Fabry-Perot Interferometer

1. Fabry-Perot Interferometer Fri. Nov. 15, 2002

2. Spectroscopy applications: Fabry- Perot Interferometer • Assume we have a monochromatic light source and we obtain a fringe pattern in the focal plane of a lens • Now plot IT along any radial direction • Let IMAX=IM • The fringes have a finite width as we scan order m m-1 m-2 m-3

3. Fabry-Perot Interferometer • Full width at half maximum = FWHM, is defined as the width of the fringe at I=(½)IM • Now we need to specify units for our application • Let us first find  such that I = ½ IM

4. Fabry-Perot Interferometer • We want, • This gives,

5. Fabry-Perot Interferometer m m-1 I= ½ IM  = 2(m-1)  = 2m  = 2(m-Δm)

6. Fabry Perot Interferometer • Thus at I = ½ IM sin(/2) = sin (m – Δm)   sin Δm • Assume Δm is small and sin Δm  Δm • Thus • FWHM ~ Fraction of an order occupied by fringe

7. Fabry-Perot Interferometer • The inverse of the FWHM is a measure of the quality of the instrument. This quality index is called the finesse • It is the ratio of the separation between the fringes to the fringe width

8. Fabry-Perot Interferometer • Note that  is determined by the reflectivity • If R ~ 0.90  = 30 R ~ 0.95  = 60 R ~ 0.97  = 100 • In practive, can’t get much better than 100 since the reflectivity is limited by the flatness of the plates (and other factors of course)

9. Fabry-Perot Interferometer • Now consider the case of two wavelengths (1, 2) present in the beam • Assume 1 2 and 1< 2 • Increase 2, dashed lines shrink • e.g. order m-1 of 2 moves toward mth order of 1 • Eventually (m-1) 2=m 1 • This defines the free spectral range m m-2 m-1 2 1 2 1 2 1

10. Fabry-Perot Interferometer • m(2-1)= 2 or mΔ =  • ΔFSR = /m • Now since • We have, • e.g. =500 nm, d = 5mm, n=1  ΔFSR= 25(10-2)mm = 0.25Å

11. Fabry-Perot Interferometer (Alternate units – wavenumbers cm-1) Define wavenumber as, Small differences are given by, Thus, which is a very convenient form.

12. Fabry-Perot Interferometer (Alternate units – wavenumbers cm-1) The corresponding frequency is, and frequency interval is, Suppose d = 5 mm and n = 1

13. Fabry-Perot Interferometer – Spectral Resolution. Rayleigh criterion Two peaks will be considered to be resolvable if their separation is greater than their FWHM – that is 2Δm This will give a dip which is about 80 % of maximum in most cases. • If two wavelengths are very close, their fringe maxima may overlap • The two peaks will now be distinguishable if the two wavelengths are too close m1 m2 (m + Δm)1 (m - Δm)2

14. Fabry-Perot Interferometer – Spectral resolution • A separation of 2Δm corresponds to minimum separation between the wavelength components • When expressed in angstroms or cm-1, the fringe width obtained is the minimum wavelength separation, • Express as

15. Fabry-Perot Interferometer – Spectral Resolving power The spectral resolving power, For Fabry – Perot

16. Fabry-Perot Applets and information on the web • http://www.physics.uq.edu.au/people/mcintyre/applets/optics/fabry.html • http://www.phys.uit.no/skibotn/fpi/ • http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/fabry.html