- **summary_type**: concept
- **word_count**: 170
The intensity at any point in YDSE depends on the phase difference φ = 2π$\Delta x$/λ. For equal intensity sources: I = \$4I_{0}$c$os^{2}$(φ/2). Important benchmarks: at φ = 0 (center), I = \$4I_{0}$; at φ = π/2 (path diff λ/4), I = \$2I_{0}$; at φ = 2π/3 (path diff λ/3), I = $I_{0}$; at φ = π (path diff λ/2), I = 0. For unequal slit widths w_{1} and w_{2}, intensity is proportional to slit width ($I \propto w$), so $I_{1}$/$I_{2}$ = w_{1}/w_{2}. The fringe visibility V = (I_max - I_min)/(I_max + I_min) = 2√($I_{1}$$I_{2}$)/($I_{1}$ + $I_{2}$) ranges from 0 (no fringes) to 1 (perfect dark minima). Maximum visibility occurs when $I_{1}$ = $I_{2}$. Energy is conserved — the average intensity over the full pattern equals $I_{1}$ + $I_{2}$, with energy merely redistributed from minima to maxima.$
Part of JOP-02 — Wave Optics: YDSE, Diffraction & Polarization
YDSE Intensity and Phase Analysis
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