Topic: Electromagnetics
A monochromatic plane wave of wavelength 500 μm is propagating in the direction as shown in the figure below. $\vec{E}_i, \vec{E}_r$ and $\vec{E}_t$ incident, reflected and transmitted electric field vectors associated with the wave. The expression for $\vec{E}_t$ and $\vec{E}_r$ are
- A. $\frac{E_0}{\sqrt{2}}(\hat{a}_x-\hat{a}_y)e^{-j\frac{2\pi \times 10^4(x+y)}{5\sqrt{2}}}$ V/m and $0.10\frac{E_0}{\sqrt{2}}(\hat{a}_x+\hat{a}_y)e^{-j\frac{2\pi \times 10^4(x-y)}{5\sqrt{2}}}$ V/m
- B. $\frac{E_0}{\sqrt{2}}(\hat{a}_x-\hat{a}_y)e^{-j\frac{2\pi \times 10^4(x+y)}{5\sqrt{2}}}$ V/m and $-0.10\frac{E_0}{\sqrt{2}}(\hat{a}_x+\hat{a}_y)e^{-j\frac{2\pi \times 10^4(x-y)}{5\sqrt{2}}}$ V/m
- C. $\frac{E_0}{\sqrt{2}}(\hat{a}_x+\hat{a}_y)e^{-j\frac{2\pi \times 10^4(x+y)}{5\sqrt{2}}}$ V/m and $\frac{E_0}{\sqrt{2}}(\hat{a}_x-\hat{a}_y)e^{-j\frac{2\pi \times 10^4(x+y)}{5\sqrt{2}}}$ V/m
- D. $\frac{E_0}{\sqrt{2}}(\hat{a}_x-\hat{a}_y)e^{-j\frac{2\pi \times 10^4(x-y)}{5\sqrt{2}}}$ V/m and $\frac{E_0}{\sqrt{2}}(\hat{a}_x+\hat{a}_y)e^{-j\frac{2\pi \times 10^4(x-y)}{5\sqrt{2}}}$ V/m
Answer: B. $\frac{E_0}{\sqrt{2}}(\hat{a}_x-\hat{a}_y)e^{-j\frac{2\pi \times 10^4(x+y)}{5\sqrt{2}}}$ V/m and $-0.10\frac{E_0}{\sqrt{2}}(\hat{a}_x+\hat{a}_y)e^{-j\frac{2\pi \times 10^4(x-y)}{5\sqrt{2}}}$ V/m
Explanation: Based on the Fresnel equations for reflection and transmission coefficients and given permittivities, $E_t=\frac{E_0}{\sqrt{2}}(\hat{a}_x-\hat{a}_y)e^{-j\frac{2\pi \times 10^4(x+y)}{5\sqrt{2}}}$ and $E_r=0.1\frac{E_0}{\sqrt{2}}(-\hat{a}_x-\hat{a}_y)e^{-j\frac{2\pi \times 10^4(x-y)}{5\sqrt{2}}}$. Note: The official key maps to (a), but mathematical derivation yields (b).