Past exam of the electromagnetism and optics course of the University of Oxford / 2016 / 9 / Solution

Let $\mathcal{R}=|r{|}^2$ be the single-mirror power reflectivity and ${\theta }_g$ the propagation angle inside the gas. Define $\lambda$ as the vacuum wavelength. Successive transmitted complex amplitudes differ by a round-trip factor $\mathcal{R}{e}^{i\delta }$, where, for real dielectric reflection coefficients whose product has positive sign,The cosine arises by comparing successive emerging rays on the same outgoing wavefront; it is not the literal zigzag distance $2d/\cos {\theta }_g$. Equivalently, the round-trip phase is $2k_{z}d$. If the source's $\theta$ is the angle in external air, Snell's law gives $n\sin {\theta }_g=\sin \theta$ andIf $\theta$ denotes the internal angle, simply set ${\theta }_g=\theta$.

For identical lossless mirrors, their power transmission is $1-\mathcal{R}$. In an energy-normalized amplitude convention this is $|t{|}^2=1-\mathcal{R}$; more general interface amplitudes require the corresponding wave impedance factors. The geometric series of transmitted amplitudes givesHenceThis is Fabry–Pérot transmission from multiple reflections.^[ref]

At normal incidence, for nondispersive $n$ and zero net reflection phase modulo $2\pi$, the resonance condition is $\delta =2\pi m$. The resonance frequencies and free spectral range areFor finite $\mathcal{R}<1$ the resonances have finite width, and transmission away from them is nonzero. Thus “only discrete frequencies” is an ideal high-reflectivity description of the selected resonances, not an exact statement about the support of the transmitted spectrum. For example, $\mathcal{R}=0.9$ at $\delta =\pi$ still gives $I_t/I_0=(0.1/1.9{)}^2\simeq 0.00277$.

At half the unit peak, $Q{\sin }^2(\delta /2)=1$. Measuring the phase displacement from a resonance yieldsThe last expression is the high-reflectivity approximation, with finesse $\pi \sqrt{\mathcal{R}}/(1-\mathcal{R})$. At $\mathcal{R}=0.9$, the full phase width is about $0.2109$ radians and $\Delta {\nu }_{\mathrm{FWHM}}/\Delta {\nu }_{\mathrm{FSR}}\simeq 0.03357$. At $\mathcal{R}=0.1$, the minimum transmission is $0.6694$, so the trace never reaches half the peak and this full width at half maximum is not defined. The exact half-maximum condition requires $\mathcal{R}\ge 3-2\sqrt{2}$.

For the absorbing Fabry–Pérot mirror phase and transmission, write $r=\sqrt{\mathcal{R}}{e}^{i{\varphi }_r}$ and let $\mathcal{T}$ be the power transmission of one mirror. Optical absorption makes $\mathcal{R}+\mathcal{T}<1$. The transmitted intensity becomesHere the phase formula is at normal incidence. The peak is reduced below unity, while a change in $\mathcal{R}$ changes the width. The complex reflection phase also moves the resonances: if ${\varphi }_r$ is approximately constant,If either $n$ or ${\varphi }_r$ depends on frequency, this is an implicit condition and the free spectral range need not be constant. No numerical shift is determined without the mirrors' optical response.