Past exam of the optics and electrodynamics course of the University of Cambridge / 2018 / 1

The London penetration depth measures the density-to-mass ratio of carriers participating in superconducting screening. With electron-counted density $n_s$ and the stipulated bare electron mass $m_e$,Using $m_e=9.109\times 10^{-31}\mathrm{kg}$, $e=1.602\times 10^{-19}\mathrm{C}$, ${\mu }_0\simeq 4\pi \times 10^{-7}\mathrm{H}{\mathrm{m}}^{-1}$ and ${\lambda }_L=22\times 10^{-9}\mathrm{m}$ givesThis is the requested conduction-electron density within the assumed local London equation model. More precisely, it is the superconducting carrier density from London screening; identifying it with all conduction Electrons assumes that they participate in screening at the measurement temperature. No temperature-dependent normal-carrier fraction is supplied.

If the condensate is instead counted as Cooper pairs, the pair mass is $2m_e$ and charge magnitude is $2e$, giving pair density $n_{\mathrm{pair}}=n_s/2\simeq 2.9\times 10^{28}{\mathrm{m}}^{-3}$. Keeping Electron and pair counting consistent avoids a spurious factor of two.

At normal incidence, continuity of tangential wave phase makes both internal wavevectors normal to the parallel faces. For the ordinary wave in a uniaxial optical medium, the electric field is an eigenpolarization of the permittivity tensor, so $\mathbf{D}\parallel \mathbf{E}$. ThusIts energy path is also normal to the faces. The extraordinary wave differs because $\mathbf{D}\perp \mathbf{k}$ but $\mathbf{E}$ need not be perpendicular to $\mathbf{k}$. In a nonmagnetic crystal, $\mathbf{H}\parallel \mathbf{B}$ and $\mathbf{S}=\mathbf{E}\times \mathbf{H}$ is perpendicular to $\mathbf{E}$. The vectors $\mathbf{D},\mathbf{E},\mathbf{k},\mathbf{S}$ lie in the same meridional plane, with $\mathbf{D}\perp \mathbf{k}$ and $\mathbf{E}\perp \mathbf{S}$. Therefore the electric-displacement angle from birefringent walk-off equals the angular separation between $\mathbf{S}$ and $\mathbf{k}$.

The observed lateral separation is accumulated over thickness $L$, so $\Delta =L\tan {\alpha }_e$. With $L=10\mathrm{mm}$ and $\Delta =0.1\mathrm{mm}$,The two emergent rays are parallel because their outgoing wavevectors are both normal to the faces in air; their internal energy paths can nevertheless be displaced by birefringent walk-off. The angle quoted is the positive angle between the field vectors, so the unspecified lateral direction does not affect its value. The unpolarized light excites both eigenpolarizations; it does not require their energy paths to coincide.

First obtain the particle speed from kinetic energy rather than treating $1\mathrm{GeV}$ as its total energy. With Electron rest energy $m_e{c}^2=0.510999\mathrm{MeV}$,For Cherenkov radiation, the photon propagation direction makes angle ${\theta }_C$ to the particle trajectory, with $\cos {\theta }_C=1/(n\beta )$. The speed correction is negligible at three significant digits, but the Cherenkov photon-cone and wavefront-cone angles must be distinguished.

If the given opening $\alpha$ is the full cone of photon directions, ${\theta }_C=\alpha /2=48.75^{\circ }$. The result isIf instead $\alpha$ denotes the full opening of the conical wavefront, its half-angle is complementary to the photon direction: ${\theta }_C=90^{\circ }-\alpha /2=41.25^{\circ }$. That convention givesThe original PDF supplies no cone diagram or definition distinguishing these geometries. Thus the literal full photon-cone interpretation gives $1.52$, while $1.33$ follows from the wavefront interpretation. It would be incorrect to interchange sine and cosine without stating which cone is meant.