라우에식 -> 브레그식 과정중 이해 안되는거 있는데요...

If $\mathbfNO$ is the reciprocal lattice vector, we know $\mathbf{G}\cdot (\mathbf{a}+\mathbf{b}+\mathbf{c})=2\pi (h+k+l)$ . The Laue equations specify $\mathbf{\Delta k}\cdot (\mathbf{a}+\mathbf{b}+\mathbf{c})=2\pi (h+k+l)$ . Whence we have $\mathbf{\Delta k}=\mathbf{G}$ or $\mathbf{k}_o - \mathbf{k}_i = \mathbf{G}$ .

From this we get the diffraction condition:

$\begin{align} \mathbf{k}_0 - \mathbf{k}_i &= \mathbf{G}\\ (\mathbf{k}_i + \mathbf{G})^2 &= \mathbf{k}_0^2\\ {k_i}^2 + 2\mathbf{k}_i\cdot\mathbf{G} + G^2 &= {k_0}^2 \end{align}$

Since $(\mathbfNO$ (considering elastic scattering) and $\mathbf{G} = -\mathbf{G}$ (a negative reciprocal lattice vector is still a reciprocal lattice vector):

$2\mathbfNO$ .

The diffraction condition $\;2\mathbfNO$ reduces to the Bragg law $\;2d\sin\theta =n\lambda$

$\;2\mathbfNO$ 여기서 어떻게 $\;2d\sin\theta =n\lambda$ 이식이 되조 ㅠ .ㅠ 물리학 고수 이신 오유 형님 갈켜주십쇼 ㅠㅠ.