In a single slit diffraction pattern, the distance between the plane of the slit and screen is 1.3 m . The width of the slit is 0.65 mm and the second maximum is formed at the distance of 2.6 mm from the centre of the screen. The wavelength of light used is

Step-by-step solution

Single-slit diffraction secondary maxima satisfy a _m = (m + 1 2 ) for m = 1, 2, 3, , where a is the slit width and the wavelength. For the second maximum ( m = 2 ) at small angle ₂ , ₂ y₂/D where y₂ = 2.6 10⁻³ m is the fringe position and D = 1.3 m the slit-to-screen distance. Substituting a = 0.65 10⁻³ m gives: a y₂ D = 5 2 Solving for : = 2 5 (0.65 10⁻³) (2.6 10⁻³) 1.3 Calculating yields = 0.52 10⁻⁶ m. Converting to Ångstroms ( 1 Å = 10⁻¹⁰ m): = 0.52 10⁻⁶ 10¹⁰ = 5200 Å The wavelength is therefore 5200 Å .