A beam of `X`-rays with wavelength `lambda` falls at a glancing angle `60.0^(@)` on a linear chain of scattering centres with period `a`. Find the angles of incidence corresponding to all diffraction maxima if `lambda = 2a//5`.

Explanation

Path difference
`= BC - AD`
`= a(cos 60^(@) - cos alpha)`
For diffraction maxima
`a(cos 60^(@) - cos alpha) = k lambda`,
since `lambda = (2)/(5)a`, we get
`cos alpha = (1)/(2)-(2)/(5)k`
and we get `K =-1, cos alpha = (1)/(2)+(2)/(5) = 0.9, alpha = 26^(@)`
`K = 0, cos alpha = (1)/(2) = 0.5, alpha = 60^(@)`
`k = 1, cos alpha = (1)/(2)-(2)/(5) = 0.1, alpha = 84^(@)`
`k = 2, cos alpha = (1)/(2)-(4)/(5) =-0.3, alpha = 107.5^(@)`
`k = 3, cos alpha = (1)/(2)-(6)/(5) =-0.7, alpha = 134.4^(@)`
other values of `k` are not allowed as they lead to `|cos alpha|gt1`.