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? 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. Figure 1: Diagram of the common Czerny- Turner Monochromator design

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 Entrance slit controlled by a micrometer coupled to a fiber optic. Exit where CCD detector is located Grating turret holding 3 gratings of different groove density An artists rendering of the SpectraPro 500i Spectrometer

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 = (dλ 2 (slits) + dλ 2 (resolution) + dλ 2 (line) ) ½ dλ 2 (slits) → bandpass determined by finite spectrometer slit widths and the linear dispersion of the grating. dλ 2 (resolution) → the limiting resolution of the spectrometer which incorporates system aberrations, diffraction effects, and the laser line width of our system. dλ 2 (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. dλ 2 (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. dλ 2 (slits) : The bandpass due to the slit width and the grating of the spectrometer can be calculated. dλ 2 (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 × R d where: R d 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. –R d = d /dx = (d cos  / (f × m) At small angles of diffraction (  < 20˚) then cos  –R d = 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.

Sample Bandpass Calculation 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  m? BP = W × R d d = (1 mm/ 1200 gr) × 10 6 (nm/mm) = nm/gr W = 50  m × (  m/mm) = 0.05 mm f = 500 mm R d = d / (f × m) = nm / (500 mm × 1) R d = nm/mm BP = 0.05 mm × nm/mm BP = nm

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.

Near level slope again, is this a pattern? “ Bumps ” in the slope due to hysteresis of the micrometer controlling entrance slit width

Both gratings yield a relatively constant FWHM until approximately a 40  m slit width meaning the exit slit is defined by 2 pixels of 20  m each. ☺ Setting the entrance slit smaller that 40  m will not improve resolution!

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? dλ 2 (resolution) FWHM = ( dλ 2 (slits) + dλ 2 (resolution) + dλ 2 (line ) ) ½ Does this trend extend over a wide range?

Actual FWHM compared to predicted BP calculations. 600 gr/mm Slit Width (  m) BP calc. (cm -1 )Actual FWHM (cm -1 ) at 785 nm

Take Home Messages 40  m entrance slit size for minimum resolution. Bandpass is not an accurate representation of the resolution achievable by our spectrometers. –FWHM = (dλ 2 (slits) + dλ 2 (resolution) + dλ 2 (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.

Thank You !