Practice question
Question
What is the distance of the fourth dark fringe from the central maximum in a double-slit experiment if
\( \lambda = 610 \, \text{nm} \), \( d = 0.3 \, \text{mm} \), and \( D = 1.5 \, \text{m} \)?
Explanation
**Double-slit vs single-slit** double-slit interference pattern has equally spaced bright fringes with envelope due to single-slit diffraction, single-slit central maximum width 2λ D/a, intensity of secondary maxima decreases with order, condition for coherence constant frequency and phase, path difference for bright n λ, dark (n+½)λ. Dark fringe position x_n = ((n + (1/2)) λ D/d) . For the fourth dark fringe, n = 3 . λ = 6.1 × 10⁻⁷ m , d = 3.0 × 10⁻⁴ m , D = 1.5 m . x₃ = ((3 + (1/2)) × 6.1 × 10⁻⁷ × 1.5/3.0 × 10⁻⁴) = (3.5 × 9.15 × 10⁻⁷/3.0 × 10⁻⁴)
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