Why not use all the colours in colorimetry?Is visible spectroscopy the only non-electronic method of all the spectroscopy method?Why are d-d electronic transitions forbidden and weakly absorbing? Why do they occur at all?Why does the absorbance of Ca decrease in the presence of certain metals?Nanoscopy Nobel Prize - why are all the tools in the visible light range?Why Acetone does not behave like its computational values?Why does Stokes raman not relax to the ground state while anti-Stokes does?Angle of Light Source for Absorbance SpectroscopyWhy doesn't the current stop in neon lamps when all the atoms have been ionized?Nanoparticle Alloys (Gold-Silver) not formingWhy does molecular oxygen absorb in the IR range?
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Why not use all the colours in colorimetry?
Is visible spectroscopy the only non-electronic method of all the spectroscopy method?Why are d-d electronic transitions forbidden and weakly absorbing? Why do they occur at all?Why does the absorbance of Ca decrease in the presence of certain metals?Nanoscopy Nobel Prize - why are all the tools in the visible light range?Why Acetone does not behave like its computational values?Why does Stokes raman not relax to the ground state while anti-Stokes does?Angle of Light Source for Absorbance SpectroscopyWhy doesn't the current stop in neon lamps when all the atoms have been ionized?Nanoparticle Alloys (Gold-Silver) not formingWhy does molecular oxygen absorb in the IR range?
$begingroup$
Colorimetry (and it's cousin visible-UV spectroscopy) is a form of chemical analysis that works by sending visible light of one wavelength through the sample. It then analyses how much of the light was absorbed in that process so you can determine concentration.
What I don't understand is why only one wavelength is used. Apparently, even for Infrared Spectroscopy, only one wavelength is analysed at one time yet the whole spectrum is needed. Why not send in white light (with all the wavelengths at once)? The output light could be split through dispersion and every wavelength could be analysed for its absorbance. Perhaps some wavelengths of light aren't worth measuring or too many frequencies at once cause interference?
spectroscopy uv-vis-spectroscopy
$endgroup$
add a comment |
$begingroup$
Colorimetry (and it's cousin visible-UV spectroscopy) is a form of chemical analysis that works by sending visible light of one wavelength through the sample. It then analyses how much of the light was absorbed in that process so you can determine concentration.
What I don't understand is why only one wavelength is used. Apparently, even for Infrared Spectroscopy, only one wavelength is analysed at one time yet the whole spectrum is needed. Why not send in white light (with all the wavelengths at once)? The output light could be split through dispersion and every wavelength could be analysed for its absorbance. Perhaps some wavelengths of light aren't worth measuring or too many frequencies at once cause interference?
spectroscopy uv-vis-spectroscopy
$endgroup$
$begingroup$
Diode array spectrophotometers work with white light. There are pros and cons to that method.
$endgroup$
– Karsten Theis
May 18 at 13:02
$begingroup$
There is also Fourier-transform (FT) vs scanning instruments.
$endgroup$
– Karsten Theis
May 18 at 13:03
add a comment |
$begingroup$
Colorimetry (and it's cousin visible-UV spectroscopy) is a form of chemical analysis that works by sending visible light of one wavelength through the sample. It then analyses how much of the light was absorbed in that process so you can determine concentration.
What I don't understand is why only one wavelength is used. Apparently, even for Infrared Spectroscopy, only one wavelength is analysed at one time yet the whole spectrum is needed. Why not send in white light (with all the wavelengths at once)? The output light could be split through dispersion and every wavelength could be analysed for its absorbance. Perhaps some wavelengths of light aren't worth measuring or too many frequencies at once cause interference?
spectroscopy uv-vis-spectroscopy
$endgroup$
Colorimetry (and it's cousin visible-UV spectroscopy) is a form of chemical analysis that works by sending visible light of one wavelength through the sample. It then analyses how much of the light was absorbed in that process so you can determine concentration.
What I don't understand is why only one wavelength is used. Apparently, even for Infrared Spectroscopy, only one wavelength is analysed at one time yet the whole spectrum is needed. Why not send in white light (with all the wavelengths at once)? The output light could be split through dispersion and every wavelength could be analysed for its absorbance. Perhaps some wavelengths of light aren't worth measuring or too many frequencies at once cause interference?
spectroscopy uv-vis-spectroscopy
spectroscopy uv-vis-spectroscopy
edited May 18 at 22:46
Melanie Shebel♦
3,53273273
3,53273273
asked May 18 at 9:05
John HonJohn Hon
152116
152116
$begingroup$
Diode array spectrophotometers work with white light. There are pros and cons to that method.
$endgroup$
– Karsten Theis
May 18 at 13:02
$begingroup$
There is also Fourier-transform (FT) vs scanning instruments.
$endgroup$
– Karsten Theis
May 18 at 13:03
add a comment |
$begingroup$
Diode array spectrophotometers work with white light. There are pros and cons to that method.
$endgroup$
– Karsten Theis
May 18 at 13:02
$begingroup$
There is also Fourier-transform (FT) vs scanning instruments.
$endgroup$
– Karsten Theis
May 18 at 13:03
$begingroup$
Diode array spectrophotometers work with white light. There are pros and cons to that method.
$endgroup$
– Karsten Theis
May 18 at 13:02
$begingroup$
Diode array spectrophotometers work with white light. There are pros and cons to that method.
$endgroup$
– Karsten Theis
May 18 at 13:02
$begingroup$
There is also Fourier-transform (FT) vs scanning instruments.
$endgroup$
– Karsten Theis
May 18 at 13:03
$begingroup$
There is also Fourier-transform (FT) vs scanning instruments.
$endgroup$
– Karsten Theis
May 18 at 13:03
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Why not send in white light (with all the wavelengths at once)?
This is certainly and routinely done today. The main thing is how much price are we willing to pay? In a rigorous sense, colorimetry refers to the fact that we use optical filters to isolate wavelengths rather than monochromators. When the latter are used, you call them spectrophotometers. Basically, the question is referring to scanning instruments versus diode array or multiplexing instruments. Both exist today and both have their advantages and disadvantages.

In scanning spectrophotometers, one wavelength passes through the sample at a time. In diode array instruments, all light passes through the sample at once and it is dispersed later. The dispersed light falls on an array of detectors (thousands of tiny detectors, usually 1024). The detectors do not know what wavelength is falling on them, however, they are calibrated.
Similarly, if you were working in the 1970s-80s you would use a scanning infrared spectrophotometer (one wavelength at a time). One needed a prism made of salt or a very special diffraction grating to disperse the infrared radiation. This made life very difficult. Imagine working with a salt prism!
In Fourier transform infrared spectroscopy, all wavelengths are passed at once, but there is no array of detectors. Unfortunately, there is no detector which can differentiate wavelengths on the basis of their frequency and separate them. We need to modulate them in such a way that an ordinary detector can "follow" the frequency of wavelengths. Here comes in the Michelson interferometer. The raw output does not look like a spectrum at all, but it has all the information "hidden" there. You need Fourier transform to recover a typical infrared spectrum.
Now one may ask why don't we use FT for UV-Vis? The answer is that the required precision of a moving mirror is so high that it is rather difficult to make one.

$endgroup$
$begingroup$
I'm confused about your last statement. I thought an FT instrument would not require a moving mirror.
$endgroup$
– Night Writer
May 18 at 19:16
1
$begingroup$
No, the interferometer used in FTIR requires a moving mirror. It is a really fancy stuff how it moves on air bearings.
$endgroup$
– M. Farooq
May 18 at 21:39
add a comment |
$begingroup$
An explanation is provided in this abstract (1):
Fourier transform spectrometry in the UV-Vis region (FT/UV-Vis),
because it is source shot-noise limited, has a signal-to-noise ratio
(S/N) disadvantage in comparison to dispersive spectrometry,
especially with dense spectra. At the expense of poorer S/N, FT/UV-Vis
can be satisfactory for high-resolution measurements. However,
low-resolution spectroscopic studies, such as molecular absorption
measurements, are not expected to be performed advantageously by
FT/UV-Vis. The broad, dense spectra with high intensity throughout a
wide spectral range should show a significant S/N degradation, in
comparison to results with dispersive spectrometry.
Further explanation of the subject of noise and why FT does not provide an advantage here can be found in course notes made available by U. Delaware Prof. S. L. Neal:
Filters do not always improve the quality of spectral data.
Measurements that are dominated by noise from the source, e.g.,
fluorescence, are often distorted by flicker noise. Whereas white
noise is frequency independent, flicker noise is larger at low
frequencies. In fact, it is sometimes called pink noise (since low
frequencies are associated with red rather than blue light).
Filtering techniques are based on the idea that the noise does not, in
general, consist of the same frequency components as the signal. In
the case of flicker noise, this condition is not met. This is one
reason that FT-UV-VIS spectrometers are not widely used. The
improvement in signal to noise that comes from measuring the entire
spectrum many times using an interferometer during the time a single
spectrum can be measured on a dispersive (monochromator based)
spectrometer is the square root of the number of times the spectrum
was measured for shot noise limited (FT-IR) measurements. In other
words, a ten point spectrum measured by the interferometer has a
signal to noise ratio that is about three times ($sqrt10$ ) larger than the
same spectrum measured using a dispersive device because signal
averaging reduces random noise by the square root of the number of
acquisitions. (Remember this is called the multiplex or Fellgett
advantage.) This improvement is always smaller in the presence of
flicker noise, and in very crowded spectra, a multiplex disadvantage
in which the signal-to-noise ratio in the data acquired by the
interferometer is lower than in the dispersive instrument can be
observed.
Reference
Mark R. Glick, Bradley T. Jones, Ben W. Smith, and James D. Winefordner
Applied Spectroscopy Vol. 43, Issue 2, pp. 342-344 (1989)V. Saptari, Fourier-Transform Spectroscopy Instrumentation Engineering, SPIE Press, Bellingham, WA (2003).
$endgroup$
1
$begingroup$
Winefordner was a very famous spectroscopist. However, the sentence "source is shot noise limited" is not clear to me. I think he means that the interferogram will be dominated by shot noise. His student, now in his 80s, happened to be a good acquaintance/ coauthor of mine. He is still very active in science and education.
$endgroup$
– M. Farooq
May 18 at 21:51
$begingroup$
@M.Farooq To be honest I am still trying to understand the argument in the excerpts, which is why I posted them as an answer without further explanation. My understanding is the shot and white (Johnson) noise have similar frequency properties (constant over freq) but one is T dependent. On the other hand, flicker noise has a 1/f dependence and can be limiting in low frequency measurements.
$endgroup$
– Night Writer
May 20 at 16:44
$begingroup$
@M.Farooq Pretty cool that I happened accidentally upon one of your colleagues :-) (if in print)
$endgroup$
– Night Writer
May 20 at 16:45
1
$begingroup$
@M.Farooq No, but he sounds exceptional (found a webpage at Florida, chem.ufl.edu/people/name/james-winefordner) :-) And it is kind of a small world... I'll have to look up your work, any recommendations?
$endgroup$
– Night Writer
May 20 at 20:52
1
$begingroup$
His name is Thomas O'Haver (emeritus now). See Google Scholar for recent work scholar.google.com/…
$endgroup$
– M. Farooq
May 20 at 21:21
|
show 1 more comment
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2 Answers
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2 Answers
2
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
Why not send in white light (with all the wavelengths at once)?
This is certainly and routinely done today. The main thing is how much price are we willing to pay? In a rigorous sense, colorimetry refers to the fact that we use optical filters to isolate wavelengths rather than monochromators. When the latter are used, you call them spectrophotometers. Basically, the question is referring to scanning instruments versus diode array or multiplexing instruments. Both exist today and both have their advantages and disadvantages.

In scanning spectrophotometers, one wavelength passes through the sample at a time. In diode array instruments, all light passes through the sample at once and it is dispersed later. The dispersed light falls on an array of detectors (thousands of tiny detectors, usually 1024). The detectors do not know what wavelength is falling on them, however, they are calibrated.
Similarly, if you were working in the 1970s-80s you would use a scanning infrared spectrophotometer (one wavelength at a time). One needed a prism made of salt or a very special diffraction grating to disperse the infrared radiation. This made life very difficult. Imagine working with a salt prism!
In Fourier transform infrared spectroscopy, all wavelengths are passed at once, but there is no array of detectors. Unfortunately, there is no detector which can differentiate wavelengths on the basis of their frequency and separate them. We need to modulate them in such a way that an ordinary detector can "follow" the frequency of wavelengths. Here comes in the Michelson interferometer. The raw output does not look like a spectrum at all, but it has all the information "hidden" there. You need Fourier transform to recover a typical infrared spectrum.
Now one may ask why don't we use FT for UV-Vis? The answer is that the required precision of a moving mirror is so high that it is rather difficult to make one.

$endgroup$
$begingroup$
I'm confused about your last statement. I thought an FT instrument would not require a moving mirror.
$endgroup$
– Night Writer
May 18 at 19:16
1
$begingroup$
No, the interferometer used in FTIR requires a moving mirror. It is a really fancy stuff how it moves on air bearings.
$endgroup$
– M. Farooq
May 18 at 21:39
add a comment |
$begingroup$
Why not send in white light (with all the wavelengths at once)?
This is certainly and routinely done today. The main thing is how much price are we willing to pay? In a rigorous sense, colorimetry refers to the fact that we use optical filters to isolate wavelengths rather than monochromators. When the latter are used, you call them spectrophotometers. Basically, the question is referring to scanning instruments versus diode array or multiplexing instruments. Both exist today and both have their advantages and disadvantages.

In scanning spectrophotometers, one wavelength passes through the sample at a time. In diode array instruments, all light passes through the sample at once and it is dispersed later. The dispersed light falls on an array of detectors (thousands of tiny detectors, usually 1024). The detectors do not know what wavelength is falling on them, however, they are calibrated.
Similarly, if you were working in the 1970s-80s you would use a scanning infrared spectrophotometer (one wavelength at a time). One needed a prism made of salt or a very special diffraction grating to disperse the infrared radiation. This made life very difficult. Imagine working with a salt prism!
In Fourier transform infrared spectroscopy, all wavelengths are passed at once, but there is no array of detectors. Unfortunately, there is no detector which can differentiate wavelengths on the basis of their frequency and separate them. We need to modulate them in such a way that an ordinary detector can "follow" the frequency of wavelengths. Here comes in the Michelson interferometer. The raw output does not look like a spectrum at all, but it has all the information "hidden" there. You need Fourier transform to recover a typical infrared spectrum.
Now one may ask why don't we use FT for UV-Vis? The answer is that the required precision of a moving mirror is so high that it is rather difficult to make one.

$endgroup$
$begingroup$
I'm confused about your last statement. I thought an FT instrument would not require a moving mirror.
$endgroup$
– Night Writer
May 18 at 19:16
1
$begingroup$
No, the interferometer used in FTIR requires a moving mirror. It is a really fancy stuff how it moves on air bearings.
$endgroup$
– M. Farooq
May 18 at 21:39
add a comment |
$begingroup$
Why not send in white light (with all the wavelengths at once)?
This is certainly and routinely done today. The main thing is how much price are we willing to pay? In a rigorous sense, colorimetry refers to the fact that we use optical filters to isolate wavelengths rather than monochromators. When the latter are used, you call them spectrophotometers. Basically, the question is referring to scanning instruments versus diode array or multiplexing instruments. Both exist today and both have their advantages and disadvantages.

In scanning spectrophotometers, one wavelength passes through the sample at a time. In diode array instruments, all light passes through the sample at once and it is dispersed later. The dispersed light falls on an array of detectors (thousands of tiny detectors, usually 1024). The detectors do not know what wavelength is falling on them, however, they are calibrated.
Similarly, if you were working in the 1970s-80s you would use a scanning infrared spectrophotometer (one wavelength at a time). One needed a prism made of salt or a very special diffraction grating to disperse the infrared radiation. This made life very difficult. Imagine working with a salt prism!
In Fourier transform infrared spectroscopy, all wavelengths are passed at once, but there is no array of detectors. Unfortunately, there is no detector which can differentiate wavelengths on the basis of their frequency and separate them. We need to modulate them in such a way that an ordinary detector can "follow" the frequency of wavelengths. Here comes in the Michelson interferometer. The raw output does not look like a spectrum at all, but it has all the information "hidden" there. You need Fourier transform to recover a typical infrared spectrum.
Now one may ask why don't we use FT for UV-Vis? The answer is that the required precision of a moving mirror is so high that it is rather difficult to make one.

$endgroup$
Why not send in white light (with all the wavelengths at once)?
This is certainly and routinely done today. The main thing is how much price are we willing to pay? In a rigorous sense, colorimetry refers to the fact that we use optical filters to isolate wavelengths rather than monochromators. When the latter are used, you call them spectrophotometers. Basically, the question is referring to scanning instruments versus diode array or multiplexing instruments. Both exist today and both have their advantages and disadvantages.

In scanning spectrophotometers, one wavelength passes through the sample at a time. In diode array instruments, all light passes through the sample at once and it is dispersed later. The dispersed light falls on an array of detectors (thousands of tiny detectors, usually 1024). The detectors do not know what wavelength is falling on them, however, they are calibrated.
Similarly, if you were working in the 1970s-80s you would use a scanning infrared spectrophotometer (one wavelength at a time). One needed a prism made of salt or a very special diffraction grating to disperse the infrared radiation. This made life very difficult. Imagine working with a salt prism!
In Fourier transform infrared spectroscopy, all wavelengths are passed at once, but there is no array of detectors. Unfortunately, there is no detector which can differentiate wavelengths on the basis of their frequency and separate them. We need to modulate them in such a way that an ordinary detector can "follow" the frequency of wavelengths. Here comes in the Michelson interferometer. The raw output does not look like a spectrum at all, but it has all the information "hidden" there. You need Fourier transform to recover a typical infrared spectrum.
Now one may ask why don't we use FT for UV-Vis? The answer is that the required precision of a moving mirror is so high that it is rather difficult to make one.

edited May 18 at 22:47
Melanie Shebel♦
3,53273273
3,53273273
answered May 18 at 15:44
M. FarooqM. Farooq
2,970316
2,970316
$begingroup$
I'm confused about your last statement. I thought an FT instrument would not require a moving mirror.
$endgroup$
– Night Writer
May 18 at 19:16
1
$begingroup$
No, the interferometer used in FTIR requires a moving mirror. It is a really fancy stuff how it moves on air bearings.
$endgroup$
– M. Farooq
May 18 at 21:39
add a comment |
$begingroup$
I'm confused about your last statement. I thought an FT instrument would not require a moving mirror.
$endgroup$
– Night Writer
May 18 at 19:16
1
$begingroup$
No, the interferometer used in FTIR requires a moving mirror. It is a really fancy stuff how it moves on air bearings.
$endgroup$
– M. Farooq
May 18 at 21:39
$begingroup$
I'm confused about your last statement. I thought an FT instrument would not require a moving mirror.
$endgroup$
– Night Writer
May 18 at 19:16
$begingroup$
I'm confused about your last statement. I thought an FT instrument would not require a moving mirror.
$endgroup$
– Night Writer
May 18 at 19:16
1
1
$begingroup$
No, the interferometer used in FTIR requires a moving mirror. It is a really fancy stuff how it moves on air bearings.
$endgroup$
– M. Farooq
May 18 at 21:39
$begingroup$
No, the interferometer used in FTIR requires a moving mirror. It is a really fancy stuff how it moves on air bearings.
$endgroup$
– M. Farooq
May 18 at 21:39
add a comment |
$begingroup$
An explanation is provided in this abstract (1):
Fourier transform spectrometry in the UV-Vis region (FT/UV-Vis),
because it is source shot-noise limited, has a signal-to-noise ratio
(S/N) disadvantage in comparison to dispersive spectrometry,
especially with dense spectra. At the expense of poorer S/N, FT/UV-Vis
can be satisfactory for high-resolution measurements. However,
low-resolution spectroscopic studies, such as molecular absorption
measurements, are not expected to be performed advantageously by
FT/UV-Vis. The broad, dense spectra with high intensity throughout a
wide spectral range should show a significant S/N degradation, in
comparison to results with dispersive spectrometry.
Further explanation of the subject of noise and why FT does not provide an advantage here can be found in course notes made available by U. Delaware Prof. S. L. Neal:
Filters do not always improve the quality of spectral data.
Measurements that are dominated by noise from the source, e.g.,
fluorescence, are often distorted by flicker noise. Whereas white
noise is frequency independent, flicker noise is larger at low
frequencies. In fact, it is sometimes called pink noise (since low
frequencies are associated with red rather than blue light).
Filtering techniques are based on the idea that the noise does not, in
general, consist of the same frequency components as the signal. In
the case of flicker noise, this condition is not met. This is one
reason that FT-UV-VIS spectrometers are not widely used. The
improvement in signal to noise that comes from measuring the entire
spectrum many times using an interferometer during the time a single
spectrum can be measured on a dispersive (monochromator based)
spectrometer is the square root of the number of times the spectrum
was measured for shot noise limited (FT-IR) measurements. In other
words, a ten point spectrum measured by the interferometer has a
signal to noise ratio that is about three times ($sqrt10$ ) larger than the
same spectrum measured using a dispersive device because signal
averaging reduces random noise by the square root of the number of
acquisitions. (Remember this is called the multiplex or Fellgett
advantage.) This improvement is always smaller in the presence of
flicker noise, and in very crowded spectra, a multiplex disadvantage
in which the signal-to-noise ratio in the data acquired by the
interferometer is lower than in the dispersive instrument can be
observed.
Reference
Mark R. Glick, Bradley T. Jones, Ben W. Smith, and James D. Winefordner
Applied Spectroscopy Vol. 43, Issue 2, pp. 342-344 (1989)V. Saptari, Fourier-Transform Spectroscopy Instrumentation Engineering, SPIE Press, Bellingham, WA (2003).
$endgroup$
1
$begingroup$
Winefordner was a very famous spectroscopist. However, the sentence "source is shot noise limited" is not clear to me. I think he means that the interferogram will be dominated by shot noise. His student, now in his 80s, happened to be a good acquaintance/ coauthor of mine. He is still very active in science and education.
$endgroup$
– M. Farooq
May 18 at 21:51
$begingroup$
@M.Farooq To be honest I am still trying to understand the argument in the excerpts, which is why I posted them as an answer without further explanation. My understanding is the shot and white (Johnson) noise have similar frequency properties (constant over freq) but one is T dependent. On the other hand, flicker noise has a 1/f dependence and can be limiting in low frequency measurements.
$endgroup$
– Night Writer
May 20 at 16:44
$begingroup$
@M.Farooq Pretty cool that I happened accidentally upon one of your colleagues :-) (if in print)
$endgroup$
– Night Writer
May 20 at 16:45
1
$begingroup$
@M.Farooq No, but he sounds exceptional (found a webpage at Florida, chem.ufl.edu/people/name/james-winefordner) :-) And it is kind of a small world... I'll have to look up your work, any recommendations?
$endgroup$
– Night Writer
May 20 at 20:52
1
$begingroup$
His name is Thomas O'Haver (emeritus now). See Google Scholar for recent work scholar.google.com/…
$endgroup$
– M. Farooq
May 20 at 21:21
|
show 1 more comment
$begingroup$
An explanation is provided in this abstract (1):
Fourier transform spectrometry in the UV-Vis region (FT/UV-Vis),
because it is source shot-noise limited, has a signal-to-noise ratio
(S/N) disadvantage in comparison to dispersive spectrometry,
especially with dense spectra. At the expense of poorer S/N, FT/UV-Vis
can be satisfactory for high-resolution measurements. However,
low-resolution spectroscopic studies, such as molecular absorption
measurements, are not expected to be performed advantageously by
FT/UV-Vis. The broad, dense spectra with high intensity throughout a
wide spectral range should show a significant S/N degradation, in
comparison to results with dispersive spectrometry.
Further explanation of the subject of noise and why FT does not provide an advantage here can be found in course notes made available by U. Delaware Prof. S. L. Neal:
Filters do not always improve the quality of spectral data.
Measurements that are dominated by noise from the source, e.g.,
fluorescence, are often distorted by flicker noise. Whereas white
noise is frequency independent, flicker noise is larger at low
frequencies. In fact, it is sometimes called pink noise (since low
frequencies are associated with red rather than blue light).
Filtering techniques are based on the idea that the noise does not, in
general, consist of the same frequency components as the signal. In
the case of flicker noise, this condition is not met. This is one
reason that FT-UV-VIS spectrometers are not widely used. The
improvement in signal to noise that comes from measuring the entire
spectrum many times using an interferometer during the time a single
spectrum can be measured on a dispersive (monochromator based)
spectrometer is the square root of the number of times the spectrum
was measured for shot noise limited (FT-IR) measurements. In other
words, a ten point spectrum measured by the interferometer has a
signal to noise ratio that is about three times ($sqrt10$ ) larger than the
same spectrum measured using a dispersive device because signal
averaging reduces random noise by the square root of the number of
acquisitions. (Remember this is called the multiplex or Fellgett
advantage.) This improvement is always smaller in the presence of
flicker noise, and in very crowded spectra, a multiplex disadvantage
in which the signal-to-noise ratio in the data acquired by the
interferometer is lower than in the dispersive instrument can be
observed.
Reference
Mark R. Glick, Bradley T. Jones, Ben W. Smith, and James D. Winefordner
Applied Spectroscopy Vol. 43, Issue 2, pp. 342-344 (1989)V. Saptari, Fourier-Transform Spectroscopy Instrumentation Engineering, SPIE Press, Bellingham, WA (2003).
$endgroup$
1
$begingroup$
Winefordner was a very famous spectroscopist. However, the sentence "source is shot noise limited" is not clear to me. I think he means that the interferogram will be dominated by shot noise. His student, now in his 80s, happened to be a good acquaintance/ coauthor of mine. He is still very active in science and education.
$endgroup$
– M. Farooq
May 18 at 21:51
$begingroup$
@M.Farooq To be honest I am still trying to understand the argument in the excerpts, which is why I posted them as an answer without further explanation. My understanding is the shot and white (Johnson) noise have similar frequency properties (constant over freq) but one is T dependent. On the other hand, flicker noise has a 1/f dependence and can be limiting in low frequency measurements.
$endgroup$
– Night Writer
May 20 at 16:44
$begingroup$
@M.Farooq Pretty cool that I happened accidentally upon one of your colleagues :-) (if in print)
$endgroup$
– Night Writer
May 20 at 16:45
1
$begingroup$
@M.Farooq No, but he sounds exceptional (found a webpage at Florida, chem.ufl.edu/people/name/james-winefordner) :-) And it is kind of a small world... I'll have to look up your work, any recommendations?
$endgroup$
– Night Writer
May 20 at 20:52
1
$begingroup$
His name is Thomas O'Haver (emeritus now). See Google Scholar for recent work scholar.google.com/…
$endgroup$
– M. Farooq
May 20 at 21:21
|
show 1 more comment
$begingroup$
An explanation is provided in this abstract (1):
Fourier transform spectrometry in the UV-Vis region (FT/UV-Vis),
because it is source shot-noise limited, has a signal-to-noise ratio
(S/N) disadvantage in comparison to dispersive spectrometry,
especially with dense spectra. At the expense of poorer S/N, FT/UV-Vis
can be satisfactory for high-resolution measurements. However,
low-resolution spectroscopic studies, such as molecular absorption
measurements, are not expected to be performed advantageously by
FT/UV-Vis. The broad, dense spectra with high intensity throughout a
wide spectral range should show a significant S/N degradation, in
comparison to results with dispersive spectrometry.
Further explanation of the subject of noise and why FT does not provide an advantage here can be found in course notes made available by U. Delaware Prof. S. L. Neal:
Filters do not always improve the quality of spectral data.
Measurements that are dominated by noise from the source, e.g.,
fluorescence, are often distorted by flicker noise. Whereas white
noise is frequency independent, flicker noise is larger at low
frequencies. In fact, it is sometimes called pink noise (since low
frequencies are associated with red rather than blue light).
Filtering techniques are based on the idea that the noise does not, in
general, consist of the same frequency components as the signal. In
the case of flicker noise, this condition is not met. This is one
reason that FT-UV-VIS spectrometers are not widely used. The
improvement in signal to noise that comes from measuring the entire
spectrum many times using an interferometer during the time a single
spectrum can be measured on a dispersive (monochromator based)
spectrometer is the square root of the number of times the spectrum
was measured for shot noise limited (FT-IR) measurements. In other
words, a ten point spectrum measured by the interferometer has a
signal to noise ratio that is about three times ($sqrt10$ ) larger than the
same spectrum measured using a dispersive device because signal
averaging reduces random noise by the square root of the number of
acquisitions. (Remember this is called the multiplex or Fellgett
advantage.) This improvement is always smaller in the presence of
flicker noise, and in very crowded spectra, a multiplex disadvantage
in which the signal-to-noise ratio in the data acquired by the
interferometer is lower than in the dispersive instrument can be
observed.
Reference
Mark R. Glick, Bradley T. Jones, Ben W. Smith, and James D. Winefordner
Applied Spectroscopy Vol. 43, Issue 2, pp. 342-344 (1989)V. Saptari, Fourier-Transform Spectroscopy Instrumentation Engineering, SPIE Press, Bellingham, WA (2003).
$endgroup$
An explanation is provided in this abstract (1):
Fourier transform spectrometry in the UV-Vis region (FT/UV-Vis),
because it is source shot-noise limited, has a signal-to-noise ratio
(S/N) disadvantage in comparison to dispersive spectrometry,
especially with dense spectra. At the expense of poorer S/N, FT/UV-Vis
can be satisfactory for high-resolution measurements. However,
low-resolution spectroscopic studies, such as molecular absorption
measurements, are not expected to be performed advantageously by
FT/UV-Vis. The broad, dense spectra with high intensity throughout a
wide spectral range should show a significant S/N degradation, in
comparison to results with dispersive spectrometry.
Further explanation of the subject of noise and why FT does not provide an advantage here can be found in course notes made available by U. Delaware Prof. S. L. Neal:
Filters do not always improve the quality of spectral data.
Measurements that are dominated by noise from the source, e.g.,
fluorescence, are often distorted by flicker noise. Whereas white
noise is frequency independent, flicker noise is larger at low
frequencies. In fact, it is sometimes called pink noise (since low
frequencies are associated with red rather than blue light).
Filtering techniques are based on the idea that the noise does not, in
general, consist of the same frequency components as the signal. In
the case of flicker noise, this condition is not met. This is one
reason that FT-UV-VIS spectrometers are not widely used. The
improvement in signal to noise that comes from measuring the entire
spectrum many times using an interferometer during the time a single
spectrum can be measured on a dispersive (monochromator based)
spectrometer is the square root of the number of times the spectrum
was measured for shot noise limited (FT-IR) measurements. In other
words, a ten point spectrum measured by the interferometer has a
signal to noise ratio that is about three times ($sqrt10$ ) larger than the
same spectrum measured using a dispersive device because signal
averaging reduces random noise by the square root of the number of
acquisitions. (Remember this is called the multiplex or Fellgett
advantage.) This improvement is always smaller in the presence of
flicker noise, and in very crowded spectra, a multiplex disadvantage
in which the signal-to-noise ratio in the data acquired by the
interferometer is lower than in the dispersive instrument can be
observed.
Reference
Mark R. Glick, Bradley T. Jones, Ben W. Smith, and James D. Winefordner
Applied Spectroscopy Vol. 43, Issue 2, pp. 342-344 (1989)V. Saptari, Fourier-Transform Spectroscopy Instrumentation Engineering, SPIE Press, Bellingham, WA (2003).
edited May 19 at 5:57
answered May 18 at 19:44
Night WriterNight Writer
3,914527
3,914527
1
$begingroup$
Winefordner was a very famous spectroscopist. However, the sentence "source is shot noise limited" is not clear to me. I think he means that the interferogram will be dominated by shot noise. His student, now in his 80s, happened to be a good acquaintance/ coauthor of mine. He is still very active in science and education.
$endgroup$
– M. Farooq
May 18 at 21:51
$begingroup$
@M.Farooq To be honest I am still trying to understand the argument in the excerpts, which is why I posted them as an answer without further explanation. My understanding is the shot and white (Johnson) noise have similar frequency properties (constant over freq) but one is T dependent. On the other hand, flicker noise has a 1/f dependence and can be limiting in low frequency measurements.
$endgroup$
– Night Writer
May 20 at 16:44
$begingroup$
@M.Farooq Pretty cool that I happened accidentally upon one of your colleagues :-) (if in print)
$endgroup$
– Night Writer
May 20 at 16:45
1
$begingroup$
@M.Farooq No, but he sounds exceptional (found a webpage at Florida, chem.ufl.edu/people/name/james-winefordner) :-) And it is kind of a small world... I'll have to look up your work, any recommendations?
$endgroup$
– Night Writer
May 20 at 20:52
1
$begingroup$
His name is Thomas O'Haver (emeritus now). See Google Scholar for recent work scholar.google.com/…
$endgroup$
– M. Farooq
May 20 at 21:21
|
show 1 more comment
1
$begingroup$
Winefordner was a very famous spectroscopist. However, the sentence "source is shot noise limited" is not clear to me. I think he means that the interferogram will be dominated by shot noise. His student, now in his 80s, happened to be a good acquaintance/ coauthor of mine. He is still very active in science and education.
$endgroup$
– M. Farooq
May 18 at 21:51
$begingroup$
@M.Farooq To be honest I am still trying to understand the argument in the excerpts, which is why I posted them as an answer without further explanation. My understanding is the shot and white (Johnson) noise have similar frequency properties (constant over freq) but one is T dependent. On the other hand, flicker noise has a 1/f dependence and can be limiting in low frequency measurements.
$endgroup$
– Night Writer
May 20 at 16:44
$begingroup$
@M.Farooq Pretty cool that I happened accidentally upon one of your colleagues :-) (if in print)
$endgroup$
– Night Writer
May 20 at 16:45
1
$begingroup$
@M.Farooq No, but he sounds exceptional (found a webpage at Florida, chem.ufl.edu/people/name/james-winefordner) :-) And it is kind of a small world... I'll have to look up your work, any recommendations?
$endgroup$
– Night Writer
May 20 at 20:52
1
$begingroup$
His name is Thomas O'Haver (emeritus now). See Google Scholar for recent work scholar.google.com/…
$endgroup$
– M. Farooq
May 20 at 21:21
1
1
$begingroup$
Winefordner was a very famous spectroscopist. However, the sentence "source is shot noise limited" is not clear to me. I think he means that the interferogram will be dominated by shot noise. His student, now in his 80s, happened to be a good acquaintance/ coauthor of mine. He is still very active in science and education.
$endgroup$
– M. Farooq
May 18 at 21:51
$begingroup$
Winefordner was a very famous spectroscopist. However, the sentence "source is shot noise limited" is not clear to me. I think he means that the interferogram will be dominated by shot noise. His student, now in his 80s, happened to be a good acquaintance/ coauthor of mine. He is still very active in science and education.
$endgroup$
– M. Farooq
May 18 at 21:51
$begingroup$
@M.Farooq To be honest I am still trying to understand the argument in the excerpts, which is why I posted them as an answer without further explanation. My understanding is the shot and white (Johnson) noise have similar frequency properties (constant over freq) but one is T dependent. On the other hand, flicker noise has a 1/f dependence and can be limiting in low frequency measurements.
$endgroup$
– Night Writer
May 20 at 16:44
$begingroup$
@M.Farooq To be honest I am still trying to understand the argument in the excerpts, which is why I posted them as an answer without further explanation. My understanding is the shot and white (Johnson) noise have similar frequency properties (constant over freq) but one is T dependent. On the other hand, flicker noise has a 1/f dependence and can be limiting in low frequency measurements.
$endgroup$
– Night Writer
May 20 at 16:44
$begingroup$
@M.Farooq Pretty cool that I happened accidentally upon one of your colleagues :-) (if in print)
$endgroup$
– Night Writer
May 20 at 16:45
$begingroup$
@M.Farooq Pretty cool that I happened accidentally upon one of your colleagues :-) (if in print)
$endgroup$
– Night Writer
May 20 at 16:45
1
1
$begingroup$
@M.Farooq No, but he sounds exceptional (found a webpage at Florida, chem.ufl.edu/people/name/james-winefordner) :-) And it is kind of a small world... I'll have to look up your work, any recommendations?
$endgroup$
– Night Writer
May 20 at 20:52
$begingroup$
@M.Farooq No, but he sounds exceptional (found a webpage at Florida, chem.ufl.edu/people/name/james-winefordner) :-) And it is kind of a small world... I'll have to look up your work, any recommendations?
$endgroup$
– Night Writer
May 20 at 20:52
1
1
$begingroup$
His name is Thomas O'Haver (emeritus now). See Google Scholar for recent work scholar.google.com/…
$endgroup$
– M. Farooq
May 20 at 21:21
$begingroup$
His name is Thomas O'Haver (emeritus now). See Google Scholar for recent work scholar.google.com/…
$endgroup$
– M. Farooq
May 20 at 21:21
|
show 1 more comment
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$begingroup$
Diode array spectrophotometers work with white light. There are pros and cons to that method.
$endgroup$
– Karsten Theis
May 18 at 13:02
$begingroup$
There is also Fourier-transform (FT) vs scanning instruments.
$endgroup$
– Karsten Theis
May 18 at 13:03