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 slits S₃ and S₄ are located symmetrically at a distance y = z 2 from the central axis. The path difference for the waves reaching S₃ from S₁ and S₂ is given by: x = y d D = z d 2D The corresponding phase difference at S₃ is: = 2 x = z d D The intensity of light at S₃ (and by symmetry, at S₄ ) is: I' = I₀ ^2 ( 2 ) = I₀ ^2 ( z d 2 D ) where I₀ is the maximum intensity of the interference pattern from S₁ and S₂ . Since S₃ and S₄ are symmetrically placed with respect to S₁ and S₂ , the waves arriving at S₃ and S₄ a