In a single-slit diffraction experiment, the first minimum is observed 3.0 mm from the centre on a screen 1.2 m from the slit. If the wavelength is 500 nm, what is the slit width?
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Question

In a single-slit diffraction experiment, the first minimum is observed 3.0 mm from the centre on a screen 1.2 m from the slit. If the wavelength is 500 nm, what is the slit width?

G1 Physics · Accepted Answer

For the first minimum, a sinθ = λ. For small angles, sinθ ≈ tanθ ≈ y ÷ D. Therefore, a(y ÷ D) = λ. Hence, a = λD ÷ y. Substitute: λ = 500×10⁻⁹ m, D = 1.2 m, y = 3.0×10⁻³ m. Thus, a = 500×10⁻⁹×1.2 ÷ (3.0×10⁻³) = 2.0×10⁻⁴ m = 0.20 mm. Correct option: B.