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

2. 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 I T along any radial direction Let I MAX =I M The fringes have a finite width as we scan mm-1m-2 order m-3

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

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

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

6. 6 Fabry Perot Interferometer Thus at I = ½ I M 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. 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. 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. 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-1 m-2 2 1 2 1 2 1

10. 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. 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. 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. 13 Fabry-Perot Interferometer – Spectral Resolution. 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 Rayleigh criterion 2Δm 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. m 1 m 2 (m + Δm) 1 (m - Δm) 2

14. 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. 15 The spectral resolving power, For Fabry – Perot Fabry-Perot Interferometer – Spectral Resolving power

16. 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