Practice question
Question
What is the distance of the first bright fringe from the central maximum in a double-slit experiment if
\( \lambda = 620 \, \text{nm} \), \( d = 0.3 \, \text{mm} \), and \( D = 1.2 \, \text{m} \)?
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) . For the first bright fringe, n = 1 . λ = 6.2 × 10⁻⁷ m , d = 3.0 × 10⁻⁴ m , D = 1.2 m . x₁ = (1 × 6.2 × 10⁻⁷ × 1.2/3.0 ×
Discussion
Comments
Please log in to join the discussion.
Login to commentNo comments yet. Be the first to start the discussion.