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In the Young's double slit experiment using a monochromatic light of wavelength λ , the path difference ( in terms of an integer n ) corresponding to any point having half the peak intensity is

Options

  1. A2 n + 1 λ 2
  2. B2 n + 1 λ 4
  3. C2 n + 1 λ 8
  4. D2 n + 1 λ 16

Correct answer

B. 2 n + 1 λ 4

Step-by-step solution

I max = 4 I ​ ∴ 2 I = 2 I + 2 I cos θ cos θ = 0 ⇒ θ = 2 n + 1 π 2 2 π λ x = 2 n + 1 π 2 x = 2 n + 1 λ 4

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