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

mikioneliir
2022-09-24
Answered

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

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Klecanlh

Answered 2022-09-25
Author has **11** answers

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

$d\mathrm{sin}\theta =p{\lambda}_{1}=(p+1){\lambda}_{2}.$

Hence

$\frac{1}{{\lambda}_{1}}=\frac{p}{d\mathrm{sin}\theta},\phantom{\rule{0ex}{0ex}}\frac{1}{{\lambda}_{2}}=\frac{p+1}{d\mathrm{sin}\theta}.$

Define

${\mathrm{\Delta}}_{\mathrm{F}\mathrm{S}\mathrm{R}}=\frac{1}{{\lambda}_{2}}-\frac{1}{{\lambda}_{1}}=\frac{1}{d\mathrm{sin}\theta}.$

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

$2d\mathrm{cos}(\theta )=p{\lambda}_{1}=(p+1){\lambda}_{2}.$

Hence we get

${\mathrm{\Delta}}_{\mathrm{F}\mathrm{S}\mathrm{R}}=\frac{1}{{\lambda}_{2}}-\frac{1}{{\lambda}_{1}}=\frac{1}{2d\mathrm{cos}\theta}.$

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

${\mathrm{\Delta}}_{\mathrm{F}\mathrm{S}\mathrm{R}}=\frac{1}{2d}.$

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 ${\mathrm{\Delta}}_{\mathrm{F}\mathrm{S}\mathrm{R}}$ 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.

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

$d\mathrm{sin}\theta =p{\lambda}_{1}=(p+1){\lambda}_{2}.$

Hence

$\frac{1}{{\lambda}_{1}}=\frac{p}{d\mathrm{sin}\theta},\phantom{\rule{0ex}{0ex}}\frac{1}{{\lambda}_{2}}=\frac{p+1}{d\mathrm{sin}\theta}.$

Define

${\mathrm{\Delta}}_{\mathrm{F}\mathrm{S}\mathrm{R}}=\frac{1}{{\lambda}_{2}}-\frac{1}{{\lambda}_{1}}=\frac{1}{d\mathrm{sin}\theta}.$

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

$2d\mathrm{cos}(\theta )=p{\lambda}_{1}=(p+1){\lambda}_{2}.$

Hence we get

${\mathrm{\Delta}}_{\mathrm{F}\mathrm{S}\mathrm{R}}=\frac{1}{{\lambda}_{2}}-\frac{1}{{\lambda}_{1}}=\frac{1}{2d\mathrm{cos}\theta}.$

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

${\mathrm{\Delta}}_{\mathrm{F}\mathrm{S}\mathrm{R}}=\frac{1}{2d}.$

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 ${\mathrm{\Delta}}_{\mathrm{F}\mathrm{S}\mathrm{R}}$ 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.

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