Discussion
Consider the superposition of two electromagnetic waves with their electric field vectors given by \(\vec{E}_{1}(z,t) = \hat{\imath} A_{1} \cos(kz - \omega t)\) and \(\vec{E}_{2}(z,t) = \hat{\jmath} A_{2} \sin(kz - \omega t + \phi)\), where \(A_{1}\) and \(A_{2}\) are the amplitudes, \(k\) is the wavenumber, \(\omega\) is the angular frequency, and \(\phi\) is the relative phase. Which of the following options represents a resultant elliptically polarized wave with its semi-major axis either along \(\hat{\imath}\) or \(\hat{\jmath}\)?
Answer: A
The wave is elliptically polarised with its axes along \(\hat{\imath}\) and \(\hat{\jmath}\) when \(\phi = 0\) and \(A_1 \neq A_2\).
Write \(\theta = kz - \omega t\). With \(\phi = 0\) the components are \(E_x = A_1 \cos\theta\) and \(E_y = A_2 \sin\theta\), so
\[ \frac{E_x^{2}}{A_1^{2}} + \frac{E_y^{2}}{A_2^{2}} = 1 \]
The tip of the field traces an ellipse whose semi-axes \(A_1\) and \(A_2\) lie along \(\hat{\imath}\) and \(\hat{\jmath}\). With \(A_1 \neq A_2\) it is a true ellipse; with \(A_1 = A_2\) it becomes a circle, so the \(\phi = 0\), \(A_1 = A_2\) option describes circular polarisation.
With \(\phi = \pi/2\), \(\sin(\theta + \pi/2) = \cos\theta\), so both components oscillate in phase and the wave is linearly polarised, whether or not the amplitudes are equal.
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