What exactly do we mean by 'Free spectral range'?

Question

While reading about Fabry perot interferometers, we conclude that transmission can only happen when twice optical length of the cavity is equal to an integer multiple of the wavelength of the incident light. Elaborating further, the source converts this same statement in the form of 'cavity's free spectral range . I understood how the formula is derived but I did not get the physical meaning of what exactly is the free spectral range. Any help will be appreciated.

Answer 1

EDIT: I just noticed you inquired about the Fabry-Perot interferometer and my answer focused on the Fabry-Perot cavity as a laser cavity. The explanations down here however applies in the same way to the interferometer.

A Fabry-Perot cavity supports modes whose wavelength is a multiple of twice the length of the cavity, thus fulfilling the $2\pi$-phase criterion.

This means that several modes of the cavity exists and not just the center wavelength. The free spectral range (FSR) of the cavity is thus the spectral distance between supported modes in the cavity. The gain medium in the laser then would typically in conventional laser stretch and cover several of these modes. If the laser is investigated with an OSA or a Fabry-perot interferometer, these modes can be detected.

The FSR is for a simple Fabry-Perot laser then defined as $\frac{c}{2L}$ where c is the speed of light in the medium of propagation (adjusted for refractive index n).

Answer 2

The term 'free spectral range' can be applied to any spectroscopic instrument, but it is especially relevant to instruments such as diffraction grating and Fabry Perot etalon where the instrument produces the spectrum two or more times (as a function of some parameter such as angle or distance) and there is thus a risk that one copy of the spectrum overlaps another, causing ambiguity.

For the diffraction grating the $p$'th order diffraction peak for light of wavelength $\lambda_1$ will appear at the same angle as the the $(p+1)$'st order diffraction peak of light of wavelength $\lambda_2$ when $$ d \sin \theta = p \lambda_1 = (p+1) \lambda_2. $$ Hence $$ \frac{1}{\lambda_1} = \frac{p}{d \sin \theta},\\ \frac{1}{\lambda_2} = \frac{p+1}{d \sin \theta}. $$ Define $$ \Delta_{\rm FSR} = \frac{1}{\lambda_2} - \frac{1}{\lambda_1} = \frac{1}{d \sin \theta}. $$ This is the free spectral range for the diffraction grating.

For the Fabry Perot the $p$'th order interference peak for wavelength $\lambda_1$ overlaps the $(p+1)$'st order for $\lambda_2$ when $$ 2 d \cos(\theta) = p \lambda_1 = (p+1) \lambda_2. $$ Hence we get $$ \Delta_{\rm FSR} = \frac{1}{\lambda_2} - \frac{1}{\lambda_1} = \frac{1}{2 d \cos \theta}. $$ This is the free spectral range of the Fabry Perot etalon, used at normal incidence, when observing the output at angles near $\theta$. The case $\theta = 0$ is often quoted, and then we get $$ \Delta_{\rm FSR} = \frac{1}{2 d}. $$ The point is that if you shine a spectrum of light into the instrument, then as long as the range of wavenumbers in the spectrum is less than $\Delta_{\rm FSR}$ then you have a good chance of interpreting the output correctly. But if the spectrum is wider than this then the overlapping interference orders make it harder to interpret the output.

(To convert wavenumber to frequency, multiply by $c$.)

Considered as a spectroscopic instrument, the Fabry Perot etalon has excellent resolution but a low free spectral range. For this reason, one often uses it in conjunction with a narrow bandpass filter so as to reduce the width of the spectrum sent as input to the etalon.