GATE EC 2013 Electromagnetics Solution Q 3

# GATE EC 2013 Electromagnetics Solution Q 3

A monochromatic plane wave of wavelength Î» = 600 Î¼m 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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