Question 7 of 8: Interference of two crossed plane waves

Question 7: Interference of two crossed plane 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 equal-strength linearly polarised plane waves cross at a shallow angle and interfere. The stated vanishing of the total magnetic field at the origin is not decoration — it is the datum that fixes the polarisation of both waves.

Find. (i) the rms amplitude of the total electric field at the origin, and (ii) at least one point where the total magnetic field is a maximum.

Approach. Test the two possible linear polarisations against the stated null, keep the one that can produce it, then add the two phasors as functions of position to read off both answers.

1. Set up the geometry. Writing $\alpha = 30^\circ$ for the half-angle, the two unit propagation vectors are $$\hat{k}_{1} = \sin\alpha\,\hat{x} + \cos\alpha\,\hat{y}, \qquad \hat{k}_{2} = -\sin\alpha\,\hat{x} + \cos\alpha\,\hat{y}$$ so both waves travel generally in $+y$ but carry opposite $x$ components of wavenumber. The wavelength is $\lambda = c/f = 3.00\ \text{cm}$ and $k = 2\pi/\lambda$.
2. Use the null to identify the polarisation. Each wave is linearly polarised, so its $\mathbf{E}$ is either along $\hat{z}$ (perpendicular to the plane containing both beams) or lies in the $x$–$y$ plane. If both were $z$-polarised, their magnetic fields would be $\hat{k}\times\hat{z}$, whose $y$ components are $\mp\sin\alpha$ and cannot cancel unless the $x$ components reinforce — the two requirements are contradictory, so no choice of relative sign gives a null. If instead both $\mathbf{E}$ vectors lie in the plane, then $\mathbf{H} = (1/\eta_0)\hat{k}\times\mathbf{E}$ is purely along $\hat{z}$ for both, and two collinear vectors can cancel. Hence the waves are polarised in the plane of propagation and are in antiphase.
3. Scale the fields from the given power density. For a uniform plane wave in free space $S = E_{\text{rms}}^{2}/\eta_0$, so each wave has $$E_{1} = \sqrt{S_1\eta_0} = \sqrt{(5\times10^{-6})(376.8)} = 43.4\ \text{mV/m (rms)}, \qquad H_{1} = \frac{E_1}{\eta_0} = 115.2\ \mu\text{A/m}$$
4. Add the electric fields at the origin. The in-plane unit polarisation vectors, each perpendicular to its own $\hat{k}$, are $\hat{e}_1 = \cos\alpha\,\hat{x}-\sin\alpha\,\hat{y}$ and $\hat{e}_2 = \cos\alpha\,\hat{x}+\sin\alpha\,\hat{y}$. With the antiphase relation forced by the null, $$\mathbf{E}_{\text{tot}}(0) = E_1\hat{e}_1 - E_1\hat{e}_2 = -2E_1\sin\alpha\,\hat{y} = -E_1\hat{y}$$ because $2\sin 30^\circ = 1$ exactly. The two $x$ components cancel and the two $y$ components add, leaving $$\boxed{E_{\text{rms}}(0) = 43.4\ \text{mV/m}}$$ numerically equal to a single wave — a coincidence of the 30° geometry, not a general result.
5. Build the magnetic standing wave. Both magnetic fields point along $\hat{z}$ and carry opposite signs, so $$\mathbf{H}_{\text{tot}}(\mathbf{r}) = \frac{E_1}{\eta_0}\,\hat{z}\,e^{-jk y\cos\alpha}\left(e^{+jkx\sin\alpha} - e^{-jkx\sin\alpha}\right) = 2j\frac{E_1}{\eta_0}\sin(kx\sin\alpha)\,e^{-jky\cos\alpha}\,\hat{z}$$ The $y$ dependence is a pure travelling phase, so the pattern stands only along $x$, exactly as the question's hint says.
6. Locate the first maximum. The magnitude $|H| = 2H_1|\sin(kx\sin\alpha)|$ peaks when $$kx\sin\alpha = \frac{\pi}{2} \;\Longrightarrow\; x = \frac{\lambda}{4\sin\alpha} = \frac{3.00}{4(0.5)} = \boxed{1.50\ \text{cm}}$$ for any $y$ and any $z$: the maxima form the whole plane $x = 1.50\ \text{cm}$. Successive maxima are $\lambda/(2\sin\alpha) = 3.00\ \text{cm}$ apart, so $x = \pm 1.50,\ \pm 4.50\ \text{cm}, \ldots$ all qualify, and the peak value is $2H_1 = 230.4\ \mu\text{A/m}$ rms.
7. Follow the electric field away from the origin. Repeating the sum at a general point gives $$\mathbf{E}_{\text{tot}} = -2E_1\left[j\cos\alpha\sin(kx\sin\alpha)\,\hat{x} + \sin\alpha\cos(kx\sin\alpha)\,\hat{y}\right]e^{-jky\cos\alpha}$$ which reduces to $-E_1\hat{y}$ at the origin, as found above. Its magnitude at $\alpha = 30^\circ$ is $2E_1\sqrt{\tfrac{3}{4}\sin^2(kx\sin\alpha) + \tfrac{1}{4}\cos^2(kx\sin\alpha)}$, which is not constant: its minimum $E_1 = 43.4\ \text{mV/m}$ lies on the $|H|$ nulls (including the origin) and its maximum $2E_1\cos\alpha = 75.2\ \text{mV/m}$ lies on the $|H|$ maxima such as $x = 1.50\ \text{cm}$. The $x$ and $y$ components are in time quadrature, so between those planes the total electric field is elliptically polarised; only the $y$ component, which is collinear for both waves, forms a true standing pattern.