• Tags: YDSE, intensity, phase
• Difficulty: Moderate

The intensity at any point on the screen depends on the phase difference φ = 2π$\Delta x$/λ = 2πdy/(λD). For equal slit intensities $I_{0}$: I = 4$I_{0}$c$os^{2}$(φ/2). Maximum intensity = 4$I_{0}$ (at φ = 0, ±2π, ...), minimum intensity = 0 (at φ = ±π, ±3π, ...). For unequal intensities $I_{1}$ and $I_{2}$: I = $I_{1}$ + $I_{2}$ + 2√(I_{1}$$I_{2})cosφ. Then I_max = (√$I_{1}$ + √$I_{2}$)^{2} and I_min = (√$I_{1}$ - √$I_{2}$)^{2}. The ratio I_max/I_min = ((√$I_{1}$ + √$I_{2}$)/(√$I_{1}$ - √$I_{2}$))^{2} is a frequently tested quantity. Note that intensity is proportional to the square of the amplitude, and slit width is proportional to intensity (wider slit = brighter).