Bragg Reflection While reading about the experiment of x-ray spectrum and Bragg reflection, I saw somewhere that it is more accurate to measure the crystal lattice constant when considering the largest measurable Bragg angle. That choosing this angle minimize the error in the calculation of the lattice parameter.
My question is why is that true? And is there a proof to support it?
 A: The Bragg equation is $2d \sin \theta=n \lambda$, so you write
$d={n \lambda \over 2 \sin \theta}$
if you measure the angle $\theta$ you get a measurement of the lattice spacing $d$. You presumably know $\lambda$ very accurately so the uncertainty on the measurement of $d$ comes from the uncertainty on the measurement of $\theta$. (This assumes your equipment measures $\theta$ - it may measure $\tan \theta$ but this makes no difference to the argument.)
In such experiments $\theta$ is generally very small, so $1/\sin\theta$ is very large, and small changes in $\theta$ lead to big changes in $1/sin \theta$. You can show this for yourself by plugging a few values, or use the error formula
$\sigma_d=|\left( {\partial d \over \partial \theta}\right)| \sigma_\theta=\left( {n \lambda \cos \theta \over 2 sin^2 \theta} \right) \sigma_\theta=\left({2d^2 n \lambda \cos\theta \over n^2 \lambda^2 }\right)\sigma_\theta={2d^2 \cos\theta \over n\lambda}\sigma_\theta\approx{2d^2\over n \lambda} \sigma_\theta $   
Showing that $\sigma_d$ is proportional to $1/n$. At higher orders you measure a larger angle and the error you get from $1/\theta$ is smaller.
