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 / S3 / 1 / 3 / 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 $2\pi n$, givingUsing this printed relation as the intended transverse diffraction grating model, the nonrelativistic neutron wavelength at $0.188\mathrm{eV}$ is $\lambda =6.5964\times 10^{-11}\mathrm{m}$. Hence the first-order data giveGeometry 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 $d\sin 0=n\lambda$ 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.