Past exam of the physics A course of the University of Cambridge / 2018 / Paper 2 / Section D / D13

The convolution of two aperture functions in one dimension isWith Fourier transform convention $\hat{f}(q)=\int f(x)e^{-iqx}dx$, the convolution theorem statesThe first identity follows by substituting the convolution integral, putting $v=x-u$, and separating the resulting double integral into $\int f(u)e^{-iqu}du$ times $\int g(v)e^{-iqv}dv$. For an absolutely integrable pair the interchange is justified by absolute convergence; finite optical apertures satisfy suitable integrability assumptions.

In Fraunhofer diffraction, the observed field is the Fourier transform of the transmitted complex aperture field. For two identical slits, let $g(x)$ be one slit and use a pair of Dirac delta distributions to locate the centres:Consequently $\hat{t}(q)=2\hat{g}(q)\cos (qD/2)$. For a rectangular slit, $\hat{g}=a\mathrm{sinc}(qa/2)$, giving a single-slit diffraction envelope times the double-slit interference factor in intensity. A $\pi$ phase reversal replaces the plus between the deltas by a minus, giving $\sin (qD/2)$ instead, as in B7.

More generally, an $N$-slit array is a single-aperture function convolved with a finite comb. Its field contains the factorThis separates the narrow diffraction grating maxima from the individual-aperture envelope. The same reasoning holds in two transverse dimensions. Multiplying an existing aperture by an additional finite mask instead convolves their field transforms and can broaden features. These are operations on complex fields: intensity is the squared modulus of the final field and cannot generally be substituted into the field-level convolution theorem.

In thin-film interference, coherent beams reflected at the two faces of a transparent film overlap. Let the film have refractive index $n_f$, thickness $t$, and internal ray angle ${\theta }_f$, determined by Snell's law. For vacuum wavelength ${\lambda }_0$, the propagation part of the relative phase isAlthough the internal round-trip ray length is $2t/\cos {\theta }_f$, the two emergent rays must be compared on the same outgoing wavefront. Subtracting their external lateral path contribution gives the effective optical path difference $2n_{f}t\cos {\theta }_f$, explaining the cosine rather than its reciprocal.

Reflection also changes phase. At normal incidence in lossless positive-index media, the electric-field Fresnel equations give $r_{ij}=(n_i-n_j)/(n_i+n_j)$: reflection towards higher index reverses sign, whereas reflection towards lower index does not. If the two interfering reflections differ by one such phase reversal, their total relative phase is ${\delta }_{\mathrm{prop}}+\pi$ modulo $2\pi$. Then reflected maxima and minima obeyIf there are zero or two reversals, the conditions interchange. These bright/dark conditions assume overlapping beams with adequate temporal optical coherence; complete cancellation further requires equal amplitudes. In transmitted light, a lossless complete film has complementary intensity variations because reflected plus transmitted power is conserved.

For multiple internal reflections at normal incidence, summing the geometric series gives the exact film reflection amplitudeThis includes both the amplitude mismatch and repeated-beam interference. A normal-incidence antireflection coating on substrate index $n_s>n_0$ can cancel reflection at one wavelength when $n_f=\sqrt{n_0{n}_s}$ and $t={\lambda }_0/(4n_f)$: both interface reflections reverse sign and the film's round-trip propagation adds $\pi$, with equal interface amplitudes. A varying thickness or angle selects different maxima for different wavelengths, producing the colours of transparent films in white light. Broad bandwidth and large incidence angles generally prevent perfect cancellation across all wavelengths.

A dispersion relation specifies the allowed angular frequency $\omega (k)$ of a wavevector $k$. A single harmonic wave $A\cos (kx-\omega t)$ has constant-phase surfaces moving at the phase velocity,For two nearby components, use the identityThe slowly varying envelope moves at $\Delta \omega /\Delta k$. Taking a narrowband limit gives the group velocity,More generally, expand $\omega (k)$ around the central wavevector of a wavepacket. The linear term translates its envelope at $v_g$; the quadratic and higher terms produce group velocity dispersion and packet distortion. This interpretation needs a sufficiently narrow spectrum and weak enough absorption and distortion for an identifiable envelope to persist.

For light in a transparent material, $k=n(\omega )\omega /c$, soThe derivative is evaluated at the wavepacket's central frequency and ${\lambda }_0$ is the vacuum wavelength. In a nondispersive medium with constant $n$, the two velocities coincide and a packet can propagate without dispersion. In normal optical dispersion $dn/d{\lambda }_0<0$, the displayed model gives $v_g<v_p$.

The two velocities can differ even without material dispersion: a waveguide mode with ${\omega }^2={\omega }_c^2+c^2{k}^2$ has $v_p=\omega /k>c$ and $v_g=c^{2}k/\omega <c$, with $v_p{v}_g=c^2$. The phase velocity is not a signal speed. Near strongly absorbing resonances an anomalous derivative can also make the group velocity exceed $c$ or become negative; reshaping then prevents its simple identification with information transport. Relativistic causality constrains the signal front, not the speed of an arbitrary moving phase or distorted intensity maximum.