GATE EC 2013 Electromagnetics Solution Q 3 - Grad Plus

# GATE EC 2013 Electromagnetics Solution Q 3

A monochromatic plane wave of wavelength λ = 600 μm is propagating in the direction as shown in figure below. Ei, Er and Et denote incident, reflected and transmitted electric field vectors associated with the wave.

The expression for {\overrightarrow E}_r

(a) 0.23\frac{E_0}{\sqrt2}\left({\widehat a}_x+{\widehat a}_z\right)e^{-j\frac{\pi\times10^4\left(x-z\right)}{3\sqrt2}}V/m

(b) -\frac{E_0}{\sqrt2}\left({\widehat a}_x+{\widehat a}_z\right)e^{j\frac{\pi\times10^4z}3}V/m

(c) 0.44\frac{E_0}{\sqrt2}\left({\widehat a}_x+{\widehat a}_z\right)e^{-j\frac{\pi\times10^4\left(x-z\right)}3}V/m

(d) \frac{E_0}{\sqrt2}\left({\widehat a}_x+{\widehat a}_z\right)e^{-j\frac{\pi\times10^4\left(x-z\right)}3}V/m

Ans : (a)

Explanation

Reflection coefficient is given as,

\tau_p=\frac{\eta_2\cos\theta_t-\eta_1\cos\theta_i}{\eta_2\cos\theta_t+\eta_1\cos\theta_i} \eta_1=\eta_0\;and\;\eta_2=\frac{\eta_0}{\sqrt{4.5}} \tau_p=\frac{\cos19.2^0-\sqrt{4.5}\cos45^0}{\cos19.2^0+\sqrt{4.5}\cos45^0}=-0.23

From the diagram the reflected E has propagation vector and components can be written as

E_r=\left(-0.23\right)\frac{E_0}{\sqrt2}\left(-{\widehat a}_x-{\widehat a}_z\right)e^{-j\frac{\pi\times10^4\left(x-z\right)}{3\sqrt2}} =\left(0.23\right)\frac{E_0}{\sqrt2}\left({\widehat a}_x+{\widehat a}_z\right)e^{-j\frac{\pi\times10^4\left(x-z\right)}{3\sqrt2}}
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