fix minus signs in Braggs law derivation

src/10_xray.md

@@ -582,7 +582,7 @@ $$
 Using $\mathbf{k} \cdot \mathbf{G} = \lvert \mathbf{k} \rvert \lvert \mathbf{G} \rvert \cos(\phi)$,  we obtain
 
 $$
-\left| \mathbf{G} \right| = 2 \lvert \mathbf{k} \rvert \cos (\phi).
+\left| \mathbf{G} \right| = -2 \lvert \mathbf{k} \rvert \cos (\phi).
 $$
 
 Note, $\phi$ is the angle between the vector $\mathbf{k}$ and $\mathbf{G}$, which is not the same as the angle between the incoming and scattered waves $\theta$.
@@ -594,17 +594,17 @@ It turns out that Miller planes are normal to the reciprocal lattice vector $\ma
 Substituting the expression for $\lvert \mathbf{G} \rvert$ into the expression for the distance between Miller planes we get:
 
 $$ 
-d_{hkl} \cos (\phi) = \frac{\pi}{\lvert \mathbf{k} \rvert}.
+d_{hkl} \cos (\phi) = -\frac{\pi}{\lvert \mathbf{k} \rvert}.
 $$ 
  
 We know that $\lvert \mathbf{k} \rvert$ is related to the wavelength by $\lvert \mathbf{k} \rvert = 2\pi/\lambda$.
 Therefore, we can write the equation above as
 
 $$ 
-2 d_{hkl} \cos (\phi) = \lambda.
+2 d_{hkl} \cos (\phi) = -\lambda.
 $$ 
 
-Lastly, we express the equation in terms of the deflection angle through the relation $\phi = \theta - \pi/2$.
+Lastly, we express the equation in terms of the deflection angle through the relation $\phi = \theta + \pi/2$.
 With this, one can finally derive **Bragg's Law**:
 
 $$