(5-1) ① You can see on the diagram below that the wavelength (λ) and the distance between the layers of atoms (d) form two of the sides of a right angled triangle, as highlighted by the thicker lines on the diagram. The picture below picks out this triangle so that we can focus more easily on the geometry.
Embedded Files
Making use of the right angles, it’s clear that the angle between the incident or reflected x-ray and the planes (θ) is equal to the angle (also, therefore, marked as θ) at the top of the right-angled triangle.
This allows us to use trigonometry to derive the relationship between d, θ and the extra distance (Δ) travelled by the reflected wave along path A
Δ = d sinθ
Since, by symmetry, there will be the same pathlength difference on the incident side, the total pathlength difference between the waves following paths A and B is 2Δ. Hence we can write:
Pathlength difference = 2d sinθ.
② In order for the reflected waves emerging along lines A and B to be still in phase, the pathlength difference must equal a whole number of wavelengths. Therefore the condition for an intensity maximum is:
nλ = 2d sinθ where n is a whole number
In turn this means that sin θ = nλ/d, which means that a series of intensity maxima at different angles θ will result.
③ If the waves were out of phase as they emerged, destructive interference and loss of intensity would result. This will be maximal when one is phase-shifted by exactly half a wavelength relative to the other. This means that the pathlength difference must be an odd number of half-wavelengths:
(n - ½) λ = 2d sinθ
Page updated
Google Sites
Report abuse