Question 3 of 8: Polarisation of two counter-propagating orthogonal waves

Question 3: Polarisation of two counter-propagating orthogonal waves (20 marks)

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. Two plane waves of the same frequency share a line of propagation but travel along it in opposite senses. Their electric fields are mutually perpendicular — one vertical, one horizontal — and both are transverse to that line.

Find. The polarisation state and the rms electric field at an in-phase point, and the polarisation state $3/8$ cm further along the line of propagation.

Approach. Convert each power density to an rms field, combine the two perpendicular components at the in-phase point, then track how counter-propagation makes their relative phase advance with position and re-read the state at the offset point.

1. Recover the component fields. For a plane wave in free space $S = E_{rms}^{2}/\eta_0$, so $$E_v = \sqrt{S_v\eta_0} = \sqrt{9 \times 377} = 58.25\ \text{V/m},\qquad E_h = \sqrt{S_h\eta_0} = \sqrt{3 \times 377} = 33.63\ \text{V/m}$$
2. (i) State the polarisation at the in-phase point. Two perpendicular components oscillating in step always sum to a straight line, so the field is linearly polarised at $$\tau = \arctan\!\frac{E_v}{E_h} = \arctan\!\frac{58.25}{33.63} = \boxed{60.00^{\circ}\ \text{above horizontal}}$$
3. (ii) Combine them for the rms amplitude. Perpendicular components add in quadrature, so $$E_{rms} = \sqrt{E_v^{2} + E_h^{2}} = \sqrt{(S_v + S_h)\eta_0} = \sqrt{12 \times 377} = \boxed{67.26\ \text{V/m}}$$
4. Find how fast the relative phase slips. The two waves travel in opposite directions, so along the common axis their phases go as $-ky$ and $+ky$. The difference therefore advances at $2k$ — twice the rate of a single wave — and over the stated offset $$\Delta\phi = 2k\,\Delta y = \frac{4\pi}{\lambda}\,\Delta y = \frac{4\pi}{3.00}\times\frac{3}{8} = \frac{\pi}{2} = 90.0^{\circ}$$
5. (iii) Read the new state. Perpendicular components in exact quadrature trace an ellipse whose axes lie along the components themselves. The amplitudes are unequal, so the figure is elliptical rather than circular, with $$\text{axial ratio} = \frac{E_v}{E_h} = \sqrt{3} = \boxed{1.732,\ \text{major axis vertical}}$$

The offset of $3/8$ cm is exactly $\lambda/8$, and it is the doubling of the phase-slip rate that turns an eighth of a wavelength into a full quarter-cycle of relative phase. The same doubling means the polarisation state repeats every $\lambda/4$ rather than every $\lambda$, running linear at $60^{\circ}$, elliptical, linear at $-60^{\circ}$, elliptical of the opposite handedness, and back.

Handedness needs a stated reference. Taking $+y$ as the direction in which the vertically polarised wave travels, the horizontal component leads the vertical one at the offset point, so the tip turns from horizontal toward vertical — a left-handed ellipse with respect to $+y$, and therefore right-handed with respect to the horizontally polarised wave’s own direction of travel. Because the question does not say which wave defines “the propagation direction”, both descriptions are correct and only the reference must be quoted with the answer.

One further point is worth making explicitly, because it separates this problem from an ordinary standing-wave problem: the two waves are orthogonally polarised, so they cannot interfere. There is no intensity standing wave anywhere, each component keeps a constant amplitude at every point, and the rms resultant is $67.26\ \text{V/m}$ at the offset point just as it is at the in-phase point. Only the traced figure changes.