Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2012 / S18 / 2 / 3 / Solution

Apart from the finite forward limit, amplitude zeros obey $\sin x-x\cos x=0$, or $\tan x=x$, where $x=q{r}_0$. The first positive root is $x_1\simeq 4.49341$. Since the accessible range is $0\le x\le 2k{r}_0$,Thus there are no zeros for $k<k_c$; the first enters at the largest momentum transfer, $\theta =\pi$, when $k=k_c$.

For larger roots, $x_n$ lies just below $a_n=\pi (n+\frac{1}{2})$, $n=1,2,\dots$. Expanding $\tan (a_n-\epsilon )\simeq 1/\epsilon$ gives $x_n=a_n-1/a_n+O(a_n^{-3})$. Keeping its leading term yields the requested diffraction estimateExactly, replace the numerator $\pi (n+\frac{1}{2})$ by the corresponding root $x_n$. The displayed approximation is a large-$q{r}_0$ root approximation: taking $k$ large at fixed low order does not make $x_1$ equal to $3\pi /2$. For example, the exact first angle at high $k$ is approximately $4.49341/(k{r}_0)$, not $4.71239/(k{r}_0)$.