Observe the setup illustrated in the figure. Through a certain mechanism, the separation between the slits S₃ and S₄ can be adjusted. The intensity is recorded at the point P , which lies on the common perpendicular bisector of S₁ S₂ and S₃ S₄ . When z = D 2d , the intensity observed at P is I . Calculate this intensity when z equals 2 D d .

Step-by-step solution

The path difference for waves reaching S₃ from S₁ and S₂ is x₃ = d(z/2) D = dz 2D . The corresponding phase difference is = 2 x₃ = dz D . By symmetry, S₃ and S₄ are in phase with equal amplitude proportional to ( 2 ) = ( dz 2 D ) . At point P on the perpendicular bisector of S₃ S₄ , constructive interference occurs. The resultant intensity is I_P(z) = K ^2 ( dz 2 D ) . Given I_P = I at z = D 2d : I = K ^2 ( d 2 D D 2d ) = K ^2 ( 4 ) = K 2 K = 2I For z = 2D d , the intensity is: I_P = 2I ^2 ( d 2 D 2D d ) = 2I ^2( )