consistent units: nm → m, mm → m.
  • Distinguish double slit from diffraction grating:
    • double slit: fringe spacing x ≈ λ D/a
    • grating: use d sin θ = nλ for principal maxima
Link: interference conditions

See: Interference Pattern.

3. Detailed Explanations

A. Why the single slit is used

The single slit ensures both slits are illuminated by waves coming from (approximately) the same wavefront, so the two slits act as coherent sources.

B. Why fringes are equally spaced (small-angle)

At a point on the screen at small angle θ, the path difference between the two slits is approximately: Δ x ≈ a sin θ ≈ aθ

Bright fringes occur when Δ x = nλ: aθ ≈ nλ ⇒ θ ≈ nλ/a

For small angles, the distance on the screen is y ≈ Dθ, so: yₙ ≈ Dnλ/a

Fringe spacing is x = yₙ₊₁-yₙ: x ≈ (λ D)/a

C. What the pattern looks like (monochromatic vs white light)

Young double-slit geometry and fringe spacingA monochromatic source illuminates two close slits separated by a. Rays reach a distant screen a distance D away, where equally spaced bright and dark fringes form.monochromatic sourceS₁S₂slit separation aθscreen distance Dxbright-to-bright fringe spacing
Scroll diagram horizontally to read all labels.
Double-slit fringes are equally spaced; the central bright fringe is the zero-order maximum and a finite slit width provides an overall diffraction envelope.

With white light, the central fringe is white and the side fringes separate into colours because different wavelengths have different fringe spacings.

D. Effect of changing parameters

From x = λ D/a:

  • increase λ → fringes further apart
  • increase D → fringes further apart
  • increase a → fringes closer together

If the region between slits and screen is filled with a medium of refractive index n, wavelength decreases (λ = λ₀/n), so the fringes become narrower.

4. Common Mistakes

  • Using x = λ D/a for diffraction gratings (use the grating equation instead).
  • Forgetting the “small-angle / distant screen” condition.
  • Mixing up a (slit separation) with slit width.
  • Forgetting unit conversions (nm and mm).

5. Exam Tips

  • Write the formula first, then substitute with units, then final line.
  • If the question gives fringe spacing in mm, convert to metres before solving for λ.
  • Use n = 0 for the central maximum if you use yₙ positions.

6. Worked Examples

Modelled example 1

Find fringe spacing

Core

Problem

Monochromatic light of wavelength 520 nm illuminates a double slit with separation a = 0.20 mm. The screen is D = 2.0 m away. Find the fringe spacing x.
Study the worked solution

  1. Convert to consistent units

    Method

    Express wavelength and slit separation in metres.

    Reason

    The powers of ten in x = λ D/a must refer to a single unit system.

    Working

    λ = 520 × 10⁻⁹ m, a = 0.20 × 10⁻³ m

  2. Apply the small-angle relation

    Method

    Use x = λ D/a.

    Reason

    The screen is distant and the double-slit fringe-spacing model is stated for the small-angle geometry.

    Working

    x = ((520 × 10⁻⁹)(2.0))/(0.20 × 10⁻³) = 5.2 × 10⁻³ m

  3. Report a useful unit

    Method

    The fringe spacing is 5.2 mm.

    Reason

    Millimetres match the scale of separation between adjacent screen fringes.

    Working

    5.2 × 10⁻³ m = 5.2 mm

Guided practice 2

Find wavelength from measured spacing

About 4 min

Problem

In a YDSE setup, a = 0.30 mm and D = 1.5 m. The measured fringe spacing is x = 3.0 mm. Find λ in nanometres.

Try this before viewing the solution

Unit: nm

Hints

Hint 1: isolate wavelength
Rearrange to λ = ax/D before substituting.
View solution step by step

  1. Rearrange before substituting

    Method

    λ = ax/D.

    Reason

    Wavelength is the required quantity.

    Working

    λ = ax/D

  2. Calculate in metres

    Method

    λ = 6.0 × 10⁻⁷ m.

    Reason

    Both a and x are converted from millimetres to metres.

    Working

    λ = ((0.30 × 10⁻³)(3.0 × 10⁻³))/1.5 = 6.0 × 10⁻⁷ m

  3. Convert to nanometres

    Method

    λ = 600 nm.

    Reason

    One nanometre is 10⁻⁹ m.

    Working

    6.0 × 10⁻⁷ m = 600 × 10⁻⁹ m

Common misconception 3

Find slit separation

Find and correct the mistake

Learner claim

Monochromatic light of wavelength 650 nm produces fringe spacing x = 4.0 mm on a screen D = 1.8 m away. A learner substitutes 650 and 4.0 directly and says a = 293 m. Diagnose the method and find the slit separation.

Try this before viewing the solution

Unit: mm

View solution step by step

  1. Identify the unit error

    Method

    The bare values 650 and 4.0 cannot be combined with D in metres.

    Reason

    They represent nanometres and millimetres, so the omitted powers of ten change the result by orders of magnitude.

    Working

    650 nm = 650 × 10⁻⁹ m; 4.0 mm = 4.0 × 10⁻³ m

  2. Use the slit-separation relation

    Method

    a = 2.93 × 10⁻⁴ m.

    Reason

    Rearranging the double-slit formula gives a = λ D/x.

    Working

    a = ((650 × 10⁻⁹)(1.8))/(4.0 × 10⁻³) = 2.93 × 10⁻⁴ m

  3. Convert and interpret

    Method

    The slit separation is 0.293 mm.

    Reason

    This is a plausible sub-millimetre separation; a is centre-to-centre slit separation, not slit width.

    Working

    2.93 × 10⁻⁴ m = 0.293 mm

Examiner practice 4

Find screen distance

3 marks

Examination question

Light of wavelength 500 nm passes through a double slit with separation a = 0.25 mm. The fringe spacing is x = 5.0 mm. Find the slit-to-screen distance D. [3 marks]

Try this before viewing the solution

Unit: m

View solution step by step

  1. Rearrange the relation

    1 mark

    Method

    D = ax/λ.

    Reason

    Screen distance is the required unknown.

    Working

    D = ax/λ

  2. Convert and substitute

    1 mark

    Method

    Use a = 0.25 × 10⁻³ m, x = 5.0 × 10⁻³ m and λ = 500 × 10⁻⁹ m.

    Reason

    Consistent SI units prevent a scale error.

    Working

    D = ((0.25 × 10⁻³)(5.0 × 10⁻³))/(500 × 10⁻⁹)

  3. State the result

    1 mark

    Method

    D = 2.5 m.

    Reason

    The value is much larger than the sub-millimetre slit separation, consistent with the distant-screen approximation.

    Working

    D = 2.5 m

Challenge 5

Position of a given order

Minimal support

Independent transfer

In a YDSE, the fringe spacing is x = 2.4 mm. How far from the central maximum is the fourth bright fringe?

Try this before viewing the solution

Unit: mm

Hints

Hint 1: count from zero order
Use yₙ ≈ nx with n = 4; do not count the central maximum as the first-order fringe.
View solution step by step

  1. Connect order to position

    Method

    The fourth bright fringe is four spacings from the centre.

    Reason

    Bright-fringe positions obey yₙ ≈ nx, with the central maximum at n = 0.

    Working

    y₄ ≈ 4x

  2. Calculate

    Method

    The distance from the centre is 9.6 mm.

    Reason

    Multiply the adjacent-fringe spacing by the order.

    Working

    y₄ ≈ 4(2.4 mm) = 9.6 mm

7. Mind Stretchers

Mind stretcher 1: Medium insertedExtension

The YDSE is carried out in air, then the space is filled with glass of refractive index n = 1.5.

How does the fringe spacing change?

Show Answer

In a medium, λ = λ₀/n. Since x = λ D/a, the fringe spacing becomes: x_glass = (1/n)xₐᵢᵣ = (1/1.5)xₐᵢᵣ So fringes get narrower by a factor of 1.5.

Mind stretcher 2: Checking the “small-angle” assumptionExtension

Using the values from Worked Example A, estimate the angle to the fourth bright fringe and comment on whether the small-angle approximation is reasonable.

Show Answer

From Worked Example A, x = 5.2 mm and D = 2.0 m, so: y₄ ≈ 4x = 20.8 mm = 2.08 × 10⁻² m θ ≈ y₄/D = (2.08 × 10⁻²)/2.0 = 1.04 × 10⁻² rad ≈ 0.60° This is a small angle, so using sin θ ≈ tan θ ≈ θ is reasonable here.

Mind stretcher 3: Optional (Enrichment)Extension

A. Diffraction envelope (why fringes fade)

Real slits have finite width, so each slit also diffracts. The double-slit fringes sit inside a single-slit diffraction envelope, which is why outer fringes are dimmer.

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027