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1.  A grating is used in first order, with the spectrum observed at an angle of 15o with respect to the grating face. If the spectrograph has a camera of focal length 300 mm, how many lines per mm are required to give a reciprocal linear dispersion of 20 Å/mm?

The reciprocal linear dispersion is given by the formula:

dλ / dx = d cos / n Fcam.

Rearranging for d gives

d = (dλ / dx) . (n Fcam / cos ) = 20 x 1 x 300 / cos (15o) = 6212 Å = 6.212 x 10-7 m, i.e. 1610 lines/mm.

Note how the units of d are given in Å, as the reciprocal linear dispersion is given in Å/mm and the camera focal length in mm.

2. A grating of width 50 mm and 300 lines/mm is used to produce a second-order spectrum. What is the limiting spectral resolution in Å at a wavelength of 550 nm?

The limiting spectral resolution is given by the formula:

R = Nn = λ / Δλ.

We require Δλ:

Δλ = λ / Nn = (550 x 10-9) / (50 x 300 x 2) = 1.8 x 10-11 m = 0.18 Å.

3. A spectrograph has a reciprocal linear dispersion of 66 Å/mm and a diffraction-limited spectral resolution of 3.3 Å. What are the maximum projected slit width and detector pixel size required to exploit this resolution?

The projected slit width that matches the diffraction-limited spectral resolution = 3.3 / 66 = 0.05 mm = 50 μm, which is the maximum permissable size.

Nyquist sampling theory tells us that at least two detector pixels are required across one resolution element to optimally sample the spectrum, so the pixel size must be less than 25 μm.


4. A spectrograph is being designed for a 2 m f/10 telescope, which is located at a site where the typical seeing is 1". What ratio of camera to collimator focal lengths would you recommend using if you have a detector with a pixel size of 25 μm?

If the typical seeing is 1", then it would be best to use a slit width of approximately this size so as to obtain reasonable throughput whilst still ensuring that the resolution of the spectrograph is defined by the slit.

For optimal sampling, the 1" slit should project to at least 2 pixels on the detector. The platescale of the telescope is 206265 / 20000 = 10.3 "/mm, so the physical width of the slit is approximately 100 μm, which must project to 2 x 25 = 50 μm. Hence the magnification of the spectrograph must be 0.5, implying that the ratio of camera to collimator focal lengths must be 0.5.


5. What changes would you have to make to the spectrograph in question 4 above if you wanted to use it on an 8 m f/10 telescope?

The collimator and camera in a spectrograph work as a re-imager, changing the focal length (and hence platescale) delivered by the telescope to a different value at the detector. The focal length of the telescope plus collimator/camera system, Fsys, is given by:

Fsys = Ftel M,

where M is the magnification, given by M = Fcam / Fcoll. Using this equation, we derived the following relation in example problem 2 in imagers:

Fsys = Dtel fcam.

Fsys determines the platescale at the detector, i.e. the number of arcseconds per pixel. A well designed spectrograph will project a slit width equal to the seeing to two pixels on the detector. Assuming the 8 m and 2 m telescopes we are considering experience the same seeing, and both use CCDs with the same pixel size, this means that Fsys for the two telescopes must be equal, i.e.:

Fsys8m = Fsys2m = Dtel8m fcam8m = Dtel2m fcam2m = 8 fcam8m = 2 fcam2m.

Hence:

fcam8m = fcam2m / 4.

This means that the focal ratio of the spectrograph camera must be reduced by a factor of 4 when moving the spectrograph from a 2 m telescope to an 8 m telescope. Since fcam8m = Fcam8m / Dcam8m, this can be achieved either by decreasing the focal length of the camera by a factor of 4 or increasing the diameter of the camera by a factor of 4. The former is not an option as changing the focal length of the camera would change the magnification and hence the size of the projected slit on the detector. Hence the only option is to increase the diameter of the camera or, strictly speaking, the light beam entering the camera, by a factor of 4. This means that the diameter of the grating and collimator must also increase by a factor of 4, since all lie in the same collimated beam. Hence spectrographs on the world's largest telescopes are enormous and very expensive instruments.


©Vik Dhillon, 29th November 2012