GATE EC 2012 Electromagnetics Solution Q 5

GATE EC 2012 Electromagnetics Solution Q 5

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

(a) \left(-8{\widehat a}_x-6{\widehat a}_y-5{\widehat a}_z\right)e^{j\left(\omega t+3x+4y\right)}\;V/m

(b) \left(-8{\widehat a}_x+6{\widehat a}_y-5{\widehat a}_z\right)e^{j\left(\omega t+3x+4y\right)}\;V/m

(c) \left(-8{\widehat a}_x-6{\widehat a}_y-5{\widehat a}_z\right)e^{j\left(\omega t-3x-4y\right)}\;V/m

(d) \left(8{\widehat a}_x+6{\widehat a}_y-5{\widehat a}_z\right)e^{j\left(\omega t-3x-4y\right)}\;V/m

Ans : c)

Explanation

We know that according to Boundary conditions for Conductors normal and tangential components of reflected E would be zero if the charge density is zero. In other words the Electric wave is totally reflected and cancel out the incident Electric field components. So the reflected wave components are given as

\overrightarrow E=\left(-8{\widehat a}_x-6{\widehat a}_y-5{\widehat a}_z\right)

Also the incident wave is travelling along negative X direction and meets the conductor medium at x = 0. So the reflected wave must be travel back along the positive X direction. In other words, the propagation vector should have negative x.

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