GATE EC 2013 Electromagnetics Solution Q 2

GATE EC 2013 Electromagnetics Solution Q 2

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.

GATE EC 2013 Electromagnetics Solution img1 - Grad Plus

The angle of incidence \theta_i and the expression for {\overrightarrow E}_i are

(a) 60^0and\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) 45^0and\frac{E_0}{\sqrt2}\left({\widehat a}_x+{\widehat a}_z\right)e^{-j\frac{\pi\times10^4z}3}V/m

(c) 45^0and\frac{E_0}{\sqrt2}\left({\widehat a}_x-{\widehat a}_z\right)e^{-j\frac{\pi\times10^4z}3}V/m

(d) 60^0and\frac{E_0}{\sqrt2}\left({\widehat a}_x-{\widehat a}_z\right)e^{-j\frac{\pi\times10^4\;z}3}V/m

Ans : (c)

Explanation

The refractive index is given as:-

R.I=\frac{\sin\theta_i}{\sin\theta_t}\;\;\;\;Also\;\;\;\;R.I=\frac{v_1}{v_2}=\sqrt{\frac{\varepsilon_{r2}}{\varepsilon_{r1}}} \therefore \sqrt{\in_{r_1}}\sin\theta_i=\sqrt{\in_{r_2}}\sin\theta_t 1\sin\theta_i=\sqrt{4.5}\sin\left(19.2^0\right) \sin\theta_i=2.12\times0.3289 \sin\theta_i=0.697 \theta_i=\sin^{-1}\left(0.697\right) \theta_i=44.2\cong45^0

For finding incidence electric field, we would require propagation vector and field components. From the diagram we can write as follows:-

{\overrightarrow E}<em>i=E</em>{i0}\left(\cos\theta_i{\widehat a}_x-\sin\theta_i{\widehat a}_z\right)e^{-j\beta}\left(x\sin\theta_i+z\cos\theta_i\right) {\overrightarrow E}_i=E_0(\cos45^0{\widehat a}_x-\sin45^0{\widehat a}_z)e^{-1}\left(\frac{2\pi}{600\times10^{-6}}\right)(x\sin45^0+z\cos45^0) {\overrightarrow E}_i=E_0\frac1{\sqrt2}\left({\widehat a}_x-{\widehat a}_z\right)e^{\frac{-j\pi\times10^4}3\frac1{\sqrt2}\left(x+z\right)} {\overrightarrow E}_i=\frac{E_0}{\sqrt2}\left({\widehat a}_x-{\widehat a}_z\right)e^{-\frac{j\pi\times10^4\left(x+z\right)}{3\sqrt2}}V/m

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