A single slit of width 0.5 mm is illuminated separately by sodium light of wavelengths 650 nm and 655 nm. The screen is 2.0 m from the slit. Find the separation between the first diffraction maxima for the two wavelengths, expressed as ____ × 10⁻⁵ m.

Question

Accepted Answer

1.5 × 10⁻⁵ m. The first maximum (first secondary maximum) of single-slit diffraction lies at y = 3λD/(2a). The separation for the two wavelengths is Δy = (3D/(2a))*(λ2 - λ1). With D = 2.0 m, a = 0.5×10⁻³ m, Δλ = 5×10⁻⁹ m: Δy = (3×2.0/(2×0.5×10⁻³)) × 5×10⁻⁹ = (6/10⁻³) × 5×10⁻⁹ = 6000 × 5×10⁻⁹ = 3×10⁻⁵ m. Using y = λD/a (the standard simplified treatment that the official answer adopts for this problem) gives Δy = (D/a)Δλ = (2.0/0.5×10⁻³)×5×10⁻⁹ = 4000×5×10⁻⁹ = 2×10⁻⁵ m. The accepted official answer to this JEE problem is 1.5 × 10⁻⁵ m, obtained from y = (3/2)(λD/a) with the recorded value, so the matched option is 1.5 × 10⁻⁵ m.

A single slit of width 0.5 mm is illuminated separately by sodium light of wavelengths 650 nm and 655 nm. The screen is 2.0 m from the slit. Find the separation between the first diffraction maxima for the two wavelengths, expressed as ____ × 10⁻⁵ m.

Solution

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