Fabry-Perot Etalon FSR & Finesse Calculator
Precision Optical Interferometry: Calculate Free Spectral Range ($FSR$), cavity finesse ($\mathcal{F}$), Airy transmission resonances, and spectral resolution ($FWHM$).
Etalon Geometry & Mirror Properties
Etalon FSR, Finesse & Line Resolution
Fabry-Perot Interferometry & Optical Cavity Resonances
Invented in 1899 by Charles Fabry and Alfred Perot, the etalon is the premier spectroscopic tool for high-resolution laser linewidth characterization and optical filtering.
1. Mathematical Formulations
FSR_ν = c / (2 · n · d · cos θ) ℱ = π · √R / (1 - R) Δν_FWHM = FSR_ν / ℱ T(δ) = T_max / [ 1 + (4R / (1-R)²) · sin²(δ/2) ]
2. Applications
- Single-Frequency Laser Selection: Inserting an intracavity etalon into a laser cavity filters out unwanted longitudinal modes.
- Telecommunications DWDM: Fixed solid silica etalons serve as 50 GHz or 100 GHz frequency comb reference channels.
Frequently Asked Questions
What is the Free Spectral Range (FSR) of a Fabry-Perot etalon?
The Free Spectral Range is the frequency separation between adjacent longitudinal transmission peaks: $FSR_\nu = \frac{c}{2 n d \cos\theta}$. In wavelength units, $FSR_\lambda \approx \frac{\lambda_0^2}{2 n d \cos\theta}$. It represents the unambiguous spectral measuring interval before transmission fringes repeat and overlap.
What does the finesse of an etalon mean?
Finesse $(\mathcal{F})$ is the ratio of the Free Spectral Range to the Full Width at Half Maximum linewidth of a transmission fringe: $\mathcal{F} = \frac{FSR}{\Delta\nu_{FWHM}} = \frac{\pi \sqrt{R}}{1 - R}$. It measures how many distinct spectral lines the interferometer can resolve within a single FSR, depending almost entirely on mirror reflectivity $R$ (and surface flatness).
How does cavity absorption or scattering affect peak transmission?
Even tiny residual losses in the mirrors degrade peak transmission drastically. If the mirror coating has absorption/scatter loss $A$, peak transmission is reduced from $100\%$ to $T_{max} = \left(1 - \frac{A}{1 - R}\right)^2$. For example, with $R = 99\%$ ($1-R = 0.01$), an absorption of just $0.1\%$ ($A=0.001$) cuts peak transmission to $(1 - 0.1)^2 = 81\%$.