In a single slit diffraction pattern, the distance between the plane of the slit and the screen is 1.4 m. The width of the slit is 0.66 mm. The second maximum is formed at the distance of 2.8 mm from the center of the screen. The wavelength of light used is

Step-by-step solution

The position of the n -th secondary maximum in a single slit diffraction pattern is given by: y_n = (2n + 1) D 2a For the second maximum, n = 2 : y₂ = 5 D 2a Rearranging for the wavelength : = 2a y₂ 5D Given values are: D = 1.4 m a = 0.66 mm = 0.66 10⁻³ m y₂ = 2.8 mm = 2.8 10⁻³ m Substituting the values into the formula: = 2 0.66 10⁻³ 2.8 10⁻³ 5 1.4 = 3.696 10⁻⁶ 7 = 0.528 10⁻⁶ m = 5280 10⁻¹⁰ m = 5280 Answer: 5280 Å