example problems |
1. |
A star has a measured V-band magnitude of 20.0. How many
photons per second are detected from this star by a 4.2 m telescope with an
overall telescope/instrument/filter/detector efficiency of 30%?
The relation between fluxes and magnitudes is given by the equation: m_{1}-m_{2} = -2.5 log_{10}(F_{1}/F_{2}). We can see from table 1 that a V = 0 star has a monochromatic flux of F_{λ} = 3.61 x 10^{-11} W m^{-2} nm^{-1}. Substituting these values into the equation above, we obtain: 20-0 = -2.5 log_{10}(F_{1}/3.61 x 10^{-11}), which gives F_{1} = 3.61 x 10^{-19} W m^{-2} nm^{-1} at the effective wavelength of the V band, 550 nm. The number of photons s^{-1} m^{-2} nm^{-1} is then given by dividing this flux by the energy of a single 550 nm photon, i.e. N_{1} = F_{1} λ / h c = 1.0 photons s^{-1} m^{-2} nm^{-1}. The total number of photons detected by a 4.2 m telescope in the V band can be obtained by multiplying this number by the collecting area of the telescope, the bandpass of the filter and the efficiency of the telescope/instrument/filter/detector: N_{1} = 1.0 x x 2.1^{2} x 86 x 0.3 = 357 photons s^{-1}. |
2.^{ } |
A non-variable star is tracked across the sky on a
photometric night. At a zenith distance of 30^{o}, 100,000 photons
are detected from the star in a 1 minute integration in the
V band, which drops to 88,000 photons at a zenith distance of
60^{o}. What is the extinction coefficient in the
V band on the night in question? You may assume that the signal
from the sky has been subtracted from these measurements.
The instrumental magnitude of the star at each zenith distance is given by m_{inst} = -2.5 log_{10} (F/t_{exp}), and these can be corrected for extinction using the equation: m_{inst} = m_{inst0} + k sec z. Hence, we can write: m_{inst1} = -2.5 log_{10} (100,000/60) = m_{inst0} + k sec 30, and m_{inst2} = -2.5 log_{10} (88,000/60) = m_{inst0} + k sec 60. Subtracting these two equations to eliminate m_{inst0} and rearranging for k, we obtain: k = 0.16 magnitudes per airmass in the V band. |
3.^{ } |
The star Vega is observed on the same night with the same
equipment at a zenith distance of 45^{o}. If 1 x 10^{7}
photons are detected from Vega in a 10 s exposure, what is the V-band
magnitude of the star in question 2? The instrumental magnitude of Vega at a zenith distance of 45^{o} is: m_{inststd} = -2.5 log_{10} (F/t_{exp}) = -2.5 log_{10} (1 x 10^{7}/10) = -15. The above-atmosphere instrumental magnitude of Vega is then given by: m_{inst0std} = m_{inst} - k sec z = -15 - 0.16 sec 45 = -15.2. In comparison, the above-atmosphere instrumental magnitude of the star in question 2 is: m_{inst0} = m_{inst1} - k sec z = -2.5 log_{10} (100,000/60) - 0.16 sec 30 = -8.2. Table 1 tells us that Vega has a magnitude of V = 0. Hence the zero point of the telescope/instrument/filter/detector combination is m_{zp} = m_{std} - m_{inst0std} = 0 - (-15.2) = 15.2 The calibrated V-band magnitude of the star in question 2 is then given by: m = m_{zp} + m_{inst0} = 15.2 + (-8.2) = 7. |
4. |
Using the same equipment as described in question 1, if the sky has
a brightness of V = 19.9 magnitudes per square arcsecond, and
the extraction aperture has a radius of 2.5", how many photons per
second from the sky are recorded in the aperture?
The solution to this question follows the method used in question 1, but remembering that the sky is not a point source like a star. Hence, unlike a star, the larger the aperture used to measure the sky, the higher the total number of photons that will be detected from the sky. For this reason, sky magnitudes are quoted in per square arcsecond units and the area of the aperture used to extract the signal from the CCD image must be taken into account. The relation between fluxes and magnitudes is given by the equation: m_{1}-m_{2} = -2.5 log_{10}(F_{1}/F_{2}). We can see from table 1 that a V = 0 star has a monochromatic flux of F_{λ} = 3.61 x 10^{-11} W m^{-2} nm^{-1}. Substituting these values into the equation above, we obtain: 19.9-0 = -2.5 log_{10}(F_{1}/3.61 x 10^{-11}), which gives F_{1} = 3.96 x 10^{-19} W m^{-2} nm^{-1} arcsecond^{-2} from the sky at the effective wavelength of the V band, 550 nm. The number of photons s^{-1} m^{-2} nm^{-1} arcsecond^{-2} is then given by dividing this flux by the energy of a single 550 nm photon, i.e. N_{1} = F_{1} λ / h c = 1.1 photons s^{-1} m^{-2} nm^{-1} arcsecond^{-2}. The total number of photons per second from the sky detected in the extraction aperture by a 4.2 m telescope in the V band can be obtained by multiplying the number above by the collecting area of the telescope, the bandpass of the filter, the efficiency of the telescope/instrument/filter/detector, and the area of the aperture: N_{1} = 1.1 x ( x 2.1^{2}) x 86 x 0.3 x ( x 2.5^{2})= 7720 photons s^{-1}. |