# Slit/Pinhole Flux Calculator

### Abstract

In this Post we provide a calculator to compute the amount of flux of the image of a star passing through a pinhole or slit in a conventional spectrograph.

The calculator is embedded in an excel document that can be downloaded as SlitPinholeFluxCalculator.xlsx or SlitPinholeFluxCalculator.xls. It is also available as a Java applet that can be executed here (it requires Java plug-in’s in your browser) or downloaded (including Java sources) here.

### Description

The width of a slit in a spectrograph compromises the amount of luminous flux on the detector against the resolving power of the spectrograph. On the one hand, wider is the slit of the spectrograph higher the number of photons on the detector. Therefore higher will be the signal to noise ratio of the recorded spectrum. On the other hand, wider is the slit, lower will be the resolving power of the spectrograph. This trade-off is affected as well by the local seeing. Better is the seeing, narrow can be slit.

The same trade-off appears when optical fibres are used to link telescopes to spectrographs. Circular apertures would give the impression that they just increase flux losses with respect to the slits. especially when the diameter equals the slit width. However, this flux reduction is compensated by an increase of the resolving power of the spectrograph. Indeed, the full width half maximum (FWHM) of the point spread function (PSF) on the detector is smaller for a fibre than for a slit. The gain in resolution with a fibre is around the same as the reduction in luminous flux. Empirically is about 20% in both cases.

The flux distribution of the image of a star on the focal plane of the telescope is defined by the convolution of the telescope aberrations and by the seeing function. If the aberrations are small, the image of a star may be simulated with good approximation to a bi-dimmensional Gaussian function (Figure 1a). The size of this gaussian function is basically defined by the FWHM (Full Width Half Maximum) of the seeing function($\varnothing s$) and scaled by the scale plate ($sp$)of the telescope.

At the telescope focal plane we can project either, a pinhole or a slit, and center our gaussian star representation on them. Figure 1b shows the top view (from telescope beam) of our gaussian star centered in a pinhole, while Figure 1c shows the same view with a slit. Figure 1d would represent the flux coming through the pinhole and entering the instrument (view from behind), while Figure 1e would represent the flux through the slit.

The calculators compute the efficiency ratio (partial flux/total flux) for both alternatives: pinhole and slit.

Figure 1. a) Simulation of a star by a gaussian function. b) Projection of the star on a slit (view from telescope). c) Same as b) but on a pinhole. d) and e) are views from the spectrograph side

### Input parameters:

$\varnothing s$ Seeing FWHM in arcseconds

$\varnothing t$ Telescope diameter in cm

$F$ Telescope Focal Number (F/#)

$\varnothing f$ Pinhole diameter in um

$d$ Slit width in um

$L$ Slit lenght in um

### Equations:

Plate Scale (in microns per arcsec)

$ps=\pi \cdot F\cdot \varnothing t\cdot \frac{{{10}^{4}}}{180\cdot 3600}$

Practical example:

Let’s assume a telescope of 30 cm of diameter ($\varnothing t$) with an aperture of F/10 ($F$). The scale plate ($ps$) will be $\pi \cdot 10\cdot \ 30 \cdot \frac{{{10}^{4}}}{180\cdot 3600}$ or 14.5444 microns per arcsec.

If your seeing were 3.4377 arcsecs ($\varnothing s$) , the size of a gaussian star (FWHM) at the telescope focal plane would be of 50 um ($ps \cdot \varnothing s$).

If the star is then centered in a pinhole of also 50 um ($\varnothing f$), the flux through the pinhole will be 50% according to Figure 3.

However, if the same star is centered on a slit with a width of 50 um ($d$) and a length of 100 um ($L$) , the flux will be 74.7%. If the slit were infinity long, the flux would be 76% (Figure 2). It is clear that the flux through the slit will be higher than the one through a pinhole with a diameter the same width ($\varnothing f$) ; the open area is definitely larger .

The relationships are not directly proportional. If the seeing is very smaller than the apertures then both, pinhole and slit, will have similar throughput (Figure 3). On the other hand, if the seeing is very big, the slit will be always more efficienct than the pinhole, especially for high lengths.

With the excel calculator you can also generate charts comparing the flux efficiency of a slit versus a pinhole in a given telescope and for different seeing values. For instance the curves in Figure 3 & 4 were obtained for a 50 cm, F/10 telescope, the slit was set to 25×125 um and the pinhole to 50 um diameter. It shows that for a seeing bigger than 2 arcsecs, less than 50% of the flux of the star will pass to the spectrograph! On the other hand, is the seeing is good (less than 2 arcsec) the pinhole will be more efficienct than the slit (in this particular example!)

Figure 4 shows the ratio between both efficiencies. However both charts show results valid only for this configuration of telescope, slit and pinhole. Try other configurations, and you will learn quickly where the photons go and which is the best compromised solution for your telescope.

Figure 3. A particular example were a pinhole may be more efficienct than a slit

Figure 4. Ratio between the curves in Figure 3

### Conclusions

We provided a calculator to compute the flux passing through a slit or a pinhole as a function of telescope parameters and seeing value. They are available as Excel spreadsheets or as a Java applet.

At same apertures the slit is more efficient than the pinhole. However the resolving power of the spectrograph will be higher for a pinhole than the slit. The ratio is about the same in efficiency. If you want to keep the same resolving power, a pinhole may be more efficient than a slit when the seeing is very small.

The larger the focal length of a telescope is, so is the plate scale, and the bigger your star will appear at the focal plane. To drive, e.g. FHWM (50%) of the target’s flux into your instrument, its aperture (slit or pinhole) have to stay proportional to the plate scale. As the resolving power of an spectrograph is inversally proportional to its aperture, either you lower the resolution or you increase the focal length, thus the size of your instrument. There are no magic solutions: big telescopes require big (and expensive) instruments.

Authors:

G. Avila* , C. Guirao*

*European Southern Observatory, Karl-Schwarzschild-Str. 2, 85748 Munich, Germany

Acknowledgments:

María Victoria Ortiz (Java developer)

Pascal Ballester* (Maths designer)

Nicolas Benes (Java Applet developer)

Jesús Rodriguez* (Web designer)

*European Southern Observatory, Karl-Schwarzschild-Str. 2, 85748 Munich, Germany