Calculating the Polarization for Dual-beam Data

This section gives a mathematical description of the calculation of the degree and orientation of the polarization for dual-beam data, based on the observed intensities. It is assumed that any required corrections (such as flat-fielding, sky-subtraction, etc), have already been applied.

Each target exposure measures the components of the incoming light polarized in two different orthogonal directions (depending on the orientation of the half-wave plate). If the symbol $I_{\alpha}$ is used to represent the intensity of the component polarized at an angle $\alpha$ to the reference direction, then in each exposure the $O$ ray image records $I_{\alpha}$ and the $E$ ray image records $I_{\alpha+90}$.

The first exposure ($T_{0}$) is taken with the half-wave plate in its 0 degrees position. The $O$ ray image will then record the intensity $I_{0}$ and the $E$ ray image will record the intensity $I_{90}$. Malus' law gives these intensities as:

$$\begin{eqnarray*} I_{0} & = & I_{p}.\cos^{2}\theta + \frac{I_{u}}{2} \\ I_{90... ...frac{I_{u}}{2} \\ & = & I_{p}.\sin^{2}\theta + \frac{I_{u}}{2} \end{eqnarray*}$$

Here, $I_{p}$ and $I_{u}$ are the polarized and unpolarized intensities in the incoming light, and $\theta$ is the angle between the plane of polarization and the reference direction (i.e. the 0 degrees position). The total intensity $I$ is the sum of $I_{p}$ and $I_{u}$, and can be found as follows:

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

Thus, summing the $O$ and the $E$ ray images gives the total intensity image.

The half-wave plate is now rotated by 22.5 degrees and another exposure ($T_{22.5}$) is taken. Rotating the half-wave plate by 22.5 degrees is equivalent to rotating the analyser by 45 degrees, and so the $O$ and $E$ ray images now record the intensities $I_{45}$ and $I_{135}$, where:

$$\begin{eqnarray*} I_{45} & = & I_{p}.\cos^{2}(45 - \theta) + \frac{I_{u}}{2} \\... ...{u}}{2} \\ & = & I_{p}.\sin^{2}(45 - \theta) + \frac{I_{u}}{2} \end{eqnarray*}$$

Again, the sum of the $O$ and $E$ ray intensities ( $I_{45}+I_{135}$) gives the total intensity $I$.

The mathematical description of polarization can be simplified by using the quantities $Q$ and $U$ defined as:

$$\begin{eqnarray*} Q & = & I_{p}.\cos 2\theta \\ U & = & I_{p}.\sin 2\theta \end{eqnarray*}$$

Together with the total intensity, $I$, these quantities are known as Stokes parameters¹². Using these definitions, the polarized intensity, $I_{p}$, is:

$$\begin{eqnarray*} I_{p} & = & \sqrt{ Q^{2} + U^{2} } \end{eqnarray*}$$

and the orientation of the plane of polarization is:

$$\begin{eqnarray*} \theta & = & 0.5.\arctan (U/Q) \end{eqnarray*}$$

The degree of polarization, $p$, is the ratio of polarized to total intensity, $I_{p}/I$. Using the expressions for $I_{0}$ and $I_{90}$ above, it can be seen that:

$$\begin{eqnarray*} I_{0} - I_{90} & = & I_{p}.\cos^{2}\theta - I_{p}.\sin^{2}\theta \\ & = & I_{p}.\cos 2\theta \\ & = & Q \end{eqnarray*}$$

Likewise,

$$\begin{eqnarray*} I_{45} - I_{135} & = & I_{p}.\cos^{2}(45 - \theta) - I_{p}.\s... ...}.\cos (90 - 2\theta) \\ & = & I_{p}.\sin 2\theta \\ & = & U \end{eqnarray*}$$

Thus, using the four intensities $I_{0}$, $I_{45}$, $I_{90}$ and $I_{135}$ (obtained on two exposures with half-wave plate positions 0 degrees and 45 degrees), both $Q$ and $U$ can be found, together with two independent estimates of $I$. This allows the polarized intensity, the degree of polarization and the orientation of the plane of polarization to be found using only two exposures. However, it is usually advisable to obtain additional exposures at half-wave plate positions of 45 degrees and 67.5 degrees in order to correct for any difference in the sensitivity of the two channels of the polarimeter (such as may be produced for instance by a polarized flat-field).