Theory — Superposition and Interference
Superposition of Waves
When two waves pass through the same point, the total displacement is the sum of the individual displacements. Where two crests meet, they add to make a bigger crest (constructive interference); where a crest meets a trough, they cancel (destructive interference). For a steady pattern the two sources must be coherent — same frequency and a fixed phase relationship.
Path Difference
Whether a point lies on a bright (constructive) or dark (destructive) region depends on the difference in distance from the two sources — the path difference Δ = |r₂ − r₁|. If the two sources start in phase, the waves arrive in step wherever the path difference is a whole number of wavelengths, and exactly out of step wherever it is an odd number of half-wavelengths.
Destructive (dark): Δ = (m + ½)·λ (m = 0, 1, 2, …)
Nodal and Antinodal Lines
Joining all the points with the same path difference traces out curves (hyperbolas). The antinodal lines (Δ = mλ) are where the waves always reinforce — these are the bright bands. The nodal lines (Δ = (m+½)λ) are where the waves always cancel — these stay calm and dark. Along the perpendicular bisector of the two sources the path difference is zero, so the central line is always a strong constructive (m = 0) band.
The Double-Slit Experiment
Passing one wave through two narrow slits a distance d apart turns the slits into two coherent sources. On a screen a distance L away, the path difference to a point a height y above the centre is approximately d·sinθ ≈ d·y/L. Setting this equal to the interference conditions gives evenly spaced bright fringes.
Fringe spacing: Δy = λL / d
Constructive
Waves arrive in phase. Crest meets crest. Path difference = mλ. Amplitudes add → bright band / loud sound / antinode.
Destructive
Waves arrive out of phase. Crest meets trough. Path difference = (m+½)λ. Amplitudes cancel → dark band / quiet spot / node.
| Change | Effect on fringe spacing Δy | Why |
|---|---|---|
| Longer wavelength λ | Δy increases | Δy ∝ λ |
| Larger slit separation d | Δy decreases | Δy ∝ 1/d |
| Greater screen distance L | Δy increases | Δy ∝ L |
Instructions — Running the Virtual Experiment
The Ripple Tank tab builds physical intuition; the Double-Slit Measurements tab provides the quantitative verification. Record every reading in your lab notebook.
Simulation — Two-Source Interference & the Double Slit
Controls
Controls
Measurement
| λ (nm) | d (mm) | L (m) | Δy measured (mm) | Δy·d/L recovered λ (nm) |
|---|---|---|---|---|
| No readings yet — choose a wavelength and click "Record reading". | ||||
Team Questions
Example Lab Report
Sample report demonstrating the expected format and level of detail. Use as a guide for your own submission.
Wave Interference: Two-Source Patterns and the Double-Slit Relation
Physics | Section: [Your Section] | Date: [Date]
Lab Members: [Names of all members present]
Purpose
To observe the interference pattern produced by two coherent sources, to relate the bright and dark regions to the path difference between the sources, and to verify the double-slit fringe relation Δy = λL/d by measuring the fringe spacing for a range of wavelengths.
Theory
Two coherent sources interfere constructively where the path difference Δ is a whole number of wavelengths and destructively where it is an odd number of half-wavelengths. For a double slit of separation d viewed on a screen a distance L away (with L ≫ d), the bright fringes occur at y_m = mλL/d, so adjacent fringes are evenly spaced by Δy.
Δy = λL / d → λ = Δy·d / L
Rearranging the fringe equation gives λ = Δy·d/L, so multiplying the measured fringe spacing by d/L should recover the wavelength used — an internal check on every measurement.
Calculations — Sample: λ = 550 nm, d = 0.20 mm, L = 2.0 m
Fringe spacing: Δy = λL/d = (550 × 10⁻⁹)(2.0)/(0.20 × 10⁻³) = 1.10 × 10⁻⁶ / 2.0 × 10⁻⁴ = 5.5 × 10⁻³ m = 5.5 mm
Recovered wavelength: λ = Δy·d/L = (5.5 × 10⁻³)(0.20 × 10⁻³)/(2.0) = 5.5 × 10⁻⁷ m = 550 nm ✓
Third-order fringe (m = 3): y₃ = 3·Δy = 3(5.5) = 16.5 mm from centre
Results Table — Fringe Spacing vs. Wavelength (d = 0.20 mm, L = 2.0 m)
| λ set (nm) | Δy measured (mm) | Δy·d/L recovered λ (nm) |
|---|---|---|
| 450 | 4.50 | 450 |
| 500 | 5.00 | 500 |
| 550 | 5.50 | 550 |
| 600 | 6.00 | 600 |
| 650 | 6.50 | 650 |
Probe checks in the ripple tank confirmed the interference conditions: antinodal (bright) points gave Δ/λ ≈ 0, 1, 2 …, while nodal (dark) points gave Δ/λ ≈ 0.5, 1.5, … . The central band lay on the perpendicular bisector (Δ = 0). Slit-separation and screen-distance trials confirmed Δy ∝ 1/d and Δy ∝ L.
Discussion
The fringe spacing increased in direct proportion to the wavelength: 450 nm gave 4.50 mm and 650 nm gave 6.50 mm, a ratio of fringe spacings (1.444) that matches the ratio of wavelengths (650/450 = 1.444). For every wavelength, multiplying the measured fringe spacing by d/L recovered the wavelength exactly, confirming Δy = λL/d. Increasing the slit separation crowded the fringes together (Δy ∝ 1/d) and moving the screen farther away spread them apart (Δy ∝ L), as the relation predicts.
The ripple-tank observations explained why these fringes appear. Bright fringes are points where the path difference from the two slits is a whole number of wavelengths, so the waves arrive in phase and reinforce; dark fringes occur at half-integer path differences, where the waves cancel. The central bright fringe sits where the two paths are equal, on the perpendicular bisector of the slits.
Conclusion
The experiment confirmed the principle of superposition and the double-slit fringe relation. Constructive and destructive interference were tied directly to path differences of whole and half wavelengths, and the measured fringe spacing obeyed Δy = λL/d, with Δy·d/L recovering each set wavelength. The proportionalities Δy ∝ λ, Δy ∝ L, and Δy ∝ 1/d were all verified.
Practice Questions
Show all work and include units in your answers.