The two orthogonal components carry power independently, so their contributions simply add:

\(P_{avg}=\dfrac{E_{x}^{2}+E_{y}^{2}}{2\eta_{0}}\)

Substituting, with the free-space intrinsic impedance \(\eta_{0}=377\ \Omega\):

\(E_{x}^{2}+E_{y}^{2}=1.5^{2}+3^{2}=2.25+9=11.25\)

\(P_{avg}=\dfrac{11.25}{2\times377}=\dfrac{11.25}{754}=0.0149\ \text{W/m}^{2}\approx15\ \text{mW/m}^{2}\)

— option 2.

Why the 75° phase makes no difference to the power. This is the point the question is testing. The phase angle determines the polarisation — here elliptical, since the components are unequal and out of phase — and it decides the shape and tilt of the ellipse the field vector traces. But power is the sum of the two components' squares, and since the x and y directions are orthogonal there is no cross term to depend on the relative phase. The same two amplitudes at 0° would give a linearly polarised wave, and at 90° with equal amplitudes a circular one, yet the power would be identical.

Where the factor of 2 comes from. The amplitudes given are peak values, so the RMS value of each is \(E/\sqrt{2}\) and the time average of \(E^{2}\) is \(E^{2}/2\). Omitting it would give 30 mW/m² — which is precisely what option 3 offers, and the most likely error.

A note on the printed exponent. The wave is stated to travel in the +z direction, so the phase term should read \(\omega t-\beta z\) rather than \(\omega t-\beta x\). The slip does not affect the answer, since the power depends only on the amplitudes.

Hence, the average power per unit area is 15 mW/m².