In YDSE using monochromatic visible light, the distance between the plane of sli

Initially, point P is directly in front of the upper slit, so the vertical displacement from the central maximum is y = d/2 = 1 mm. For small angles, the path difference is Δ = d y / D = d^2/(2D). Given D = 1.7 m, and that a maximum (constructive interference) is observed, we have Δ = nλ, so λ = d^2/(2nD). After moving the screen 50 cm closer, D' = 1.2 m, so the new path difference is Δ' = d^2/(2D'). The point P now lies between the third and fourth minima, so 2.5λ < Δ' < 3.5λ, and the intensity is one-fourth of the maximum, indicating a phase difference of 2π/3 or 4π/3 (mod 2π), which gives Δ' = (k + 1/3)λ or (k + 2/3)λ. Only n = 2 satisfies the range condition (1.7647 < n < 2.4706). Substituting n = 2 gives λ = d^2/(4D) = (4 × 10^{-6})/(4 × 1.7) = 5.88 × 10^{-7} m ≈ 5.9 × 10^{-7} m, which matches the intensity condition within approximation. This wavelength is in the visible range, corresponding to option C.