Quiz 1.11

For the given choice of Fig. (1.3) we have $\uvec{k}=\uvec{z}$, $\uvec{p}=\uvec{x}\cos\psi+\uvec{y}\sin\psi$ and $\uvec{q}=-\uvec{x}\sin\psi+\uvec{y}\cos\psi$ Eq. (1.31) is written

$$\vec{E}=\uvec{x}\alpha(\cos\psi\cos\beta\cos\phi-\sin\psi\sin\beta\sin\phi)+ \uvec{y}\alpha(\sin\psi\cos\beta\cos\phi+\cos\psi\sin\beta\sin\phi)$$

Identifying the $\uvec{x}$ and $\uvec{y}$ coordinates with Eq. (1.35)

$$E_{0x}\cos\delta_x =\alpha\cos\psi\cos\beta, E_{0x}\sin\delta_x =\alpha\sin\psi\sin\beta, \tan\delta_x =\tan\psi\tan\beta$$

$$E_{0y}\cos\delta_y =\alpha\sin\psi\cos\beta, E_{0y}\sin\delta_y =-\alpha\cos\psi\sin\beta, \tan\delta_y =-\frac{\tan\beta}{\tan\psi}$$

and Eq. (1.36) are easily deduced. Using Eq. (1.33) and Eq. (1.36), the two first Stocke's parameters are found to be $I=E_{0x}^2+E_{0y}^2$ and $Q=E_{0x}^2-E_{0y}^2$.

Now, we note that $U=-V/\tan\delta$ and using Eq. (1.34), we find $U^2=4E_{0x}^2E_{0y}^2\cos^2\delta$ and $V^2=4E_{0x}^2E_{0y}^2\sin^2\delta$. The sign is checked using Eq. (1.33) and the two last Stocke's parameters are found to be $U=2E_{0x}E_{0y}\cos\delta$ $V=-2E_{0x}E_{0y}\sin\delta$.