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A plane wave of wavelength \(\lambda\) is travelling in a direction making an angle \(30^\circ\) with the positive x-axis and \(90^\circ\) with the positive y-axis. The electric field of the plane wave can be represented as \((E_0\) is a constant\().

1. \(\mathbf{E}=\hat{y}E_0 e^{j\left(\omega t-\frac{\sqrt{3}\pi}{\lambda}x-\frac{\pi}{\lambda}z\right)}\)
2. \(\mathbf{E}=\hat{y}E_0 e^{j\left(\omega t-\frac{\pi}{\lambda}x-\frac{\sqrt{3}\pi}{\lambda}z\right)}\)
3. \(\mathbf{E}=\hat{y}E_0 e^{j\left(\omega t+\frac{\sqrt{3}\pi}{\lambda}x+\frac{\pi}{\lambda}z\right)}\)
4. \(\mathbf{E}=\hat{y}E_0 e^{j\left(\omega t-\frac{\pi}{\lambda}x+\frac{\sqrt{3}\pi}{2\lambda}z\right)}\)

Correct answer: \(\mathbf{E}=\hat{y}E_0 e^{j\left(\omega t-\frac{\sqrt{3}\pi}{\lambda}x-\frac{\pi}{\lambda}z\right)}\)

Solution

For a plane wave, the phase term is \(\omega t-\mathbf{k}\cdot\mathbf{r}\). The given direction implies the propagation vector has components proportional to \(\cos 30^\circ\) along x and zero along y, with the remaining component along z. Substituting the corresponding phase constants gives the stated expression.

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