A Young's double slit experiment is set up with identical sources emitting monochromatic light of wavelength . A thin transparent sheet of refractive index 1.5 is introduced directly in front of one of the slits. Due to this, the intensity at the geometrical center of the screen, which was originally a maximum, reduces to 75 % of the maximum intensity. The minimum thickness of the transparent sheet is:

Step-by-step solution

The intensity I in a Young's double slit experiment is given by: I = I_ max ^2 ( 2 ) Given that the new intensity at the center is 75 % of the maximum intensity: 0.75 I_ max = I_ max ^2 ( 2 ) ^2 ( 2 ) = 3 4 ( 2 ) = 3 2 For the minimum thickness, we consider the smallest positive phase difference: 2 = 6 = 3 The corresponding path difference x is: x = 2 = 2 ( 3 ) = 6 When a transparent sheet of thickness t and refractive index is introduced, the additional optical path difference created is: x = ( - 1)t Substituting