Q: How do I calibrate my narrow band photometry?
A: Here are three methods using the F656N filter as an example.
METHOD # 1. You can use PHOTFLAM, obtained from the header or from SYNPHOT.
PHOTFLAM = 1.471 E-16 in ergs/s/cm2/Å for an object with 1 DN/s.
(Obtained using the bandbpar command with SYNPHOT. Note that the value of PHOTFLAM in the header may be off by a few percent; see the CAUTION at the bottom of this item for a discussion and the SYNPHOT commands needed to fix the problem).
The value of PHOTFLAM assumes a continuum source, and then divides by the "width" of the filter to convert to per angstrom. The F656N filter can be approximated by a rectangular filter curve with peak system efficiency 0.10379 (this is QTmax from Table 6.1 of version 4 of the WFPC2 Instrument Handbook, or you can just estimate it from the throughput curve on page 217) and a 28.33 angstrom width. Note that the width quoted in Table 6.1 and Figure A2 (i.e., 21.4 angstroms) is not quite right for our purposes. You can see from page 217 that the area under the filter curve is larger than the area in the rectangle defined by a width of 21.4 angstroms. 28.33 is the value of RECTW (rectangular width) for the filter, also obtained from running bandpar in SYNPHOT.
Hence, a continuum source would actually produce:
1.471 E-16 X 28.33 Å = 4.2 E -15 ergs/s/cm2 for 1 DN/s.
Now the detector doesn't care whether the real source is in 1 Angstrom or spread over 28 Å, 1 DN/s is still 4.2 E -15 ergs/s/cm2, so that is the number to use for your line flux calculation.
Note that if your line is at a wavelength that is not near the peak you should make a simple linear correction (QTpeak/QTactual), based either on an eyeball estimate from the filter curve, or using SYNPHOT (see CAUTIONS below). Similarly, if your line is very broad you may want to use SYNPHOT to convolve it with the filter transmission to obtain a better estimate of the effective QT.
METHOD # 2. Another approach is to use the Holtzman article (PASP, 107, 1087).
F = (DN/s) X (14/GR) X E / (A X QT)
for DN/s = 1:
= 1 X 14/2 X (6.626 E-27 X 2.998 E10 / 6563 E-8) / (3.1416 X 1202 X .10379) = 7 X 3.027 E-12 / 4695 = 4.5 E-15
which agrees reasonably well with the estimate from method # 1.
See the article for definitions. Using gain = GR = 2 for this example, appropriate for most science observations taken with gain = 7 setup rather than gain = 14. Exact value of GR depends on chip.
METHOD # 3. The third, and probably easiest method is to use the Exposure Time Calculator found at:
Under point sources, and emission line, click on emission line, enter 1.0e-16 (no spaces) for the line flux, click on H6563 for line, click on F656N for filter, enter 1000 second for time, click on calculate, scroll down under intermediate results and find that this gives .168 electrons per second. For a gain of 7 this translates to .168 / 7 = .0240 DN/s for a 1.0 E -16 flux, or inverting, 1 DN/s = 4.2 E-15, again similar to above.
Note: Page 112 of the WFPC2 Instrument Handbook shows an example similar to ours but gets 0.155 electron/s rather than our value of 0.168 electrons/s mentioned above. This was due to some confusion about whether a 10% aperture correction was needed (see the Holtzman article for a discussion of aperture corrections). The value of 0.168 supersedes the Handbook example.
1. A common problem people run into is that the default spectral resolution for SYNPHOT is often much too coarse. For example, for the F656N filter the default is 10 Å. A related problem is that the wavelength table extends beyond the filter transmission curve in some cases. Hence, the value of PHOTFLAM in the header is often off by a few percent (only F437N, F656N and F658N are off by more than 2%). To fix this run the following SYNPHOT commands (adjusted to your filter):
genwave fine_wave 6520 6600 1 bandpar wfpc2,1,a2d7,f656N wavetab="fine_wave"
The thing called URESP in the output is the value of PHOTFLAM.
Now if you want to see the throughput at a particular wavelength, say 6550, use:
bandpar wfpc2,1,f656N refwave=6550 wavetab="fine_wave"(Note that the gain, a2d7, has been taken out.)
In this case TLAMBDA = 0.058437 and TPEAK = 0.11306, so if we have a very narrow line peaked at 6550 we should multiply PHOTFLAM*RECTW by QTpeak/QTlambda = 0.11306/0.058437 = 1.93 (see below for concern about possibility of shift in filter response). You might note that TPEAK is about 10% higher than listed above. We believe this is due to some confusion about when the 10% aperture correction gets made. In any case, this is irrelevent for this part of the calculation since the ratio is the only thing we need. This ratio is called QTactual/QTmax in method # 1 above.
2. It is not uncommon for filters to shift toward the blue 5 - 10 Å with time. This was seen for the WFPC1 filters for example. We are in the process of measuring this for the WFPC2 filters. If you see any indications of such a shift in your data please contact the WFPC2 team.