Question 3 of 8: Polarization of Two In-Phase, Counter-Propagating Orthogonally Polarized Plane Waves

Question 3: Polarization of Two In-Phase, Counter-Propagating Orthogonally Polarized Plane Waves (equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $f=10$ GHz waves in free space ($\eta_0=120\pi\approx376.7\ \Omega$); wave 1 travels in $+x$, $\hat z$-polarized (vertical), $S_1=9$ W/m²; wave 2 travels in $-x$, $\hat y$-polarized (horizontal), $S_2=3$ W/m²; at a reference point $P_0$ the two fields are in phase.

Find. (i) polarization at the in-phase point $P_0$; (ii) the RMS $E$-field amplitude there; (iii) the polarization at $P_0+\Delta x$.

Approach. Get each wave's RMS field from $S=E_{rms}^2/\eta_0$; at $P_0$ the two orthogonal, in-phase components add as a fixed-ratio vector (linear polarization); moving $\Delta x$ along the propagation axis shifts the two counter-propagating waves' phases in opposite senses, so recompute the relative phase there and classify the resulting polarization.

1. RMS field of each wave. Using $S=E_{rms}^2/\eta_0$, $$E_{1,rms}=\sqrt{S_1\eta_0}=\sqrt{9\times376.7}=58.23\ \text{V/m (vertical, }\hat z),$$ $$E_{2,rms}=\sqrt{S_2\eta_0}=\sqrt{3\times376.7}=33.62\ \text{V/m (horizontal, }\hat y).$$
2. (i) Polarization at the in-phase point. At $P_0$ both components oscillate as $\cos(\omega t)$ with no relative phase, so the resultant field vector $\mathbf E=E_{2,rms}\hat y+E_{1,rms}\hat z$ (in RMS terms) stays on a fixed line in the $y$-$z$ plane at all times — this is linear polarization, tilted from the horizontal ($\hat y$) axis by $$\theta=\tan^{-1}\!\left(\frac{E_{1,rms}}{E_{2,rms}}\right)=\tan^{-1}\!\left(\frac{58.23}{33.62}\right)=\boxed{60.0^\circ}.$$
3. (ii) Resultant RMS amplitude. The two component fields are orthogonal, so their RMS magnitudes add in quadrature: $$E_{rms}=\sqrt{E_{1,rms}^2+E_{2,rms}^2}=\sqrt{9\eta_0+3\eta_0}=\sqrt{12\times376.7}=\boxed{67.24\ \text{V/m}}.$$
4. (iii) Relative phase 3/8 cm away. Free-space wavelength $\lambda=c/f=(3\times10^8)/(10^{10})=0.03$ m $=3$ cm, so $k=2\pi/\lambda=209.4$ rad/m. Wave 1 ($e^{j(\omega t-kx)}$) loses phase $k\,\Delta x$ moving to $+\Delta x$; wave 2 ($e^{j(\omega t+kx)}$, travelling $-x$) gains phase $k\,\Delta x$ over the same move. The phase between the two components therefore opens up by $$\Delta\phi=2k\,\Delta x=2\left(\frac{2\pi}{0.03}\right)(0.00375)=\frac{\pi}{2}\ (90^\circ),$$ using $\Delta x=3/8\ \text{cm}=0.00375$ m (exactly $\lambda/8$ per wave).
5. Classify the new polarization. Two orthogonal field components of unequal amplitude with a $90^\circ$ relative phase, aligned with the principal ($y,z$) axes, describe an elliptically polarized wave whose axes are the horizontal and vertical directions, with axial ratio $$AR=\frac{E_{1,rms}}{E_{2,rms}}=\sqrt{\frac{S_1}{S_2}}=\sqrt{3}=\boxed{1.73:1}$$ (major axis vertical). The sense of rotation (right- or left-hand) is set by which component leads: moving in $+x$, wave 1 (vertical) is retarded and wave 2 (horizontal) is advanced, so the horizontal component leads the vertical by $90^\circ$.