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In Young's double slit experiment, the ratio of maximum and minimum intensities in the fringe system is 9: 1 . The ratio of amplitudes of coherent sources is

Options

  1. A1: 1
  2. B9: 1
  3. C2: 1
  4. D3: 1

Correct answer

C. 2: 1

Step-by-step solution

The intensity I of a wave is proportional to the square of its amplitude A , so I A² . Let the amplitudes of the two coherent sources be A₁ and A₂ . The maximum intensity I_ max and minimum intensity I_ min are given by the relations I_ max = (A₁ + A₂)² and I_ min = (A₁ - A₂)² . Given the ratio I_ max I_ min = 9 1 , we have (A₁ + A₂)² (A₁ - A₂)² = 9 1 . Taking the square root on both sides, we get A₁ + A₂ A₁ - A₂ = 3 1 . Cross-multiplying gives A₁ + A₂ = 3(A₁ - A₂) , which simplifies to A₁ + A₂ = 3A₁ - 3A₂ . Rearra

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