Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2015 / S3 / 2

Using $m\simeq 4m_p$ and $T=15\mathrm{meV}$, the nonrelativistic de Broglie wavelength and speed areFor the square interferometer, the enclosed area projected into a vertical plane is $a^2\sin \theta$. The gravitational phase shift in matter-wave interferometry is $\Delta \phi =mg{a}^2\sin \theta /(\hslash v)$ to leading order in the small gravitational energy change. Thus rotation from horizontal to vertical changes the fringe count byOne way to see the phase is that the horizontal arms differ in gravitational potential energy by $mga\sin \theta$ for a flight time $a/v$. Which-path detection entangles the atom with distinguishable detector states. If those states are orthogonal and the path record is ignored, their coherence is lost and the interference fringes disappear. This is a loss of interference visibility, not a change in the atom's de Broglie wavelength.

For the macroscopic mass,The ideal fixed-momentum expression tends to infinity as $v\to 0$. An everyday localized object is not a perfectly coherent momentum eigenstate, however: its thermal and environmental momentum uncertainty remains nonzero, and decoherence rapidly suppresses spatially separated coherent alternatives. Merely setting the mean velocity to zero does not prepare an object with a macroscopic coherence length.

For a uniformly illuminated single slit of width $w$, the far-field amplitude is proportional to $\sin u/u$, with $u=\pi w\sin \vartheta /\lambda$. Its first zeros occur at $\sin \vartheta =\pm \lambda /w$. If the quoted peak width is the full width between those zeros, the small-angle geometry givesThe Fraunhofer diffraction approximation is appropriate since $w^2/(\lambda L)\ll 1$. If the supplied width instead meant a half-width or full width at half maximum, its numerical conversion factor would differ; the displayed answer states the first-zero convention explicitly.