Sanai Ball

2022-10-08

Why is a cavity's free spectral range determined by the group refractive index?
Many sources give the free spectral range of a resonator as a function of the group refractive index ${n}_{g}$ by
$\mathrm{\Delta }{\nu }_{FSR}=\frac{c}{2{n}_{g}L}.$
I have a hard time understanding intuitively why the phase refractive index n is not used here, since cavity resonances occur when a monochromatic standing wave extends over the entire length of the cavity and the phase matching condition is met after one round trip. It appears to me that in this case there are no wave packets that are subject to dispersion.

Johnny Parrish

The group refractive index is defined by
${n}_{g}=\frac{c}{{v}_{g}}$
where ${v}_{g}=\frac{\mathrm{d}\omega }{\mathrm{d}k}$ is the group velocity. $\omega$ is the angular frequency and k is the wavevector in the medium which can be expressed as $k=n\frac{\omega }{c}$ for ordinary refractive index n. This then gives
${n}_{g}=c\frac{\mathrm{d}k}{\mathrm{d}\omega }=n+\omega \frac{\mathrm{d}n}{\mathrm{d}\omega }=n+{k}_{0}\frac{\mathrm{d}n}{\mathrm{d}{k}_{0}}$
where ${k}_{0}=\frac{\omega }{c}$ is the wavevector in vacuum.
Now consider the phase acquired after one round trip in the cavity,
$\varphi =2kL=2n{k}_{0}L$
with L the cavity length. If ${k}_{0}$ changes by a small amount $\mathrm{\Delta }{k}_{0}$ then $\varphi$ changes by
$\mathrm{\Delta }\varphi =2L\left(n+{k}_{0}\frac{\mathrm{d}n}{\mathrm{d}{k}_{0}}\right)\mathrm{\Delta }{k}_{0}=2L{n}_{g}\mathrm{\Delta }{k}_{0}$
which in terms of frequency $\nu =\frac{c{k}_{0}}{2\pi }$ is
$\mathrm{\Delta }\varphi =\frac{4\pi L{n}_{g}}{c}\mathrm{\Delta }\nu .$
The free spectral range $\mathrm{\Delta }{\nu }_{FSR}$ is defined such that $\mathrm{\Delta }\varphi =2\pi$ and hence we get
$\mathrm{\Delta }{\nu }_{FSR}=\frac{c}{2{n}_{g}L}.$

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