advanced30 min studyLesson 219 in Physics

Fabry–Perot resonances and finesse

Sum repeated transmitted fields and relate intensity reflectivity, free spectral range, and resonance linewidth under explicit ideal-cavity assumptions.

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What you’ll understand

  • Distinguish a mirror amplitude coefficient from intensity reflectivity.
  • Derive the Airy transmission from a geometric field series.
  • Calculate free spectral range and approximate finesse without mixing width conventions.

Two partially reflecting mirrors form a frequency-selective instrument because transmitted light can emerge after many different round trips. Those contributions interfere, producing narrow transmission peaks when their phases agree. A Fabry–Perot therefore refines the two-beam interferometer by adding a long sequence of delayed fields. The prediction depends on mirror amplitude coefficients, while mirror specifications usually quote intensity reflectivity; confusing them changes both resonance contrast and linewidth.

A round trip multiplies an amplitude twice

Let each identical lossless mirror have intensity reflectivity R=∣r∣2R=|r|^2 and intensity transmissivity 1−R1-R, with equal external impedances and no absorption. The magnitude of the single-mirror field reflection coefficient is R\sqrt R. A complete internal round trip encounters two mirrors, so its field attenuation magnitude is ∣r1r2∣=R|r_1r_2|=R, while its surviving intensity fraction is R2R^2. These three quantities refer to different operations and should never share an unqualified symbol.

Define δ\delta as the total round-trip phase including any constant reflection phases. The first transmitted field is followed by copies multiplied by ReiδRe^{i\delta}, so the coherent field series is geometric. Its denominator becomes small near phase agreement. A symmetric lossless cavity has unit peak transmission even when each mirror transmits little individually: the steady internal field builds until the emerging fields satisfy both interference and energy conservation. Loss, unequal mirrors, or mode mismatch can reduce that peak.

IoutIin=T(δ)=(1−R)2∣1−Reiδ∣2=11+4R(1−R)2sin⁡2(δ/2),R=∣r∣2\frac{I_{\mathrm{out}}}{I_{\mathrm{in}}}=T(\delta)=\frac{(1-R)^2}{|1-Re^{i\delta}|^2}=\frac{1}{1+\frac{4R}{(1-R)^2}\sin^2(\delta/2)},\qquad R=|r|^2
T and R are intensity ratios; r is a dimensionless field reflection coefficient and delta is in radians. This Airy function assumes identical lossless mirrors, coherent steady illumination, a single matched spatial mode, equal exterior media and fixed polarization.↔ Scroll wide equations horizontally.

Spacing and linewidth are independent scales

For normal incidence in a nondispersive cavity of length LL and refractive index nn, round-trip propagation phase grows as 4πnLν/c4\pi nL\nu/c. Increasing frequency until this phase advances by 2π2\pi reaches the next longitudinal resonance. The resulting free spectral range is a frequency spacing, not a linewidth. Strong material or mirror dispersion changes the derivative of round-trip phase, so a group-delay calculation must replace the simple nondispersive expression.

Near a high-reflectivity peak, approximate the sine of the small phase detuning by its argument. The half-maximum points give a full phase width approximately 2(1−R)/R2(1-R)/\sqrt R. Dividing the 2π2\pi resonance spacing by this full width gives finesse πR/(1−R)\pi\sqrt R/(1-R). The square root appears because R is an intensity reflectivity. This is a narrow-peak approximation; using a half width as the full width introduces a separate factor-of-two error.

ΔνFSR=c2nL,F=ΔνFSRΔνFWHM≃πR1−R,ΔνFWHM≃c(1−R)2nLπR\Delta\nu_{\mathrm{FSR}}=\frac{c}{2nL},\qquad \mathcal F=\frac{\Delta\nu_{\mathrm{FSR}}}{\Delta\nu_{\mathrm{FWHM}}}\simeq\frac{\pi\sqrt R}{1-R},\qquad \Delta\nu_{\mathrm{FWHM}}\simeq\frac{c(1-R)}{2nL\pi\sqrt R}
The frequency quantities are in hertz, L is in metres, n is dimensionless and c is vacuum light speed in m/s. Finesse is dimensionless. The final two approximate formulas require sufficiently narrow resonances; the FSR formula additionally assumes normal incidence and negligible dispersion.↔ Scroll wide equations horizontally.

Longer storage narrows a resonance

High reflectivity lets many delayed copies contribute appreciably, extending the time over which the cavity remembers a coherent input. A longer temporal response corresponds to a narrower spectral response. This is a relation between the steady frequency response and the passive decay of a mode, not evidence that changing mirror reflectivity during a run implements the same stationary experiment. A rapidly scanned source may also fail to reach the steady Airy response if the scan outruns the cavity settling.

Increasing cavity length at fixed mirrors reduces both free spectral range and linewidth by the same factor, leaving ideal finesse unchanged. Increasing reflectivity instead raises finesse and narrows the peaks relative to their spacing. Finesse is therefore different from resolving power at an optical carrier, which compares that carrier frequency with a linewidth. An instrument can have closely packed resonances, and order ambiguity must still be addressed when using it to interpret an unknown spectrum.

A ten-centimetre air cavity

A symmetric lossless air-spaced Fabry–Perot has length 0.10 m0.10\,\mathrm m and each mirror has intensity reflectivity R=0.81R=0.81. Using c=3.0×108 m/sc=3.0\times10^8\,\mathrm{m/s} and n=1n=1, estimate free spectral range, finesse, and full resonance linewidth. Identify the round-trip field and intensity factors.

  1. The single-mirror reflection magnitude is 0.81=0.90\sqrt{0.81}=0.90. Two reflections give round-trip field magnitude 0.81 and round-trip surviving intensity 0.812=0.65610.81^2=0.6561, before any additional loss.

  2. The free spectral range is (3.0×108)/(2×0.10)=1.5×109 Hz(3.0\times10^8)/(2\times0.10)=1.5\times10^9\,\mathrm{Hz}. The narrow-peak finesse estimate is π(0.90)/(0.19)≃14.88\pi(0.90)/(0.19)\simeq14.88, using intensity R in the denominator.

  3. Divide the 1.5 GHz spacing by 14.88 to obtain full width approximately 1.01×108 Hz1.01\times10^8\,\mathrm{Hz}, or 101 MHz. This is the full width at half maximum, not the detuning from the centre to one half-maximum point.

The result

The ideal estimates are FSR 1.5 GHz, finesse 14.9 and FWHM 101 MHz; the round-trip field factor is 0.81 and intensity factor is 0.6561.

Explore the model

Interactive lab

Interactive lab · oscillator

A passive memory analogy

For each fresh comparison, Reset and set BOTH Damping ratio ζ and Natural frequency ω₀. Compare ζ=0.10 and ζ=0.50 with ω₀=2, predicting more visible oscillations for the smaller damping, then Play. This is a unit-mass mechanical oscillator starting at x=1 and velocity=0, not a driven optical cavity.

A passive memory analogyUnit mass, x(0) = 1 and v(0) = 0. The analytic solution includes underdamped, critically damped and overdamped cases; time is measured in seconds. Current readings: Displacement x(t): 1; Velocity v(t): 0; Regime: underdamped. All parameters are adjustable with the labeled controls below.02.557.510−1.15−0.5700.571.15Time · sDisplacementunderdamped
Displacement x(t)Current time
Displacement x(t)
1
Velocity v(t)
0
Regime
underdamped
t = 0 s

Unit mass, x(0) = 1 and v(0) = 0. The analytic solution includes underdamped, critically damped and overdamped cases; time is measured in seconds. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forLonger passive persistence motivates a narrow-resonance analogy. The lab has no mirror reflectivity, drive-frequency scan, Airy transmission, or finesse readout.

Check your understanding

Each identical mirror has intensity reflectivity R. Ignoring other losses, what is the magnitude of the intracavity field factor after one complete round trip?

Try it yourself

A symmetric lossless cavity has mirror intensity reflectivity R=0.60R=0.60. Find its antiresonant transmission at δ=π\delta=\pi. If its length doubles with the same mirrors and negligible dispersion, what happens to free spectral range and finesse?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use the exact Airy function at the midpoint between adjacent resonances. Distinguish frequency scale changes from the dimensionless mirror-dependent contrast.

Compare with the solution

At antiresonance the transmission is (1−R)2/(1+R)2=(0.40/1.60)2=1/16=0.0625(1-R)^2/(1+R)^2=(0.40/1.60)^2=1/16=0.0625. Thus 6.25 percent is transmitted, despite unit peak transmission at resonance in this ideal symmetric lossless model. Doubling length halves the free spectral range because the round-trip time doubles. The Airy curve as a function of round-trip phase is unchanged, so finesse remains unchanged and its frequency linewidth halves wherever an FWHM is defined. No high-reflectivity approximation is needed for the antiresonant transmission calculation.

Take it with you

  • Single-mirror field reflectivity, round-trip field attenuation, and intensity survival are different factors.
  • Finesse compares resonance spacing with a full width at half maximum.
  • Changing length rescales the frequency axis; changing mirror reflectivity alters the relative sharpness.

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