Past exam of the physics A course of the University of Cambridge / 2011 / Paper 1 / Section A

For weak damping, the ringdown comparison bound for two resonator modes gives $I(t)=I(0)e^{-\omega t/Q}$. Interpreting the stated loudness comparison as an intensity comparison,Therefore the information gives an upper bound, not a unique value:Equality would describe equal intensities at two seconds; the stated strict inequality allows every smaller positive $Q_L$ consistent with the weak-damping model.

By linear scalar diffraction theory, complementary aperture transmissions $t$ and $1-t$ give $U_t+U_{1-t}=U_{\mathrm{open}}$ under identical illumination. In Fraunhofer diffraction, away from the direct forward beam $U_{\mathrm{open}}=0$, so $U_{1-t}=-U_t$: the complementary screens have identical off-axis diffraction intensities. This is Babinet's principle. Their complex amplitudes, rather than their intensities, add. The direct beam is the exception; real conducting-edge vector effects are outside this ideal scalar statement.

For source and screen distances $s,r$, the paraxial optical path excess at aperture radius $u$ is $u^2(1/s+1/r)/2$. The outer boundary of Fresnel half-period zone $n$ has excess $n\lambda /2$, henceThis is the outer radius of the second zone, measured from the axis.

A conventional body-centred cubic cell contains two atoms, so its edge $a$ obeys $\rho a^3=2(23m_u)$. Taking $m_u=1.66054\times 10^{-27}\mathrm{kg}$ gives $a\simeq 4.29\times 10^{-10}\mathrm{m}$. The neighbour shells of a body-centred cubic lattice have eight nearest neighbours that are at $\sqrt{3}a/2$; the six next-nearest neighbours lie along the cell axes. Thus the next-nearest-neighbour distance is $a\simeq 0.429\mathrm{nm}$.

For a free-electron Fermi circle in a square lattice, include two spin states per wavevector. A square cell of side $a$ supplies $z/a^2$ electrons per unit area; counting the disk of occupied states givesThe first Brillouin zone is $|k_x|,|k_y|\le \pi /a$. For $z=1$, $k_F=2.507/a$, so the circular Fermi surface lies wholly inside this square and encloses half its area. The figure uses the stated free-electron approximation; the lattice symmetry alone does not prescribe an exact circular surface.

For $z=2$, $k_F=\sqrt{4\pi }/a=3.545/a$. The circle encloses one whole zone's area, crosses each face because $k_F>\pi /a$, and misses the corners because $k_F<\sqrt{2}\pi /a$. In a reduced Brillouin zone, fold the portions outside the square into neighbouring band branches. A weak periodic potential opens band gaps at crossings and deforms these pieces; divalence alone does not establish a band insulator.