Observe the setup illustrated in the figure. Through a certain mechanism, the separation between the slits S_3 and S_4 can be adjusted. The intensity is recorded at the point P , which lies on the common perpendicular

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Quantrex Academy · Accepted Answer

Correct answer: C. 2I

The path difference for waves reaching S_3 from S_1 and S_2 is Δ x_3 = (d(z/2))/(D) = (dz)/(2D) . The corresponding phase difference is Δ φ = (2π)/(λ) Δ x_3 = (π dz)/(λ D) . By symmetry, S_3 and S_4 are in phase with equal amplitude proportional to ((Δ φ)/(2)) = ((π dz)/(2λ D)) . At point P on the perpendicular bisector of S_3 S_4 , constructive interference occurs. The resultant intensity is I_P(z) = K ^2((π dz)/(2λ D)) . Given I_P = I at z = (Dλ)/(2d) : I = K ^2((π d)/(2λ D) × (Dλ)/(2d)) = K ^2((π)/(4)) = (K)/(2) K = 2I For z = (2Dλ)/(d) , the intensity is: I_P = 2I ^2((π d)/(2λ D) × (2Dλ)/(d)) = 2I ^2(π) I_P = 2I × (-1)^2 = 2I

Observe the setup illustrated in the figure. Through a certain...

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