In a double slit experiment shown in figure, when light of wavelength 400 nm is used, dark fringe is observed at P. If D = 0.2 m. the minimum distance between the slits S₁ and S₂ i $s ______$ mm.

Question

In a double slit experiment shown in figure, when light of wavelength 400 nm is used, dark fringe is observed at P. If D = 0.2 m. the minimum distance between the slits S₁ and S₂ i s ______ mm.

G1 Physics · Accepted Answer

Image For minimum path difference at P: 2√(D² + d²) - 2D = λ/2 Therefore, √(D² + d²) - D = λ/4 So, √(D² + d²) = D + λ/4 Squaring, D² + d² = D² + Dλ/2 + λ²/16 Hence, d² = Dλ/2 + λ²/16 Given, D = 0.2 m λ = 400 nm = 400 × 10⁻⁹ m d² ≈ (0.2 × 400 × 10⁻⁹)/2 = 4 × 10⁻⁸ d = 2 × 10⁻⁴ m = 0.20 mm