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 waves from S₁ and S₂ reach S₃ with a path difference x₃ = d(z/2) D = dz 2D . The phase difference at S₃ is = 2 x₃ = dz D . By symmetry, the secondary sources S₃ and S₄ are always in phase and have identical amplitudes A₃ = A₄ ( dz 2 D ) . Since point P is on the perpendicular bisector of S₃ S₄ , the waves interfere constructively, giving intensity I_P(z) = K ^2 ( dz 2 D ) . For z = D 2d , the intensity is I : I = K ^2 ( d 2 D D 2d ) = K ^2 ( 4 ) = K 2 K = 2I For z = 3D 2d , the intensity is: I_P = 2I ^2 ( d 2 D