Dual-beam Flat-field Corrections, and the E and F-Factors

This section describes the flat-fielding process for dual-beam data, and the E and F-factor corrections in mathematical terms.

Let the intensity in the $O$ and $E$ beams transmitted by the analyser be $I_{o}(\alpha)$ and $I_{e}(\alpha)$, where $\alpha$ is the effective analyser angle (i.e. twice the half-wave plate rotation angle). Malus' law gives:

$$\begin{eqnarray*} I_{o}(\alpha) & = & I_{p}.\cos^{2}( \alpha - \theta ) + \frac... ...lpha) & = & I_{p}.\sin^{2}( \alpha - \theta ) + \frac{I_{u}}{2} \end{eqnarray*}$$

where $I_{p}$ is the polarized intensity, $I_{u}$ is the unpolarized intensity, and $\theta$ is the angle between the plane of polarization and the reference direction.

The signals measured by the detector (before flat-fielding) are:

$$\begin{eqnarray*} M_{o}(\alpha) & = & S_{o}.I_{o}(\alpha).E_{\alpha} \\ M_{e}(\alpha) & = & S_{e}.I_{e}(\alpha).E_{\alpha} \end{eqnarray*}$$

where $S_{o}$ and $S_{e}$ are the sensitivities of the detector to the $O$ and $E$ rays (these are independent of $\alpha$, but vary across the detector), and $E_{\alpha}$ is an exposure factor which takes into account any differences in exposure time, sky transparency, etc. Note, it is assumed that the $O$ and $E$ ray images are in a fixed position with respect to the detector in all exposures.

For target exposure $T_{0}$ (for which $\alpha$ is zero), the transmitted intensities are denoted as $I^{t}_{o}(0)$ and $I^{t}_{e}(0)$ and the measured signals as $M^{t}_{o}(0)$ and $M^{t}_{e}(0)$.

If the polarization of the flat-field surface is spatially constant, then the measured signals in the master flat-field will be proportional to the detector sensitivity functions $S_{o}$ and $S_{e}$. If the constants of proportionality for the $O$ and $E$ ray images are $K_{o}$ and $K_{e}$, then the measured signals in the master flat-field can be written as:

$$\begin{eqnarray*} M^l_{o} & = & S_{o}.K_{o} \\ M^l_{e} & = & S_{e}.K_{e} \end{eqnarray*}$$

Target exposure $T_{0}$ is flat-fielded by dividing it by the master flat-field. Thus, the measured intensities after the flat-field correction ($M^{c}_{o}(0)$ and $M^{c}_{e}(0)$) are:

$$\begin{eqnarray*} M^{c}_{o}(0) & = & M^{t}_{o}(0) / M^{l}_{o} \\ & = & \frac{... ...{ S_{e}.K_{e} } \\ & = & \frac{ I^{t}_{e}(0).E_{0} }{ K_{e} } \end{eqnarray*}$$

The target exposure $T_{45}$ is taken with an analyser angle of 90 degrees (accomplished by rotating the half-wave plate by 45 degrees), and the corresponding $O$ and $E$ ray intensities are $I^{t}_{o}(90)$ and $I^{t}_{e}(90)$, where:

$$\begin{eqnarray*} I^{t}_{o}(90) & = & I_{p}.\cos^{2}( 90 - \theta ) + \frac{I_{... ...p}.\cos^{2}( \theta ) + \frac{I_{u}}{2} \\ & = & I^{t}_{o}(0) \end{eqnarray*}$$

In other words, exposure $T_{45}$ records the same intensities as exposure $T_{0}$, but swapped so that the $O$ ray becomes the $E$ ray, and vice versa. The measured target signals at this new analyser angle are:

$$\begin{eqnarray*} M^{t}_{o}(90) & = & S_{o}.I^{t}_{o}(90).E_{90} \\ & = & S_{... ...S_{e}.I^{t}_{e}(90).E_{90} \\ & = & S_{e}.I^{t}_{o}(0).E_{90} \end{eqnarray*}$$

These measured signals are flat-fielded to give the following corrected signals:

$$\begin{eqnarray*} M^{c}_{o}(90) & = & M^{t}_{o}(90) / M^{l}_{o}(0) \\ & = & \... ... S_{e}.K_{e} } \\ & = & \frac{ I^{t}_{o}(0).E_{90} }{ K_{e} } \end{eqnarray*}$$

To simplify the notation, put $s_{1}=M^{c}_{o}(0)$, $s_{2}=M^{c}_{e}(0)$, $s_{3}=M^{c}_{o}(90)$ and $s_{4}=M^{c}_{e}(90)$. In other words, $s_{1}$ and $s_{2}$ are the flat-fielded $O$ and $E$ ray signals from exposure $T_{0}$, and $s_{3}$ and $s_{4}$ are the corresponding signals from exposure $T_{45}$. In order to calculate the polarization we need signals which are proportional to the incoming intensities, with a common constant of proportionality. In order to achieve this, we need to estimate the ratio of the exposure factors, $E_{0}$ and $E_{90}$, and the ``F-factor'', $F$, where:

$$\begin{eqnarray*} F & = & K_{e}/K_{o} \end{eqnarray*}$$

From the above expressions for the flat-fielded signals, it can be seen that:

$$\begin{eqnarray*} F & = & \sqrt {\left( \frac {s_{1}}{s_{4}} . \frac {s_{3}}{s_{2}} \right)} \end{eqnarray*}$$

We use this value of $F$ to correct the measured $E$ ray flat-fielded signals, $s_{2}$ and $s_{4}$ to get:

$$\begin{eqnarray*} s'_{2} & = & s_{2} . F \\ & = & \frac{ I^{t}_{e}(0).E_{0} }... ...{K_{e}}{K_{o}} \\ & = & \frac{ I^{t}_{o}(0).E_{90} }{ K_{o} } \end{eqnarray*}$$

Summing the $O$ and corrected $E$ rays signals for exposure $T_{0}$ ($s_{1}$ and $s'_{2}$) gives:

$$\begin{eqnarray*} s_{1} + s'_{2} & = & \frac{ I^{t}_{o}(0).E_{0} }{ K_{o} } + ... ...(0) ).E_{0} }{ K_{o} } \\ & = & \frac{ I_{T}.E_{0} }{ K_{o} } \end{eqnarray*}$$

where $I_{T}$ is the total intensity (equal to the sum of the $O$ and $E$ ray intensities). Likewise, summing the $O$ and corrected $E$ rays signals for exposure $T_{45}$ ($s_{3}$ and $s'_{4}$) gives:

$$\begin{eqnarray*} s_{3} + s'_{4} & = & \frac{ I^{t}_{e}(0).E_{90} }{ K_{o} } + ... ...) ).E_{90} }{ K_{o} } \\ & = & \frac{ I_{T}.E_{90} }{ K_{o} } \end{eqnarray*}$$

From this, the ratio of the exposure factors $E_{0}$ and $E_{90}$ can be found by dividing these expression:

$$\begin{eqnarray*} \frac{ s_{3} + s'_{4} }{ s_{1} + s'_{2} } & = & \frac{ I_{T}.... ...rac{ K_{o} }{ I_{T}.E_{0} } \\ & = & \frac{ E_{90} }{ E_{0} } \end{eqnarray*}$$

This ratio, together with the F-factor found earlier, allow the measured signals $s_{1}$ to $s_{4}$ to be corrected so that they all have a common calibration. An identical procedure can be applied to the other pair of target exposures ($T_{22.5}$ and $T_{67.5}$), leading to estimates of their exposure factors, and another estimate of the F factor.

Note, each pair of target exposures must be flat-fielded using the same master flat-field frame.