Past exam of the optics and electrodynamics course of the University of Cambridge / 2018 / Section B / B2

The five tasks below concern a scanned Michelson interferometer. The original dashed curves are labelled visibility in arbitrary units; divide them by their maximum to obtain normalized interferometric visibility, which lies between zero and one.

A non-polarizing beam splitter divides the incident spectrum into a fixed arm and an arm with a translating mirror. Recombining their returned fields gives a detector signal as a function of mirror position $x$. The round-trip optical path difference is $D=2x$ relative to the equal-arm position, so the delay is $\tau =2x/c$ in air approximated as vacuum. Record both detector signal and calibrated position, subtract the constant background, and take a Fourier transform of the interferogram.

For matched arm polarizations and balanced throughput, write the ideal signal aswhere $S(\omega )$ is the incident spectral power density and $C$ is a constant set by the beam splitter and detector. This is the cosine-transform form of Fourier transform spectroscopy, or equivalently the Wiener-Khinchin theorem applied to delayed self-interference.

The delay-range resolution in Fourier transform spectroscopy follows because recording a finite delay range multiplies the correlation by a window. Its transform broadens each spectral line by an instrumental width of order the reciprocal delay range. A mirror travel range $L$ creates a delay range of order $2L/c$, soAt fixed wavelength, absolute wavelength resolution therefore improves inversely with travel range. The numerical coefficient depends on whether $L$ is a one-sided or total travel, on the window or apodization, and on whether resolution means first zero, half width or another criterion. The proportionality is the requested convention-independent result. Position sampling must also be fine enough to sample the fastest fringe variation without aliasing.

Let the two arms deliver intensities $I_1,I_2$ with unit Jones vectors ${\mathbf{e}}_1,{\mathbf{e}}_2$. For a perfectly coherent input, the recombined intensity under relative phase $\varphi$ isConsequently the polarization overlap in interferometric visibility isFor equal returned intensities, two linear polarizations at angle $\theta$ have overlap $\cos \theta$ and thereforeChoose one linear direction as ${\mathbf{e}}_L=(1,0{)}^T$ and a circular polarization as ${\mathbf{e}}_C=(1,\pm i{)}^T/\sqrt{2}$. Their overlap has magnitude $1/\sqrt{2}$, givingA non-polarizing beam splitter does not by itself guarantee equal arm losses; if $I_1\ne I_2$, multiply either boxed polarization factor by $2\sqrt{I_1{I}_2}/(I_1+I_2)$. Perfect temporal coherence does not compensate for polarization or intensity imbalance.

For equal line intensities, the correlation is proportional to $e^{-i{\omega }_1\tau }+e^{-i{\omega }_2\tau }=2e^{-i\bar{\omega }\tau }\cos (\Delta \omega \tau /2)$. The two-line coherence beating envelope is therefore $|\cos (\Delta \omega \tau /2)|$.

In the original first plot, the first visibility zeros are at approximately $\tau =\pm 100\mathrm{ps}$; the envelope revives at $0$ and about $\pm 200\mathrm{ps}$. HenceUse the envelope recurrence or its nulls, not the much faster carrier fringes. The visibility zeros indicate equal component intensities in the ideal model; unequal intensities would leave a nonzero minimum. The arbitrary vertical normalization does not affect the inferred separation.

Interpret the stated ${\tau }_1=1\mathrm{ns}$ as the coherence-decay time, or the mean interval between fully phase-randomizing collisions in the simple impact model of pressure broadening. Then each line has correlation $e^{-|\tau |/{\tau }_1}{e}^{-i{\omega }_j\tau }$. The collisional phase relaxation and Lorentzian linewidth transform givesThe intrinsic spectrum is the sum of two such Lorentzian line shapes, still centred $5.0\mathrm{GHz}$ apart. With equal weights and balanced arms, its normalized visibility isFor the intensity normalization of the first source plot, the corresponding model interferogram isThe constant background remains $4$; only the oscillatory contrast decays. The beating zeros and recurrence spacing remain fixed, but successive visibility revivals shrink, reaching a factor $e^{-0.2}\simeq 0.819$ at $|\tau |=200\mathrm{ps}$ and $e^{-1}$ at $1\mathrm{ns}$. Long-delay fringes disappear toward the background.

The sketch chooses a carrier frequency for legibility; the envelope and linewidth follow from the inferred spacing and specified coherence time. A finite experimental delay window additionally convolves these intrinsic Lorentzians with the instrumental line profile. If ${\tau }_1$ were instead a population lifetime, the lifetime-limited field decay rate would be $1/(2{\tau }_1)$ and the corresponding linewidth would be half the boxed value. The source does not define that time convention explicitly; pressure-collision coherence decay is the interpretation used here.

For $N$ equal-intensity lines of spacing $\Delta \omega$, sum their correlations as a geometric series:The interferogram of an equally spaced spectral multiplet has normalized visibilityIn the second original plot, full revivals occur at approximately $\tau =0,\pm 100,\pm 200\mathrm{ps}$, soBetween $0$ and $100\mathrm{ps}$ there are three zeros, near $25$, $50$ and $75\mathrm{ps}$, and two small subsidiary lobes. Since the model has $N-1$ zeros between adjacent revivals,The peak envelope of about $8$ and the intensity background of about $8$ are consistent with four equal components in the source's normalization, but the internal zero count is the useful shape-based check. The inference assumes the equal spectral weights represented by the graph; revival spacing alone does not fix the number of lines, and arbitrary unequal or hidden components would require more spectral information. Delayed self-interference measures spectral powers, not the relative phases between different source frequencies, so these revivals alone do not prove that the source emits a phase-locked pulse train.