The user needs to make a choice on two points:
When calculating the signal from the astronomical object, a detector area of size [2 pixels in the dispersion direction × 7 pixels in the direction perpendicular to the dispersion direction] is used. The fraction of energy from the source contained in 7 pixels (in the direction perpendicular to dispersion) is $E_{fs} = 80\%$.
It is necessary to specify the characteristic diameter of the source in arcseconds. When calculating the signal from the astronomical object, a detector area of size [2 pixels in the dispersion direction × 7 pixels in the direction perpendicular to the dispersion direction] is used.
Based on the specified initial parameters, as well as the known characteristics of the Spektr-UF, the signal-to-noise ratio from the astronomical object at a given exposure time $t$ is subsequently calculated:
or the exposure time required to achieve a given S/N ratio:
Where:
$B_{sky}$ - sky background, $e^{-}/s/pixel$;
$B_{det}$ - detector dark current, $e^{-}/s/pixel$;
$N_{bin}$ - total number of binned pixels obtained during the readout of information from the CCD detector (by default $N_{bin} = 1$);
$N_{read}$ - number of detector readouts (by default $N_{read} = 1$);
$R$ – readout noise, $e^{-}$;
$C$ – useful signal from the astronomical object, $e^{-}/s$, which for different tasks is calculated as follows:
Where:
$E_{f}^{spec}$ – fraction of energy contained in $N_{spix}$ in the direction perpendicular to dispersion;
$A_{eff}^{spec} = A \cdot Q_{tel}(\lambda) \cdot S_{spec}(\lambda)$, $cm^2$;
$A = \frac{SH}{100} \cdot \frac{\pi D^2}{4}$ – working telescope area, $cm^2$;
$SH$ – obscuration coefficient, %;
$D$ – diameter of the telescope primary mirror, $cm$;
$Q_{tel}(\lambda)$ – telescope throughput;
$S_{spec}(\lambda)$ – spectrograph efficiency;
$d$ – dispersion, $Å$/pixel;
$N_{\lambda pix}$ – number of pixels in the dispersion direction;
$F_{\lambda}$ - flux from the astronomical object, $[erg/(cm^2\ s\ Å)]$, specified and normalized;
$h$ - Planck's constant, $erg\ s$;
$c$ – speed of light, $Å/s$.
Where:
$scale_s$ – image scale in the direction perpendicular to dispersion, $arcsec/pixel$;
$W$ - slit width in arcseconds.
$I_{\lambda}$ – spectral energy distribution of the extended astronomical object from one square arcsecond (surface brightness), $erg/(cm^2\ s\ Å\ arcsec^2)$.
Calculation of the sky background is performed similarly to expression for the extended source.
Note: It should also be noted that for an emission spectral line, when calculating the $S/N$ ratio, it is necessary to take into account both the flux in the spectral line, $C_{line}$, and the flux in the continuum, $C_{cont}$. The first expression in this case transforms to: