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Question

In a double-slit experiment, if \( \lambda = 580 \, \text{nm} \), \( d = 0.2 \, \text{mm} \), and \( D
= 1.0 \, \text{m} \), what is the distance of the fourth bright fringe from the central maximum?

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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). Bright fringe position x_n = (n λ D/d) . For the fourth bright fringe, n = 4 . λ = 5.8 × 10⁻⁷ m , d = 2.0 × 10⁻⁴ m , D = 1.0 m

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