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In Young's double slit interference experiment, using two coherent waves of different amplitudes, the intensities ratio between bright and dark fringes is 3 . Then, the value of the ratio of the amplitudes of the wave that arrive there is

Options

  1. A3 : 1
  2. B( 3 +1 3 -1 )
  3. C1: 3
  4. D( 3 -1 3 +1 )

Correct answer

B. ( 3 +1 3 -1 )

Step-by-step solution

Let the amplitudes of the two coherent waves be A₁ and A₂ . The intensity of a wave is proportional to the square of its amplitude, I A^2 . The maximum intensity I_ max at bright fringes is given by I_ max = (A₁ + A₂)^2 and the minimum intensity I_ min at dark fringes is given by I_ min = (A₁ - A₂)^2 . Given the ratio of intensities is I_ max I_ min = 3 , we have (A₁ + A₂)^2 (A₁ - A₂)^2 = 3 . Taking the square root on both sides, we get A₁ + A₂ A₁ - A₂ = 3 . Applying componendo and dividendo, (A₁ + A₂) + (A₁ - A₂)

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