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Point Source

If we consider for example a point source with a uniform emission of magnitude 15 in the band J, being the sky airmass 1.2, and a seeing of 0.5, with the S2 objective inserted, for an integration time of 60 seconds, we could get:


\begin{displaymath}M_{J}=20\Longrightarrow Z_{J}=8.4989\Longrightarrow
F=3.17029736695\cdot 10^{-15}
~J\cdot s^{-1}\cdot m^{-2}\cdot \mu m^{-1}
\end{displaymath}

where we have used data in Tab. 2.3 and eq. (2.5), (2.6) and expressed the result in Joule units.

Using eq. (2.7):


\begin{displaymath}P=\frac{h\cdot c}{\lambda}=1.5888\cdot10^{-19} J\slash ph\hbox{ at $\lambda=1.25~\mu m$}
\end{displaymath}


\begin{displaymath}\displaystyle\left(\frac{F}{P}\right)_{Obj}=1.9954036801\cdot 10^{4}~ph\cdot s^{-1}\cdot m^{-2}\cdot\mu m^{-1}
\end{displaymath}


\begin{displaymath}\Delta_{i}=0.3\cdot 10^{-3} ~\mu m
\end{displaymath}


S=51.5676726123  m2


T=60  s


E=0.197

With the previously specified parameter we can get from formulae in Tab. 2.1:


\begin{displaymath}N_{Obj}=\frac{F\cdot \Delta_{i}\cdot T\cdot E\cdot S}{P}= 3.64877455858\cdot 10^{6} ~cnts
\end{displaymath} (2.15)

The ETC for ISAAC in imaging observation mode with the input parameter as specified above predicts $N_{ETC}=3.31584248\cdot 10^{6}$ cnts.



Pascal Ballester
1999-07-29