A diffraction pattern due to a single slit of width 0.12mm is obtained with a blue green light of wavelength 500 nm. The angular separation between central maximum and second order secondary maximum of the diffraction pattern is

Step-by-step solution

Given slit width, a = 0.12 mm = 1.2 10⁻⁴ m Wavelength of light, = 500 nm = 5 10⁻⁷ m The angular position of the n -th secondary maximum in a single slit diffraction pattern is given by: _n = (2n + 1) 2a For the second order secondary maximum, n = 2 : ₂ = 5 2a Substituting the given values: ₂ = 5 5 10⁻⁷ 2 1.2 10⁻⁴ ₂ = 25 10⁻⁷ 2.4 10⁻⁴ ₂ = 10.416 10⁻³ rad ₂ = 1.042 10⁻² rad Answer: 1.042 10⁻² rad