📏 Young's Double Slit
🔦 Diffraction Grating
🎸 Standing Waves (String)

The Principle of Superposition

When two or more waves meet at a point, the resultant displacement is the vector sum of the individual displacements of the waves. This fundamental principle explains physical phenomena like interference, standing waves, and beats.

ytotal = y₁ + y₂

1. Constructive vs. Destructive

Interference occurs when coherent waves overlap:

  • Constructive Interference: Waves arrive in phase (crest meets crest). Resultant amplitude is A₁ + A₂.
  • Destructive Interference: Waves arrive in antiphase (crest meets trough). Resultant amplitude is |A_1 - A_2|.

2. Path and Phase Difference

A path difference (Δx) between two waves from a single source results in a phase difference (φ):

φ = (Δx / λ) × 360°
Exam Tip: For light, interference only occurs if the source is coherent—meaning it has a constant phase difference and identical frequency. This is why laser light is typically used for Young's Double Slit experiments.

Deep Dive: Worked Examples

✅ Example 1: Path Difference for Sound

Two speakers are 1.0m and 1.25m from a microphone. If the wavelength is 0.5m, find the interference type.

Step 1: Path Difference (Δx) Δx = 1.25 - 1.0 = 0.25 m
Step 2: Compare to λ Δx / λ = 0.25 / 0.5 = 0.5 (or ½λ)
Step 3: Conclusion ½λ corresponds to 180° phase difference. Result = Destructive Interference.

✅ Example 2: Resultant Intensity

If two coherent waves of amplitude A interfere constructively, how does the resulting intensity compare to one wave?

Step 1: Calculate New Amplitude A_{total} = A + A = 2A
Step 2: Relation Intensity ∝ A² I_{new} ∝ (2A)² = 4A²
Step 3: Solve The intensity is 4 times greater.

✅ Example 3: Phase Change on Distance

A wave has λ = 20cm. Find the phase difference between two points 5cm apart.

Step 1: Formula φ = (Δx / λ) × 360
Step 2: Plug in values φ = (5 / 20) × 360 = 0.25 × 360
Step 3: Solve φ = 90° (or π/2 rad)

✅ Example 4: Double Slit Fringe Separation

Light of λ=600nm passes through slits 0.2mm apart. Find fringe spacing on a screen 1.5m away.

Step 1: Use y = λD / d λ = 600e-9, D = 1.5, d = 0.2e-3
Step 2: Calculation y = (600e-9 × 1.5) / 2e-4
Step 3: Solve y = 4.5e-3 m = 4.5 mm

✅ Example 5: Standing Wave Nodes

A string of length 1.2m vibrates at its 3rd harmonic. Find the distance between adjacent nodes.

Step 1: Identify wavelength 3rd harmonic means 1.5 wavelengths fit in length L. λ = L / 1.5 = 1.2 / 1.5 = 0.8m
Step 2: Node spacing Distance between nodes = λ/2
Step 3: Solve Node spacing = 0.8 / 2 = 0.4 m