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Question

What is the fringe width in a double-slit experiment if \( \lambda = 620 \, \text{nm} \), \( d = 0.5 \,
\text{mm} \), and \( D = 2.0 \, \text{m} \)?

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Explanation

**Interference** occurs when two coherent waves superpose, path difference Δ = d sinθ, bright fringe when Δ = n λ, n integer, seventh bright Δ=7λ, dark fringe Δ = (2n-1)λ/2, second dark Δ=3λ/2, third dark 5λ/2, sixth dark 11λ/2, distance of bright fringe from central y = n λ D/d, D screen distance, d slit separation, for λ=650 nm d=0.5 mm D=1.0 m fifth dark? Actually dark y=(2n-1)λ D/(2d). Fringe width β = (λ D/d) . λ = 6.2 × 10⁻⁷ m , d = 5.0 × 10⁻⁴ m , D = 2.0 m . β = (6.2 × 10⁻⁷ × 2.0/5.0 × 10⁻⁴) =

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