Light of wavelength 580 nm is incident normally on a slit of width ' a '. The distance between slit and screen is 2.5 m and the distance of the second order maximum from the centre of the screen is 14.5 mm in a diffraction pattern. The value of ' a ' is

Step-by-step solution

For a single slit diffraction pattern, the position of the n -th secondary maximum on the screen is given by: y = (2n + 1) D 2a For the second order maximum, n = 2 . Substituting this into the formula, we get: y = 5 D 2a Rearranging for the slit width a : a = 5 D 2y Substituting the given values = 580 10⁻⁹ m , D = 2.5 m , and y = 14.5 10⁻³ m : a = 5 580 10⁻⁹ 2.5 2 14.5 10⁻³ a = 7250 10⁻⁹ 29 10⁻³ a = 250 10⁻⁶ m a = 0.25 10⁻³ m Answer: 0.25 10⁻³ , m