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A plane wave propagating in air with \(\vec{E} = (8\hat{a}_x + 6\hat{a}_y + 5\hat{a}_z)e^{j(\omega t + 3x - 4y)}\) V/m is incident on a perfectly conducting slab positioned at \(x \le 0\). The \(\vec{E}\) field of the reflected wave is

1. \((−8\hat{a}_x − 6\hat{a}_y − 5\hat{a}_z)e^{j(\omega t+3x+4y)}\) V/m
2. \((−8\hat{a}_x + 6\hat{a}_y − 5\hat{a}_z)e^{j(\omega t+3x+4y)}\) V/m
3. \((−8\hat{a}_x − 6\hat{a}_y − 5\hat{a}_z)e^{j(\omega t−3x−4y)}\) V/m
4. \((−8\hat{a}_x + 6\hat{a}_y − 5\hat{a}_z)e^{j(\omega t−3x−4y)}\) V/m

Correct answer: \((−8\hat{a}_x − 6\hat{a}_y − 5\hat{a}_z)e^{j(\omega t+3x+4y)}\) V/m

Solution

For reflection from a perfect conductor, the reflected wave must make the total tangential electric field zero at the surface. The reflected wave therefore has the opposite propagation direction in the normal coordinate and an electric field that cancels the incident field at the boundary. This leads to the given reflected field expression.

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