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_
Related questions
Newton originally supported the corpuscular theory of light. The wave theory was strongly In a Young's double-slit experiment, a thin transparent sheet of thickness t and refractivCoherent sources are those that emit waves ofIf polarised light of intensity I passes through a polariser whose axis makes angle θ withAn unpolarised light of intensity I₀ passes through a polariser. The transmitted intensityWhich of the following phenomena is characteristic ONLY of transverse waves and NOT of lonThe polarising angle (Brewster angle) for glass of refractive index 1.5 isThe width of the central maximum in single-slit diffraction on a screen at distance D is