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In a Young's double-slit experiment, the interference pattern on the screen has a maximum to minimum intensity ratio of 25 : 9 . The ratio of the widths of the two slits used in the experiment is :

Options

  1. A4 : 1
  2. B5 : 3
  3. C256 : 1
  4. D16 : 1

Correct answer

D. 16 : 1

Step-by-step solution

The ratio of maximum to minimum intensity in the interference pattern is given by: I_ I_ = ( I₁ + I₂ I₁ - I₂ )^2 Given that I_ I_ = 25 9 , taking the square root on both sides yields: I₁ + I₂ I₁ - I₂ = 5 3 Cross-multiplying to solve for the ratio of the amplitudes: 3 I₁ + 3 I₂ = 5 I₁ - 5 I₂ 8 I₂ = 2 I₁ I₁ I₂ = 4 1 Squaring this ratio gives the ratio of the intensities of the two slits: I₁ I₂ = 16 1 Since the intensity of light passing through a slit is directly proportional to its width ( I w ), the ratio of the sl

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