In Young's double slit experiment, two slits are illuminated with light of wavelength 800 nm. The line joining slit A1 to point P is perpendicular to the line A1A2 (i.e., P is directly in front of A1). The first minimum is detected at P. The screen is at distance D = 5 cm from the slits. F

Question

In Young's double slit experiment, two slits are illuminated with light of wavelength 800 nm. The line joining slit A1 to point P is perpendicular to the line A1A2 (i.e., P is directly in front of A1). The first minimum is detected at P. The screen is at distance D = 5 cm from the slits. Find the slit separation d.

Accepted Answer

0.2 mm. A2P - A1P = sqrt(D²+d²) - D = lambda/2. For d << D: sqrt(D²+d²) - D ≈ d²/(2D). So d²/(2D) = lambda/2 => d² = D*lambda = 5*10⁻² * 800*10⁻⁹ = 4*10⁻⁸ m² => d = 2*10⁻⁴ m = 0.2 mm.

Solution

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