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Light of wavelength 600 n m is incident normally on a slit of width 0.2 m m . The angular width of central maxima in the diffraction pattern is (measured from minimum to minimum)

Options

  1. A6 × 10 - 3 r a d
  2. B4 × 10 - 3 r a d
  3. C2.4 × 10 - 3 r a d
  4. D4.5 × 10 - 3 r a d

Correct answer

A. 6 × 10 - 3 r a d

Step-by-step solution

λ = 600 n m = 600 × 1 0 - 9 m d = 0.2 m m = 0.2 × 1 0 - 3 m ∵ Linear width of central maximum, x = 2 θ D Where, 2 θ : Angular width of central maximum. We know that, x = 2 λ D d Angular width 2 θ . D = 2 λ D d ⇒ 2 θ = 2 λ d = 2 × 600 × 1 0 - 9 0.2 × 1 0 - 3 = 6 × 1 0 - 7 × 1 0 4 = 6 × 1 0 - 3 r a d

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