An explanation of the information contained in the above image can be gotten here.
EF Peg is a UGem type cataclysmic variable having a period of about 162.5 days. It varies between 10.7 magnitude down past 17th magnitude photographic. It's AAVSO designation is 2110+13. It's coordinates are 21 15 03 + 14 04.1.
I first imaged this variable on 10/15/01 with the image shown above. I used this AAVSO chart.
The image shown above contains several frames. The largest is the photometry frame. On the right starting from the top is the measurement frame. It is keyed to the red and yellow circles in the photometry frame. The yellow circled star marks the location of the variable. More on this in a moment. The red circles mark the location of the comparison stars. The other columns in this frame give additional information about the comp or target star.
The middle frames are not very important here.
The lower frame contains one piecof useful information, that being the point spread function, or PSF, characterized as the Full WIdth at Half Maximum or FWHM. This value is 5.5. It means that the measured star has a PSF of 5.5 arc-seconds, which is about average for the photometry that I do. The yellow cross cursor in the photometry frame marks the star used to measure the PSF.
The interesting thing about the targetr star, (marked in yellow in the photometry frame) is that the location contains two stars, not just one. The EF Peg star is very close to th comp star 124. The intensity found within the inner circle is actually light from two stars, one of which I know is 12.4 mag V. Te intensity of light of the target star must be the intensity from oth stars minus the intensity of a star having a magnitude of 12.4.
Arne Henden, from the US Naval Observatory, Flagstaff station supplied the following information for converting from intensity to and from magnitude.
Steve,
You have to subtract in intensity-space. The formula
is what you would expect:
total = intensity1 + intensity2
or
intensity1 = total - intensity2
to convert your magnitudes into intensities:
intensity = 10 ** (-0.4 * magnitude)
so for this case,
intensity1 = 10**(-0.4*11.821) - 10**(-0.4*12.4)
= 0.00001869 - 0.00001096
= 0.000007725
once you have derived the intensity of the variable, then
convert it back into a magnitude with
magnitude = -2.5 log (intensity)
= -2.5 log (0.000007725)
= 12.8
Then think about it to ensure the answer seems to be in
the right ballpark (no errors!). First, both stars are
fainter than their sum; good. Second, if both stars were
equal in magnitude, the sum would be 0.75mag brighter than
either alone. The sum is slightly less than this, so the
answer seems to be in the right ballpark.
You would get better results if you knew the comparison
star's magnitude more precisely than 12.4. Be sure that
you are using a large enough aperture to completely include
the light from both stars, and that the aperture is centered
between the two stars.
Arne
