Two coherent sources produce interference fringes. If the intensities of individual sources are I₁ = 4I₀ and I₂ = I₀, the maximum intensity in the fringe pattern is
A{'text': '5I₀', 'label': 'A'}
B{'text': '9I₀', 'label': 'B'}
C{'text': '3I₀', 'label': 'C'}
D{'text': '4I₀', 'label': 'D'}
Answer & Solution
Correct answer: B. {'text': '9I₀', 'label': 'B'}
1. Maximum intensity of interference: I_max = (√I₁ + √I₂)².
2. Amplitudes A₁ ∝ √I₁ = 2 and A₂ ∝ √I₂ = 1 (in units of √I₀).
3. So (√I₁ + √I₂)² = (2 + 1)² × I₀ = 9 I₀.
4. Minimum intensity would be (2 − 1)² × I₀ = I₀.
_Source: NCERT Class 12 Physics, Ch 10 "Wave Optics", §10.4_
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