A wave travelling in the + z -direction is the resultant of two linearly polarized components, E_{x} = 3cosωt, and Ey = 2cos(ωt + 45°). Determine
(a) the axial ratio, and
(b) the angle between the major axis of the polarization ellipse and the +x axis.
Wave is travelling in +z direction.
\begin{aligned}&E_{ x }=3 \cos \omega t \\&E_{ y }=2 \cos \left(\omega t +45^{\circ}\right) \\&t \text { varies from } 0 \text { to } T=\frac{2 \pi}{\omega} \\& OA _{\max }= OP =\text { semi major axis of ellipse. } \\& OA _{\min }= OQ \\&O A^{2}=E^{2}=E_{x}^{2}+E_{y}^{2} \\&E^{2}=(3 \cos \omega t)^{2}+\left(2 \cos \left(\omega t+45^{\circ}\right)\right)^{2}\end{aligned}
(O A)^{2}=E^{2}=6.5+4.5 \cos (2 \omega t )-2 \sin 2 \omega t (1)
O A \text { is max, when, } \frac{d E}{d t}=0 , and Solving (1)
\begin{aligned}&\tan (2 \omega t )=-4 / 9 \\&\Rightarrow \omega t =-11.98^{\circ} \\&\text { So, } E_{ x }=3 \cos (11.98)=2.934 \\&E_{ y }=2 \cos (-11.98+45)=1.676 \\&E=\sqrt{E_{x}^{2}+E_{y}^{2}}=3.378 \\&\text { at, } \omega t =78.02 \\&E_{ y }=0.6227 \text { and } E_{ y }=-1.08 \\&E=\sqrt{E_{x}^{2}+E_{y}^{2}}=1.246 \\&\text { So, } E_{\max }=3.3378 \\&E_{\min }=1.246 \\&\text { Axial ratio }=\frac{E_{\max }}{E_{\min }}=\frac{3.378}{1.246}=2.71\end{aligned}(b) \tan \theta=\frac{E_{y}}{E x} \text { at } \operatorname{Emax}
\tan \theta=\frac{1.676}{2.934} \Rightarrow \vartheta=29.73^{\circ}
