Physical intuition for FSR dependence on cavity length d in a scanning Fabry-Pérot interferometer

[eneacasucci]

I am analyzing data from a Brillouin spectroscopy experiment using a scanning Fabry-Pérot Interferometer (FPI). The source is a monochromatic laser with a fixed wavelength $\lambda$.

I have a conceptual question regarding the relationship between the mechanical piezo scan and the Free Spectral Range (FSR) in frequency.

1. Spatial Periodicity: In a scanning FPI, the resonance condition repeats whenever the mirror separation changes by $\Delta d_{scan} = \lambda/2$. This spatial interval is constant and effectively independent of the macroscopic cavity length ($d$).
2. Spectral Periodicity: The theoretical definition of FSR in frequency is $\Delta \nu_{FSR} = c/2d$. This indicates an inverse dependence on the cavity length $d$.

My Question:If the mechanical displacement required to cover one full interference order is fixed ($\lambda/2$) regardless of $d$, what is the physical intuition for why the frequency interval covered by this displacement changes with $d$?

source:

In other words, why does increasing the macroscopic cavity length $d$ cause the same physical mirror displacement ($\lambda/2$) to correspond to a smaller frequency bandwidth (GHz)? I am trying to reconcile the "mechanical" view (fixed scan length) with the "spectral" view (variable FSR).

 

[Ibix]

They're different things. In the mechanical scan you're holding the wavelength constant and varying the cavity length, while in the spectral case you're holding the cavity length constant and varying the wavelength. In either case, you get a gap between resonance peaks by calculating the condition for one more/less half wave in the cavity, but this leads in different directions because you're changing different things.

In the spatial case, you have waves of fixed wavelength so you get resonance every time you change the length of the cavity by a half wavelength. The fixed wavelength dictates the cavity length change.

In the spectral case you have a cavity of fixed length, so you get resonance when you change the input wavelength to fit in one more half wave. The fixed cavity length dictates the resonant wavelengths. It's perhaps easiest to visualise for the fundamental and first harmonic modes of a cavity, which have wavelengths $2d$ and $d$ respectively (not realistic for optical wavelengths, obviously!). The wavelength gap is obviously $d$.

 

[eneacasucci]

They're different things. In the mechanical scan you're holding the wavelength constant and varying the cavity length, while in the spectral case you're holding the cavity length constant and varying the wavelength.

Thanks for the answer. The distinction of holding different parameters fixed is clear to me now.
However, I am still confused about why, in a Brillouin experiment where we mechanically scan the cavity (varying $d$), the Free Spectral Range (FSR) is defined using the formula derived for a fixed cavity length:

$\Delta\lambda_{FSR} = \frac{\lambda_0^2}{2 \cdot n \cdot d}$

In the experiment, looking at the software, we see two intense Rayleigh peaks whose distance is identified as the FSR. Physically, these are not different wavelengths resonating in a fixed cavity; they are the same wavelength ($\lambda_0$) resonating at different cavity lengths (consecutive interference orders).

 

[A.T.]

I am still confused about why, in a Brillouin experiment where we mechanically scan the cavity (varying $d$), the Free Spectral Range (FSR) is defined using the formula derived for a fixed cavity length.

Unless you are varying d at a substantial fraction of c, you are effectively doing repeated experiments with a fixed d. A formula just states a relationship between the quantities it contains, not which of the quantities are to be changed during repeated experiments to check the formula.

 

[tech99]

If you change wavelength, the resonances actually occur at fixed intervals of frequency, so are not exactly uniform wavelength intervals.

 

[eneacasucci]

What I find hard to visualise is the fact that in the spectrum there are different Rayleigh peaks but they represent the same laser source but at different orders since d varies, and I cannot visualise what actually the FSR range is. The distance between the peaks is lambda/2 but why does it change in wavelenght changing d?

 

[tech99]

Regarding the spacing of resonance as you vary the wavelength (or frequency). Imagine a low frequency analogy. Take transmission line 5m long with a short circuits at each end. Apply some energy at frequency that can be varied from zero upwards. When the wavelength becomes 10m, corresponding to 30 MHz, we will see the first resonance, as the line is half a wavelength long. We then increase the frequency and we see the next resonance at a wavelength of 5m (60 MHz), when the line is a two half waves long. The next resonance occurs at 90 MHz, corresponding to a wavelength of 3.3m, when the line is 3 half waves long. Notice that these resonances are spaced at intervals of 30 MHz and this action continues right up in freqeuncy as high as you wish to go. So for a length of optical fibre, we can still have resonances spaced at frequency intervals corresponding to one half wavelength of fibre. For instance, 5m of fibre gives resonances every 30 MHz. The spacing of the resonances expressed as the applied wavelength is not quite linear will be uniform over a small range.

 

[eneacasucci]

thank you so much for your time.
The example was clear, but in a Fabry–Perot interferometer (also called an étalon) we change the lenght of the "trasmission line" that is the distance of the two mirros, right? We change it firstly macroscopically and then microscopically with a piezo. We see multiple peaks due to the Rayleigh scattering, whose frequency and wavelenght is the same, but we see many of them because changing the distance we reaches higher or lower orders, right?