= Solution
For a transverse periodic row of scatterers of spacing $d$, normal incidence makes their incident phases equal. Scattering at angle $\theta$ from that normal creates an outgoing path difference $d\sin\theta$ between neighbours. Constructive interference requires a phase difference $2\pi n$, giving
$$
\boxed{d\sin\theta=n\lambda.}
$$
Using this printed relation as the intended transverse <diffraction grating> model, the nonrelativistic <neutron> wavelength at $0.188\,\mathrm{eV}$ is $\lambda=6.5964\times10^{-11}\,\mathrm m$. Hence the first-order data give
$$
\boxed{d=\frac{\lambda}{\sin(0.117)}=5.651\times10^{-10}\,\mathrm m.}
$$
\b[Geometry qualification:] this is not a derivation from reflection at crystal planes parallel to the surface, as the source prose says. The ordinary <Bragg's law> for planes of normal separation $d$ is $2d\sin\theta_B=n\lambda$, where $\theta_B$ is the incident glancing angle to those planes. At normal incidence on parallel planes, $\theta_B=\pi/2$ and specular reflected rays leave along the normal, so their angle from the normal is zero; they cannot obey the printed positive-order relation $d\sin0=n\lambda$ for $\lambda>0$. This is a concrete counterexample to identifying the two geometries. The numerical spacing above is therefore conditional on the printed grating-like relation, rather than a forced assertion that the stated Bragg geometry proves it.
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