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Spectral Resolution and Spectrometers. A Brief Guide to Understanding and Obtaining the Proper Resolution of the 785 Raman System. Spectral Resolution and Spectrometers. How does a monochromator work? How to calculate spectral resolution.

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## Spectral Resolution and Spectrometers

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**Spectral Resolution and Spectrometers**A Brief Guide to Understanding and Obtaining the Proper Resolution of the 785 Raman System.**Spectral Resolution and Spectrometers**• How does a monochromator work? • How to calculate spectral resolution. • How does entrance and exit slit width effect the resolution? • What defines which slit is used to calculate resolution? • What should we report as our resolution and how do we obtain it?**How does a Monochromator work?**Figure 1: Diagram of the common Czerny-Turner Monochromator design Light (A) is focused onto an entrance slit (B) and is collimated by a curved mirror (C). The collimated beam is diffracted from a rotatable grating (D) and the dispersed beam re-focused by a second mirror (E) at the exit slit (F). Each wavelength of light is focused to a different position at the slit, and the wavelength which is transmitted through the slit (G) depends on the rotation angle of the grating.**Monochromator vs. Spectrometer**• A spectrometer is a monochromator with an array type detector and no exit slit. • By having no slit at the exit (or by having the slit all the way open), you can detect all of the wavelengths focused at the exit focal plane. Figure 2: Spectrometer with grating turret and CCD detector**785 Raman Spectrometer**Grating turret holding 3 gratings of different groove density Entrance slit controlled by a micrometer coupled to a fiber optic. An artists rendering of the SpectraPro 500i Spectrometer Exit where CCD detector is located**Calculating Spectral Resolution**• In the most fundamental sense, both bandpass and resolution are used as a measure of an instrument’s ability to separate adjacent spectral lines. • Spectral bandpass is the FWHM of the wavelength distribution passed by the exit slit. • Resolution is related to bandpass but determines whether the separation of two peaks can be distinguished. • Resolution of an instrument is limited by the FWHM of its Instrumental Profile.**FWHM of Instrumental Profile**FWHM = (dλ2(slits) + dλ2(resolution) + dλ2(line))½ dλ2(slits)→ bandpass determined by finite spectrometer slit widths and the linear dispersion of the grating. dλ2(resolution)→ the limiting resolution of the spectrometer which incorporates system aberrations, diffraction effects, and the laser line width of our system. dλ2(line) → natural line width of the spectral line being probed. This FWHM is our limit of resolution for the spectrometer.**How do you calculate the FWHM of the Instrumental Profile?**• The instrumental profile FWHM is something you can measure experimentally. • dλ2(line): By only observing the 785 laser line with the spectrometer we can eliminate the broadening of the FWHM due to the natural line width of a spectral line. • dλ2(slits): The bandpass due to the slit width and the grating of the spectrometer can be calculated. • dλ2(resolution): The limiting resolution of the spectrometer is something that you solve for knowing the other variables of the FWHM equation.**How to Calculate Bandpass**• BP = W × Rd where: Rd is reciprocal linear dispersion W is the slit width of the entrance or exit slit (which ever is larger) • The reciprocal linear dispersion represents the number of wavelength intervals (e.g., nm) contained in each interval of distance (e.g., mm) along the focal plane. • Rd = dl/dx = (d cos b) / (f × m) • At small angles of diffraction (b < 20˚) then cos b ~ 1; • Rd = d / (f × m) • BP = W × (d / (f × m)) The only thing left to do now is to determine what our slit width should be to solve for our bandpass.**Given a 1200 gr/mm grating, an angle of reflection less than**20˚, and f = 500 mm, what is the BP of a spectrometer with a slit width of 50 mm? BP = W × Rd d = (1 mm/ 1200 gr) × 106 (nm/mm) = 833.33 nm/gr W = 50 mm × 10-3 (mm/mm) = 0.05 mm f = 500 mm Rd = d / (f × m) = 833.33 nm / (500 mm × 1) Rd = 1.667 nm/mm BP = 0.05 mm × 1.667 nm/mm BP = 0.083 nm Sample Bandpass Calculation**Two Questions need to be Addressed**Question 1: • Which slit width do you use to calculate the bandpass with? • Earlier it was stated that the slit width that defines the BP is the larger of the entrance and exit slit. • Our spectrometers do not really have an exit slit, instead a CCD detector sits in the focal plane of the exit, so what defines the exit slit?!?! Question 2: • Is the bandpass a close enough estimation of the FWHM of the instrumental profile, our limiting resolution?**What defines our exit slit?**• A CCD is an array detector with each pixel acting as a tiny individual detector. • The short answer to the our question is that the size of a single pixel may define the exit slit width. But is this true?**Near level slope again, is this a pattern?**“Bumps” in the slope due to hysteresis of the micrometer controlling entrance slit width**☺ Setting the entrance slit smaller that 40 mm will not**improve resolution! Both gratings yield a relatively constant FWHM until approximately a 40 mm slit width meaning the exit slit is defined by 2 pixels of 20 mm each.**Is the bandpass a close enough estimation of the limiting**FWHM instrumental profile? • As you have already seen it isn’t, but to what extent? • What causes this difference? FWHM = (dλ2(slits) + dλ2(resolution) + dλ2(line))½ • Does this trend extend over a wide range?**Take Home Messages**• 40 mm entrance slit size for minimum resolution. • Bandpass is not an accurate representation of the resolution achievable by our spectrometers. • FWHM = (dλ2(slits) + dλ2(resolution) + dλ2(line))½ • The FWHM of the instrumental profile can be measured experimentally and should be done when conducting experiments so as to report the correct resolution achievable at that time.**Take Home Messages Cont…**• The two largest contributing factors to the broadening of the instrumental profile are: • Laser line width • Bandpass (which grating you chose controls the dispersion which dictates the bandpass) • It is also very important to note that condensed phase molecules have natural line widths much larger than either of these cases and will dominate the resolution of your spectrum • Your limiting resolution is still important when you are looking for shifts in a spectrum.

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