Past exam of the physics A course of the University of Cambridge / 2015 / Paper 2 / Section A

The photon relation gives $\lambda =hc/E=2.4797\times 10^{-12}\mathrm{m}$. The diffraction-limited angular scale of an aperture of diameter $D$ is thereforeFor a uniformly illuminated circular aperture, the Rayleigh criterion instead gives $1.22\lambda /D=3.03\times 10^{-11}\mathrm{rad}$. Both express the same requested estimate. This is an ideal diffraction limit, conditional on the effective diameter behaving as a coherent imaging aperture; photon energy and diameter alone do not specify the performance of a practical gamma-ray telescope.

At distance $L$, the small-angle approximation gives $\ell =L\theta$. Thus the transverse resolution is approximately $2.5\times 10^{10}\mathrm{m}$, or $3.0\times 10^{10}\mathrm{m}$ with the circular-aperture Rayleigh criterion. This is a transverse separation at the source, rather than a determination of its distance along the line of sight.

Let $T$ be the common tension and ${\mu }_j$ the linear mass densities. The characteristic impedance of a stretched string is $Z_j=T/c_j=\sqrt{T{\mu }_j}$. Continuity of displacement and transverse force at a massless junction gives $1+r=t$ and $Z_1(1-r)=Z_{2}t$. ConsequentlyThe reflected power is $7.18\%$. The minus sign of $r$ represents a phase reversal, not negative power. Both incident and reflected waves travel in the first string, so their power ratio is simply the squared displacement-amplitude ratio. As a check, $\mathcal{T}=(Z_2/Z_1)|t{|}^2=0.92820$ and $\mathcal{R}+\mathcal{T}=1$.

Compare a directly transmitted ray emerging at $A$ with a ray that makes one extra round trip inside the Fabry–Pérot interferometer and emerges at $C$. Both emerge parallel at angle $\theta$ to the plate normal. The extra internal distance is $2d/\cos \theta$, and their emergence points differ laterally by $AC=2d\tan \theta$. The first ray has already travelled $AC\sin \theta$ towards a common outgoing wavefront when the second reaches $C$. The air-gap optical path difference in a Fabry–Pérot interferometer follows because optical path length equals geometric path length in the air-gap approximation:The subtraction is essential: comparing distances only inside the plates would compare different outgoing wavefronts. The corresponding propagation phase difference is $2\pi \Delta /\lambda$. Coating reflection phases can add a separate phase offset; they do not change this geometric optical path difference.

Count two electron spin states in each reciprocal-space cell of volume $(2\pi {)}^3/V$. Filling a Fermi sphere at zero temperature givesThe free electron model then givesThe energy zero is the bottom of the model's conduction band. The supplied carrier density already incorporates the one-electron-per-atom assumption; it must not be multiplied by another atomic count.

Write $a=V_c^{1/3}$ for the primitive cubic lattice spacing. With wavevector convention $e^{i\mathbf{k}\cdot \mathbf{r}}$, reciprocal-lattice primitive vectors have length $2\pi /a$. The first Brillouin zone is the cube $-\pi /a\le k_j\le \pi /a$, henceIt has the same volume as any primitive reciprocal unit cell. A reciprocal coordinate defined in cycles per metre would instead give $1/V_c$; that is a different convention from this $k$-space convention.