---
title: "Conditions for constructive and destructive interference"
subject: "Engineering Physics"
chapter: 4
key: "semester_1/engineering_physics:constructive-destructive-path-difference"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/constructive-destructive-path-difference
site: BCT Engineering
---

# Conditions for constructive and destructive interference

Engineering Physics · Chapter 4: Physical Optics

- Asked in: 2074 Chaitra, 5 marks
- Video: [Constructive and Destructive interference | Physics | Khan Academy](https://www.youtube.com/watch?v=oTjTXS40pqs) — Khan Academy Physics
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/constructive-destructive-path-difference
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-4

## Question

What are constructive and destructive interference? Prove that the path difference for constructive interference is an integer multiple of λ and that for destructive interference is an odd integer multiple of λ/2.

## Short answer

Constructive interference yields maximum intensity when path difference is an integer multiple of λ, Δ = nλ. Destructive interference gives minimum intensity when Δ = (2n+1)λ/2 (odd multiple of half wavelength).

## Answer

**Constructive interference**: two coherent waves meet in phase, so their amplitudes add and the intensity is maximum (bright). **Destructive interference**: they meet out of phase, amplitudes subtract and the intensity is minimum (dark).

#### Proof
1. Let two coherent waves of amplitudes A₁ and A₂ with phase difference δ superpose. The resultant intensity is

```text
I = A₁² + A₂² + 2 A₁A₂ cos δ
```

2. A path difference Δ corresponds to a phase difference

```text
δ = (2π/λ) Δ
```

3. **Constructive**: I is maximum when cos δ = +1, so δ = 2nπ (n = 0, 1, 2 ...). Then

   (2π/λ)Δ = 2nπ → **Δ = nλ**, an integral multiple of λ (equivalently an even multiple of λ/2). I_max = (A₁ + A₂)².

4. **Destructive**: I is minimum when cos δ = −1, so δ = (2n+1)π. Then

   (2π/λ)Δ = (2n+1)π → **Δ = (2n+1) λ/2**, an odd integral multiple of λ/2. I_min = (A₁ − A₂)².

| | Phase difference | Path difference | Intensity |
|---|---|---|---|
| Constructive | 0, 2π, 4π ... | 0, λ, 2λ ... | (A₁+A₂)² |
| Destructive | π, 3π, 5π ... | λ/2, 3λ/2 ... | (A₁−A₂)² |

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. In destructive interference the waves meet:
   A. in phase, adding amplitudes
   B. out of phase, subtracting amplitudes
   C. with different frequencies
   D. with equal phase but different amplitudes
   Answer: B. Out of phase waves cancel partly or fully, giving minimum intensity. (Key point: Meaning)

2. How is the phase difference δ related to the path difference Δ?
   A. δ = (λ/2π) Δ
   B. δ = (2π/λ) Δ
   C. δ = 2πλ Δ
   D. δ = Δ/λ
   Answer: B. One wavelength of path corresponds to 2π of phase, so δ = 2πΔ/λ. (Key point: Phase-path relation)

3. Constructive interference needs cos δ = +1. What path difference results?
   A. Δ = (2n+1)λ/2
   B. Δ = nλ
   C. Δ = nλ/2 only for odd n
   D. Δ = λ/4
   Answer: B. δ = 2nπ gives Δ = nλ. (Key point: Constructive condition)

4. Destructive interference occurs at what path difference, and what is I_min?
   A. nλ; (A1 + A2)^2
   B. (2n+1)λ/2; (A1 − A2)^2
   C. (2n+1)λ/2; (A1 + A2)^2
   D. nλ; (A1 − A2)^2
   Answer: B. cos δ = −1 gives δ = (2n+1)π, so Δ = (2n+1)λ/2 and I_min = (A1 − A2)^2. (Key point: Destructive condition)

## Related in Engineering Physics

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