Practice question
Question
What is the angular position of the first minimum in a single-slit diffraction pattern if the slit
width is \( 8.0 \, \mu\text{m} \) and the wavelength is \( 640 \, \text{nm} \)?
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. First minimum occurs at sin θ = (λ/a) . λ = 6.4 × 10⁻⁷ m , a = 8.0 × 10⁻⁶ m . sin θ = (6.4 × 10⁻⁷/8.0 × 10⁻⁶) = 0.08 , θ = sin⁻¹(0.08) ≈ 4.6° . Using Δ = d sinθ, y =
Discussion
Comments
Please log in to join the discussion.
Login to commentNo comments yet. Be the first to start the discussion.