Two identical narrow slits, each of width a, form a double-slit arrangement with centre-to-centre separation d = 4a. The interference fringes are observed inside the single-slit diffraction envelope. Which interference maximum is the first one to be absent because it coincides with a diffraction minimum?
Physics · Wave Optics · diffraction of light
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Question

Two identical narrow slits, each of width a, form a double-slit arrangement with centre-to-centre separation d = 4a. The interference fringes are observed inside the single-slit diffraction envelope. Which interference maximum is the first one to be absent because it coincides with a diffraction minimum?

Physics · Wave Optics · diffraction of light

G1 Physics · Accepted Answer

Interference maxima satisfy d sinθ = nλ, where n is the interference order. Single-slit diffraction minima satisfy a sinθ = mλ, where m = 1, 2, 3, ... For a missing interference order, both conditions must hold at the same angle. Divide the two equations: d ÷ a = n ÷ m. Given, d = 4a. Therefore, n = 4m. The first diffraction minimum has m = 1. Hence the first missing interference maximum has n = 4. Therefore the answer is 4.