Past exam of the physics course of the year 1 of the natural sciences course of the University of Cambridge / 2016 / Paper 1 / Section D

Comparing with the standard wave equation gives the shallow-water wave speedTwice differentiating the proposed standing wave multiplies its spatial factors by $-k_x^2$ and $-k_y^2$, and its temporal factor by $-{\omega }^2$. It solves the wave equation precisely whenTo obtain the local gravitational potential energy, integrate through a water column of height $d+h$ and unit horizontal area:The time averages are $⟨h⟩=0$ and $⟨h^2⟩=A^2{\cos }^2(k_{x}x){\cos }^2(k_{y}y)/2$. ThusThe constant is retained because the specified zero of gravitational potential is the tank bottom. This is local time-averaged potential energy, not the sum of potential energy and kinetic energy.

The no-normal-flow wall conditions are Neumann boundary conditions. Differentiating the cosine factors at the opposite walls givesNegative integers give the same cosine modes. The formal $m=n=0$ case has $\omega =0$: it is a uniform change of the mean level, not a finite-frequency oscillation. In a closed tank with fixed water volume even that mean-level change is excluded from the oscillatory disturbance.

Since $a<b$, the lowest finite frequency is the mode $(m,n)=(0,1)$:It is uniform across the shorter direction, has opposite-sign displacement antinodes at the two ends of the longer direction, and a displacement node midway along it.

For the open-ended Bay of Fundy model, approximate the ocean mouth as a displacement node and the closed head as an antinode. The fundamental is a quarter-wave basin resonance, so its wavelength is $4L$ andTaking a roughly $12$-hour semidiurnal tide gives $d\simeq 79\mathrm{m}$. Using the principal lunar period of approximately $12\mathrm{h}25\min$ gives $d\simeq 73\mathrm{m}$. The idealized estimate is therefore about $7\text{–}8\times 10^1\mathrm{m}$. The lunar period is half of a roughly $24$-hour $50$-minute lunar day.^[ref]

This estimate treats the mouth as an ideal open boundary and replaces the real basin by a uniform depth. The supplied large tidal height motivates resonance; it is not enough by itself to determine damping or an amplification factor in this model.

Snell's law is $n_1\sin {\theta }_1=n_2\sin {\theta }_2$, with angles measured from the interface normal. The Huygens-Fresnel principle explains it by constructing the transmitted wavefront from secondary wavelets: during the same time interval, distances travelled in the two media are proportional to $c/n_1$ and $c/n_2$. The geometry of the advancing plane wavefront then gives $\sin {\theta }_1/\sin {\theta }_2=(c/n_1)/(c/n_2)=n_2/n_1$.

For incidence from index $n$ into vacuum, the transmitted angle is $\alpha =\arcsin (n\sin \theta )$. The ray bends away from the normal, so its positive deviation isFor $0<\theta <{\theta }_c=\arcsin (1/n)$,Thus the graph rises monotonically from zero to a finite endpoint with a vertical tangent:At the endpoint the transmitted ray grazes the interface. Beyond the critical angle for total internal reflection, no propagating transmitted ray exists and total internal reflection occurs.

Inside a prism, the two ray angles with the face normals sum to its apex angle. For the equilateral cross-section, they are $\theta$ and $\pi /3-\theta$. The entry deviation equals the corresponding exit deviation under ray reversal. Adding the two changes of direction therefore givesThis is symmetric about $\theta =\pi /6$. Symmetry alone gives a stationary point; to establish that it is a minimum, differentiate once more:within the transmitted-ray interval. The sum is therefore strictly convex, andThese expressions require $1<n<2$ for a non-grazing path through both faces; $n=2$ is a grazing limit, and $n>2$ admits no such two-face transmitted path. More explicitly, both internal angles must be at most ${\theta }_c$, which is possible only if $2{\theta }_c\ge \pi /3$.

Atmospheric hexagonal ice crystals can present pairs of faces with a $60^{\circ }$ angle, acting as small prisms. Their orientations turn the minimum-deviation cone around the Moon into an optical halo. No transmitted ray of this family is deflected less than the minimum deviation of a prism. Moreover, near the minimum, $\Delta -{\Delta }_{\min }$ is quadratic in the departure from the symmetric ray, so a range of incident directions is compressed into a small outgoing angular range. This caustic makes the inner rim sharp and bright, with the ray density decreasing farther outside. The common 22-degree halo has an angular radius near $22^{\circ }$ about either the Sun or Moon, hence a diameter near $44^{\circ }$.^[ref]