--- title: "Intensity distribution in interference" subject: "Engineering Physics" chapter: 4 key: "semester_1/engineering_physics:interference-intensity-resultant" source: https://www.bctengineer.com/q/semester_1/engineering_physics/interference-intensity-resultant site: BCT Engineering --- # Intensity distribution in interference Engineering Physics · Chapter 4: Physical Optics - Asked in: 2070 Asar, 5 marks; 2069 Asar, 5 marks (asked 2 times) - Video: [Intensity distribution in INTERFERENCE](https://www.youtube.com/watch?v=hC1COh3Zyws) — BTP campus - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_1/engineering_physics/interference-intensity-resultant - Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-4 ## Question What is interference? Explain the intensity distribution in interference with mathematical treatment. (Show that two coherent sources of constant phase difference δ and amplitudes A1 and A2 superimpose to give intensity I = A1² + A2² + 2A1A2 cos δ.) ## Short answer Interference is the modification of intensity when coherent waves superpose. For amplitudes A₁, A₂ and phase difference δ, resultant amplitude squared is A₁² + A₂² + 2A₁A₂ cos δ, so intensity I = A₁² + A₂² + 2A₁A₂ cos δ. Maxima at δ = 2nπ, minima at δ = (2n+1)π. ## Answer **Interference** is the modification of intensity produced when two or more coherent light waves superpose, giving maxima and minima in the resultant intensity. #### Intensity distribution ```text S1 o--------\ \--> P path difference Δ S2 o--------/ ``` 1. Let two coherent waves of amplitudes A₁ and A₂ and phase difference δ reach point P: y₁ = A₁ sin ωt, y₂ = A₂ sin(ωt + δ) 2. By the principle of superposition: y = y₁ + y₂ = A₁ sin ωt + A₂ sin(ωt + δ) 3. Expand: y = sin ωt (A₁ + A₂ cos δ) + cos ωt (A₂ sin δ) 4. Put A₁ + A₂ cos δ = A cos φ and A₂ sin δ = A sin φ, so y = A sin(ωt + φ). 5. Squaring and adding: ```text A² = (A₁ + A₂ cos δ)² + (A₂ sin δ)² = A₁² + A₂² + 2 A₁A₂ cos δ ``` 6. Intensity is proportional to the square of the amplitude, so ```text I = A₁² + A₂² + 2 A₁A₂ cos δ ``` #### Maxima and minima - **Maximum**: cos δ = 1, δ = 2nπ (path difference nλ): I_max = (A₁ + A₂)² - **Minimum**: cos δ = −1, δ = (2n+1)π (path difference (2n+1)λ/2): I_min = (A₁ − A₂)² If A₁ = A₂ = a: I = 4a² cos²(δ/2), varying between 4a² and 0. The average intensity is 2a², equal to the sum of separate intensities, so energy is conserved: it is only redistributed. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the resultant displacement at P obtained? A. By multiplying y1 and y2 B. By adding y1 + y2 (principle of superposition) C. By averaging the intensities D. By subtracting phases only Answer: B. The resultant is the sum y = A1 sin ωt + A2 sin(ωt + δ). (Key point: Superposition) 2. Which expression results for the intensity? A. A1^2 + A2^2 B. A1^2 + A2^2 + 2A1A2 cos δ C. (A1 + A2)^2 always D. A1^2 − A2^2 + 2A1A2 sin δ Answer: B. Squaring and adding the amplitude components gives A^2 = A1^2 + A2^2 + 2A1A2 cos δ, and I ∝ A^2. (Key point: Resultant intensity) 3. What are I_max and I_min? A. (A1 + A2)^2 and (A1 − A2)^2 B. A1^2 + A2^2 for both C. 2A1A2 and 0 D. (A1 − A2)^2 and (A1 + A2)^2 Answer: A. cos δ = 1 gives (A1 + A2)^2 and cos δ = −1 gives (A1 − A2)^2. (Key point: Maxima and minima) 4. For equal amplitudes a, I = 4a^2 cos^2(δ/2). What does the average intensity 2a^2 show? A. Energy is created at maxima B. Energy is destroyed at minima C. Energy is only redistributed, not lost D. Intensity is always 2a^2 Answer: C. The average equals the sum of the separate intensities, so interference redistributes energy. 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