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# A linearly polarized electromagnetic wave given as $E=E_0 \hat i \cos (kz-wt)$ is incident normally on a perfectly reflecting infinite wall at $z=a$ Assuming that the material of the wall is optically inactive , the reflected wave will be given as

$\begin{array}{1 1} \overrightarrow{E_r} = -E_0 \hat i \cos(kz -wt) \\ \overrightarrow{E_r} = -E_0 \hat i \cos(kz +wt) \\ \overrightarrow{E_r} = -E_0 \hat i \cos(kz +wt) \\ \overrightarrow{E_r} = E_0 \hat i \sin(kz -wt) \end{array}$

For the reflected wave , $z= -z$
$\hat i = -\hat i$
and additional phase of $\pi$ in the incident wave .
$E_r= E_0 (- \hat i) cos (k(-z)-wt+\pi)$
But $\cos (- \theta ) = \cos \theta$
$\overrightarrow{E_r} = E_0 \hat i \cos(kz +wt)$