In a Young’s double slit experiment, two slits are illuminated with a light of wavelength 800 nm. The line joining A₁P is perpendicular to A₁A₂ as shown in the figure. If the first minimum is detected at P, the value of slit separation a will be. The distance of screen from slits D = 5 cm:

Question

G1 Physics · Accepted Answer

For the first minimum, path difference A₂P - A₁P = λ/2. Here A₁P = D and A₂P = √(D² + a²). Thus √(D²+a²) - D = λ/2. Since a ≪ D, √(D²+a²) ≈ D + a²/(2D). Hence a²/(2D) = λ/2, so a = √(λD) = √(800 × 10⁻⁹ × 0.05) = 2 × 10⁻⁴ m = 0.2 mm.