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MHT CET202523 Apr 2025Evening ShiftPhysicsWave OpticsActual

In Young's double slit experiment, the light of wavelength ' ' is used. The intensity at a point on the screen is 'I' where the path difference is 4 . If ' I _ o ' denotes the maximum intensity then the ratio of ' I _ o ' to ' I ' is ( 45^ =1 / 2 )

Options

  1. A2:1
  2. B4:1
  3. C8:1
  4. D12:1

Correct answer

A. 2:1

Step-by-step solution

The intensity at a point on the screen in Young's double-slit experiment is given by I = I_o ^2 ( 2 ) , where I_o is the maximum intensity and is the phase difference. The phase difference relates to path difference by = 2 x . For x = 4 , we have = 2 . Substituting into the intensity formula gives I = I_o ^2 ( 4 ) . Since ( 4 ) = 1 2 , this simplifies to I = I_o ( 1 2 ) . The ratio is therefore I_o I = 2 , or equivalently 2:1 . Final answer: A

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