Superposition & Interference

You’ll be able to

Apply the principle of superposition to overlapping waves
Use path difference to distinguish constructive from destructive interference
Relate the double-slit condition to bright and dark fringes

Superposition and interference

When two waves overlap, their displacements simply add — the principle of superposition. Where two crests meet, they reinforce into a bigger crest: constructive interference. Where a crest meets a trough, they cancel: destructive interference. For a stable pattern the waves must be coherent (same frequency and a fixed phase relationship). Interference is the signature behavior of waves — particles do not add and cancel this way, so an interference pattern is proof you are dealing with a wave.

Path difference decides the outcome

What determines constructive versus destructive interference at a point is the path difference — how much farther one wave traveled than the other to reach it. If the path difference is a whole number of wavelengths (0, λ, 2λ, …), the waves arrive in phase and interfere constructively (bright/loud). If it is a half-integer number of wavelengths (½λ, 1½λ, …), they arrive out of phase and interfere destructively (dark/silent). In Young’s double-slit experiment, this produces alternating bright and dark fringes on a screen.

Interference conditions & double slit

Constructive: Δ = m·λ · Destructive: Δ = (m + ½)·λ · d·sinθ = m·λ

Δ is the path difference and m = 0, 1, 2, … is the order. For a double slit of spacing d, bright fringes appear at angles where d·sinθ = mλ.

Worked example

In a double-slit experiment, the slit separation is d = 0.10 mm and light of wavelength λ = 500 nm is used. Find sinθ for the first-order (m = 1) bright fringe.

1.Bright-fringe condition: d sinθ = mλ, so sinθ = mλ/d.
2.Convert units: d = 0.10 mm = 1.0 × 10⁻⁴ m, λ = 500 nm = 5.0 × 10⁻⁷ m.
3.Substitute (m = 1): sinθ = (5.0 × 10⁻⁷) / (1.0 × 10⁻⁴).
4.Divide: sinθ = 5.0 × 10⁻³ = 0.0050.

Answer: sinθ = 0.0050, so the first bright fringe sits at a very small angle (θ ≈ 0.29°) off the center.

Checkpoint

Two coherent waves arrive at a point with a path difference of exactly one full wavelength. The interference there is:

Tip

Translate path difference into wavelengths: whole numbers (0, 1λ, 2λ) give bright/constructive, half-numbers (½λ, 1½λ) give dark/destructive. Counting in units of λ is faster and less error-prone than working in meters.

Checkpoint

In a double-slit experiment, d = 0.10 mm and λ = 500 nm. For the first-order bright fringe (m = 1), what is sinθ?

On the exam

Convert everything to meters before using d sinθ = mλ: millimeters are 10⁻³ and nanometers are 10⁻⁹. A power-of-ten error here is the most common way to lose the double-slit point.

Answer the 2 checkpoints as you read.