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 ${I}_{o}\left(\alpha \right)$
and ${I}_{e}\left(\alpha \right)$,
where $\alpha $
is the effective analyser angle (*i.e.* twice the half-wave plate rotation angle). Malus’ law (see appendix
B) gives:

$$\begin{array}{rcll}{I}_{o}\left(\alpha \right)& =& {I}_{p}.{cos}^{2}\left(\alpha -\theta \right)+\frac{{I}_{u}}{2}& \text{}\\ {I}_{e}\left(\alpha \right)& =& {I}_{p}.{sin}^{2}\left(\alpha -\theta \right)+\frac{{I}_{u}}{2}& \text{}\end{array}$$

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:

$$\begin{array}{rcll}{M}_{o}\left(\alpha \right)& =& {S}_{o}.{I}_{o}\left(\alpha \right).{E}_{\alpha}& \text{}\\ {M}_{e}\left(\alpha \right)& =& {S}_{e}.{I}_{e}\left(\alpha \right).{E}_{\alpha}& \text{}\end{array}$$

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 ${I}_{o}^{t}\left(0\right)$ and ${I}_{e}^{t}\left(0\right)$ and the measured signals as ${M}_{o}^{t}\left(0\right)$ and ${M}_{e}^{t}\left(0\right)$.

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:

$$\begin{array}{rcll}{M}_{o}^{l}& =& {S}_{o}.{K}_{o}& \text{}\\ {M}_{e}^{l}& =& {S}_{e}.{K}_{e}& \text{}\end{array}$$

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

$$\begin{array}{rcll}{M}_{o}^{c}\left(0\right)& =& {M}_{o}^{t}\left(0\right)/{M}_{o}^{l}& \text{}\\ & =& \frac{{S}_{o}.{I}_{o}^{t}\left(0\right).{E}_{0}}{{S}_{o}.{K}_{o}}& \text{}\\ & =& \frac{{I}_{o}^{t}\left(0\right).{E}_{0}}{{K}_{o}}& \text{}\\ & & & \text{}\\ {M}_{e}^{c}\left(0\right)& =& {M}_{e}^{t}\left(0\right)/{M}_{e}^{l}& \text{}\\ & =& \frac{{S}_{e}.{I}_{e}^{t}\left(0\right).{E}_{0}}{{S}_{e}.{K}_{e}}& \text{}\\ & =& \frac{{I}_{e}^{t}\left(0\right).{E}_{0}}{{K}_{e}}& \text{}\end{array}$$

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

$$\begin{array}{rcll}{I}_{o}^{t}\left(90\right)& =& {I}_{p}.{cos}^{2}\left(90-\theta \right)+\frac{{I}_{u}}{2}& \text{}\\ & =& {I}_{p}.{sin}^{2}\left(\theta \right)+\frac{{I}_{u}}{2}& \text{}\\ & =& {I}_{e}^{t}\left(0\right)& \text{}\\ & & & \text{}\\ {I}_{e}^{t}\left(90\right)& =& {I}_{p}.{sin}^{2}\left(90-\theta \right)+\frac{{I}_{u}}{2}& \text{}\\ & =& {I}_{p}.{cos}^{2}\left(\theta \right)+\frac{{I}_{u}}{2}& \text{}\\ & =& {I}_{o}^{t}\left(0\right)& \text{}\end{array}$$

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:

$$\begin{array}{rcll}{M}_{o}^{t}\left(90\right)& =& {S}_{o}.{I}_{o}^{t}\left(90\right).{E}_{90}& \text{}\\ & =& {S}_{o}.{I}_{e}^{t}\left(0\right).{E}_{90}& \text{}\\ & & & \text{}\\ {M}_{e}^{t}\left(90\right)& =& {S}_{e}.{I}_{e}^{t}\left(90\right).{E}_{90}& \text{}\\ & =& {S}_{e}.{I}_{o}^{t}\left(0\right).{E}_{90}& \text{}\end{array}$$

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

$$\begin{array}{rcll}{M}_{o}^{c}\left(90\right)& =& {M}_{o}^{t}\left(90\right)/{M}_{o}^{l}\left(0\right)& \text{}\\ & =& \frac{{S}_{o}.{I}_{e}^{t}\left(0\right).{E}_{90}}{{S}_{o}.{K}_{o}}& \text{}\\ & =& \frac{{I}_{e}^{t}\left(0\right).{E}_{90}}{{K}_{o}}& \text{}\\ & & & \text{}\\ {M}_{e}^{c}\left(90\right)& =& {M}_{e}^{t}\left(90\right)/{M}_{e}^{l}\left(0\right)& \text{}\\ & =& \frac{{S}_{e}.{I}_{o}^{t}\left(0\right).{E}_{90}}{{S}_{e}.{K}_{e}}& \text{}\\ & =& \frac{{I}_{o}^{t}\left(0\right).{E}_{90}}{{K}_{e}}& \text{}\end{array}$$

To simplify the notation, put ${s}_{1}={M}_{o}^{c}\left(0\right)$, ${s}_{2}={M}_{e}^{c}\left(0\right)$, ${s}_{3}={M}_{o}^{c}\left(90\right)$ and ${s}_{4}={M}_{e}^{c}\left(90\right)$. 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:

$$\begin{array}{rcll}F& =& {K}_{e}/{K}_{o}& \text{}\end{array}$$

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

$$\begin{array}{rcll}F& =& \sqrt{\left(\frac{{s}_{1}}{{s}_{4}}.\frac{{s}_{3}}{{s}_{2}}\right)}& \text{}\end{array}$$

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

$$\begin{array}{rcll}{s}_{2}^{\prime}& =& {s}_{2}.F& \text{}\\ & =& \frac{{I}_{e}^{t}\left(0\right).{E}_{0}}{{K}_{e}}.\frac{{K}_{e}}{{K}_{o}}& \text{}\\ & =& \frac{{I}_{e}^{t}\left(0\right).{E}_{0}}{{K}_{o}}& \text{}\\ & & & \text{}\\ {s}_{4}^{\prime}& =& {s}_{4}.F& \text{}\\ & =& \frac{{I}_{o}^{t}\left(0\right).{E}_{90}}{{K}_{e}}.\frac{{K}_{e}}{{K}_{o}}& \text{}\\ & =& \frac{{I}_{o}^{t}\left(0\right).{E}_{90}}{{K}_{o}}& \text{}\end{array}$$

Summing the $O$ and corrected $E$ rays signals for exposure ${T}_{0}$ (${s}_{1}$ and ${s}_{2}^{\prime}$) gives:

$$\begin{array}{rcll}{s}_{1}+{s}_{2}^{\prime}& =& \frac{{I}_{o}^{t}\left(0\right).{E}_{0}}{{K}_{o}}+\frac{{I}_{e}^{t}\left(0\right).{E}_{0}}{{K}_{o}}& \text{}\\ & =& \frac{\left({I}_{o}^{t}\left(0\right)+{I}_{e}^{t}\left(0\right)\right).{E}_{0}}{{K}_{o}}& \text{}\\ & =& \frac{{I}_{T}.{E}_{0}}{{K}_{o}}& \text{}\end{array}$$

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 ${s}_{4}^{\prime}$) gives:

$$\begin{array}{rcll}{s}_{3}+{s}_{4}^{\prime}& =& \frac{{I}_{e}^{t}\left(0\right).{E}_{90}}{{K}_{o}}+\frac{{I}_{o}^{t}\left(0\right).{E}_{90}}{{K}_{o}}& \text{}\\ & =& \frac{\left({I}_{e}^{t}\left(0\right)+{I}_{o}^{t}\left(0\right)\right).{E}_{90}}{{K}_{o}}& \text{}\\ & =& \frac{{I}_{T}.{E}_{90}}{{K}_{o}}& \text{}\end{array}$$

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

$$\begin{array}{rcll}\frac{{s}_{3}+{s}_{4}^{\prime}}{{s}_{1}+{s}_{2}^{\prime}}& =& \frac{{I}_{T}.{E}_{90}}{{K}_{o}}.\frac{{K}_{o}}{{I}_{T}.{E}_{0}}& \text{}\\ & =& \frac{{E}_{90}}{{E}_{0}}& \text{}\end{array}$$

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.*