In Young's experiment the distance between the two slits is 0.4 cm and the distance of the screen from the slits is 100 cm. If the wavelength of the light is 5000 A, the distance of the fourth dark fringe from the central bright fringe is .

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(1) 4.37×10⁻² cm. Given: d = 0.4 cm = 4×10⁻³ m, D = 100 cm = 1 m, λ = 5000 A = 5×10⁻⁷ m, and the fourth dark fringe. Idea: a dark fringe needs the two paths to differ by an odd number of half wavelengths, which puts the nth dark fringe at xₙ = ((2n-1)λ D)/(2d). For the fourth dark fringe n = 4, so 2n - 1 = 7. x₄ = (7 × (5×10⁻⁷) × 1)/(2 × (4×10⁻³)) = (3.5×10⁻⁶)/(8×10⁻³). x₄ = 4.375×10⁻⁴ m. In centimetres that is 4.375×10⁻² cm, the printed 4.37×10⁻² cm.

In Young's experiment the distance between the two slits is 0.4 cm and the distance of the screen from the slits is 100 cm. If the wavelength of the light is 5000 A, the distance of the fourth dark fringe from the central bright fringe is ________.

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