Practice question

Given: In a single-slit diffraction pattern, what is the angular width of the ntral maximum if the slit width is 2.0 μm and the wavelength is 400 nm ? These values define the system as per NCERT data. Formula: Angular width of the ntral maximum 2θ = 2lambda/a. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: lambda = 4.0 × 10⁻⁷ m, a = 2.0 × 10⁻⁶ m . sin θ = lambda/a = frac4.0 × 10⁻⁷².0 × 10⁻⁶= 0.2, θ = sin^{-1(0.2) approx 11.5°, 2θ approx 23° . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.