Advantages of Fourier Transform Spectroscopy

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Advantages of Fourier Transform Spectroscopy

• Time domain spectroscopy

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Advantages of Fourier Transform Spectroscopy

• Fig. 7-40, pg. 185 "Illustrations of time doamin
  plots (a) and (b) frequency domain plots (c),
  (d), and (e)."

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Advantages of Fourier Transform Spectroscopy

• Fig. 7-41, pg. 186 "Time domain spectrum of a
  source made up of several wavelengths."

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Advantages of Fourier Transform Spectroscopy

• Michelson Interferometer
• Fig. 7-42, pg. 186 "Schematic of a Michelson
  interferometer illuminated by a monochromatic
  source."

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Advantages of Fourier Transform Spectroscopy

• Fig. 7-43, pg. 188 Comparison of interferograms
  and optical spectra.

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FT-IR
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Advantages of Fourier Transform Spectroscopy

• Throughput no slits or gratings. Full power of
  source reaches detector
• Resolution depends on length of mirror scan,
  can be very high
• Rapid data collection

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Scan Times, Frequency Domain

• Resolution elements spectral width/resolution
• IR spectrum, 500 cm-1 to 4500 cm-1
• For 3 cm-1 resolution, 4500/3 1500 elements
• For 1 cm-1 resolution, 4500/1 4500 elements
• Scan at 0.5 sec/element,1500 elements,12.5min
• Scan at 0.5 sec/element,4500 elements,37.5min
• Signal/Noise S/N
• S/N improvement n1/2 where n time per element
• To improve S/N by 3, n 9. 9 X 0.5 s 4.5 sec
• Scan time goes to 9 x 37.5 min 5.6 hrs for high
  resolution scan.

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Scan Times, Time Domain

• About 0.5 sec per scan required
• Entire spectrum collected with each scan
• Assume short data collection, 12.5 min
• 750 sec/0.5 sec per scan 1500 spectra
• S/N improvement is 15001/2 39
• Multiplex advantage due to short scan time in
  time domain spectroscopy

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Measurements in the time domain

• Light moves fast, c 3 x 108 m/s
• Consider infrared, ? 3 x 10-6 m
• 3 x 10-6 m/3 x 108 m/s 10-14 s for one
  wavelength of light to pass by a detector
• To resolve 3 x 10-6 m from 4 x 10-6 m in
  frequency domain requires time resolution of less
  than 10-14 s - too fast for any known photo
  detector!

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Measuring time versus intensity

• P(t) kcos(2p?1t) kcos(2 p?2t)
• After this is measured, P(?) is given by a
  Fourier transform of P(t).
• How do you measure P(t) when c3 x 108m/s?
• Answer Interferometry
• P(t) P(t)/k, k is called modulation factor
• f frequency at detector 2 vm/?
• vm mirror velocity, ? wavelength of light

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Retardation
Retardation d 2(M-F) dmax 0 dmin ½ ? dmin
dmax ½ ? ? 2(dmin dmax) At M F, d
0 At M F ½ ?, d - ?
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Optical frequency and detector frequency

• Detector frequency f 2vm/?
• c ??photon
• f 2vm/? (2vm/c)?photon k?photon
• Since c is large, k is small and f is an easily
  measured frequency for a simple detector

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Optical frequency and detector frequency

• detector frequency (observed) is given by f
  2vm/?. This is the frequency for observing
  fringes in the interference pattern.
• Example vm 3 cm/s, ? 3 x 10-4 cm
• f104 s-1 and 1/f 10-4 s
• For ? 3 x 10-4 cm, ?photon c/?
• ?photon (3 x 1010 cm/s)/(3 x 10-4 cm) 1014
  s-1 and 1/? 10-14 s
• Conclusion Detector needs to detect signals
  with a time resolution of about 10-4 s to
  reproduce the optical signal that would require
  time resolution of about 10-14 s

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Resolution with interferometry

• Principle one complete beat pattern must be
  scanned to fully resolve all components
• Closely spaced features have smaller differences
  in frequencies, so a long cycle is needed to see
  a full beat
• To resolve 2 features separated by ??, the length
  of the scan distance is given by
  L 1/2??
• Example For resolution of 0.1 cm-1,
  L 1/(2)(0.1 cm) 5 cm