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Linked Answer Questions 52 and 53: A monochromatic plane wave of wavelength \(\lambda = 600\,\text{nm}\) is propagating in the direction shown in the figure below. \(E_i\), \(E_r\), and \(E_t\) denote the incident, reflected, and transmitted electric field vectors associated with the wave. The angle of incidence \(\theta_i\) and the expression for \(E_i\) are:

1. 60° and \(E_0/\sqrt{2}\,\hat{a}_x\hat{a}_z\,e^{-j10^4(x+z)/(3\sqrt{2})}\) V/m
2. 45° and \(E_0/\sqrt{2}\,\hat{a}_x\hat{a}_z\,e^{-j10^4(x+z)/(3\sqrt{2})}\) V/m
3. 45° and \(E_0/\sqrt{2}\,\hat{a}_x\hat{a}_z\,e^{-j10^4(x+z)/(3\sqrt{2})}\) V/m
4. 60° and \(E_0/\sqrt{2}\,\hat{a}_x\hat{a}_z\,e^{-j10^4 z/3}\) V/m

Correct answer: 45° and \(E_0/\sqrt{2}\,\hat{a}_x\hat{a}_z\,e^{-j10^4(x+z)/(3\sqrt{2})}\) V/m

Solution

The phase factor shows the wave vector has equal components along \(x\) and \(z\), so the propagation direction makes a \(45^\circ\) angle with the normal in the \(x\)-\(z\) plane. The given field expression matches the corresponding plane-wave form with the same phase dependence.

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