Maximum intensity in Young’s double-slit experiment is I₀. Distance between slits is d = 5λ. What will be intensity in front of one of the slits when the screen is placed at D = 10d?
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Maximum intensity in Young’s double-slit experiment is I₀. Distance between slits is d = 5λ. What will be intensity in front of one of the slits when the screen is placed at D = 10d?

G1 Physics · Accepted Answer

Step 1: For point opposite one slit, Δx = √(D² + d²) − D ≈ d²/(2D) for D ≫ d. Step 2: With D = 10d, Δx = d/20 = 5λ/20 = λ/4. Step 3: Phase Δφ = 2πΔx/λ = π/2. For equal amplitudes, I = I₀ cos²(Δφ/2). Step 4: I = I₀ cos²(π/4) = I₀/2. Final answer: (C) I₀/2.