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3.3.7Tools and Inquiry — IB Physics skills

Young’s Double-Slit Experiment — IB Physics SL Revision Notes & Topical Questions

5 min read 5 parts Lesson 29 of 33 Unit quiz

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Part 1 of 5How the Pattern Forms

3.3.7 Young's Double-Slit Experiment

Light passing through two narrow slits spreads out by diffraction, and the overlapping waves interfere.

Coherent sources
  • The two slits act as sources with a constant phase difference, because they are lit by the same wavefront.
Bright fringes
  • Constructive interference occurs where the path difference is a whole number of wavelengths, nλ.
Dark fringes
  • Destructive interference occurs where the path difference is a half-integer number of wavelengths, (n + ½)λ.
Pattern
  • Alternating bright and dark fringes of equal width appear on a screen placed far from the slits.

Part 2 of 5Fringe Separation

For slits close together and a distant screen, simple geometry gives the fringe spacing.

1
Let d be the slit separation, D the distance from slits to screen, and s the distance between adjacent bright fringes.
2
For small angles the path difference to a point on the screen is approximately d multiplied by the angle to that point.
3
A bright fringe needs a path difference λ for the first order, giving an angle of about λ/d.
4
Multiplying the angle by D gives the fringe separation s = λD/d.
5
Using this, a longer wavelength, a larger D or a smaller d all make the fringes more widely spaced.

Part 3 of 5Significance and Practice

Significance and Practice
Wave evidence
  • The pattern provides strong evidence for the wave nature of light, since particles alone would not give fringes.
White light
  • Each colour gives fringes of a different spacing, so the pattern is coloured with a white central fringe.
Laser source
  • A laser gives a bright, coherent beam of one wavelength, producing clear fringes.
Measurement
  • Measuring across many fringes and dividing reduces the uncertainty in s.
ChangeEffect on fringe separation s
Increase λs increases
Increase Ds increases
Increase ds decreases

Part 4 of 5Worked Example

Question

  • Light of wavelength 600 nm passes through two slits 0.50 mm apart.
  • The screen is 2.0 m away.
  • Calculate the fringe separation.

  1. Convert unitsλ = 600 × 10⁻⁹ m, d = 0.50 × 10⁻³ m.
  2. s = λD/d = (600 × 10⁻⁹ × 2.0) / (0.50 × 10⁻³).
  3. s = 1.2 × 10⁻⁶ / 5.0 × 10⁻⁴ = 2.4 × 10⁻³ m.

Answer The fringe separation is 2.4 mm.

Part 5 of 5Exam Checklist & Tips

3.3.7 – Key Points for the Exam

  • Coherent Having a constant phase difference between the waves.
  • Path difference The difference in distance travelled by two waves reaching the same point.
  • Constructive interference Superposition giving a larger amplitude, when waves arrive in phase.
  • Fringe separation The distance between neighbouring bright or dark fringes.

Common Mistakes

  • Using the distance between the screen and a single slit incorrectly in place of D.
  • Forgetting to convert nanometres and millimetres to metres.
  • Saying the slits are sources of different frequency; they must be coherent.
  • Describing dark fringes as places where light is absorbed, when waves cancel by interference.

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