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  • A linearly polarised electromagnetic wave given as <img alt="\mathrm{E=E_0 i c o s(k z-\omega t)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BE%3DE_0%20i%20c%20o%20s%28k%20z-%

A linearly polarised electromagnetic wave given as \mathrm{E=E_0 i c o s(k z-\omega t)} is incident normally on a perfectly reflecting infinite wall at \mathrm{z=a}. Assuming that the material of the wall is optically inactive, the reflected wave will be given as

 

Option: 1

\mathrm{ E_r=-E_0 i \cos (k z-w t)} \\


Option: 2

\mathrm{ E_r=E_0 i \cos (k z+\omega t)} \\


Option: 3

\mathrm{E_r=-E_0 i \cos (k z+\omega t)}\\


Option: 4

\mathrm{=E_0 i \sin (k z-\omega t)}


Answers (1)

best_answer

When a wave is reflected from denser medium, the reflected wave is without change in type of wave but with a change in phase by \mathrm{180^{\circ} or \: \pi} radian. Therefore, for the reflected wave we use \mathrm{z=-z, i=-1} and additional phase of \mathrm{\pi} in the incident wave. The incident EM wave is, \mathrm{E=E_0 i \cos (k z-\omega t)}. The reflected EM wave is

\mathrm{\mathbf{E}_1=E_0(-i) \cos [k(-z)-\omega t+\pi]=-E_0 i \cos [-(k z+\omega t)+\pi]}

             \begin{array}{rr} =\mathrm{E_0 l \cos [-(k z+\omega t)]} & \mathrm{{[\because \cos (\theta+\pi)=-\cos \theta]}} \\ \\=\mathrm{E_0 l \cos (k z+\omega t) }& \mathrm{{[\because \cos (-\theta)=\cos \theta]}} \end{array}

 

Posted by

Ritika Jonwal

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