Practice question
Question
In a double-slit experiment, if two wavelengths \( 600 \, \text{nm} \) and \( 400 \, \text{nm} \) are
used, what is the smallest distance from the central maximum where their bright fringes coincide? (\( d
= 0.2 \, \text{mm} \), \( D = 1.0 \, \text{m} \))
Explanation
**Intensity not depend on speed** when enters denser medium because intensity I ∝ n E₀²? Actually Poynting vector S = E×H, energy density u =½ ε E², for same amplitude E₀ intensity proportional to n, but amplitude changes at interface due to reflection, total energy conserved incident = reflected + transmitted, interference does not destroy energy, it redistributes. Bright fringe position x_n = (n λ D/d) . Fringes coincide when n₁ λ₁ = n₂ λ₂ . 600 n₁ = 400 n₂ , n₂ = (3/2) n₁ . Smallest integers: n₁ = 2 , n₂ = 3 . x = (2 × 6.0 × 10⁻⁷ ×
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