A linearly polarized electromagnetic wave given as $E = E_{\circ} \hat{i} \cos (k z - ω t)$ 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

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$E_{r} = - E_{\circ} \hat{i} \cos (k z - ω t)$

$E_{r} = - E_{\circ} \hat{i} \cos (k z + ω t)$

$E_{r} = E_{\circ} \hat{i} \sin (k z + ω t)$

$E_{r} = E_{\circ} \hat{i} \cos (k z + ω t)$

answer is B.

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Detailed Solution

Since the EM wave is incident normally on the reflecting wall, it will reflect back in the same direction with $\hat{i} = - \hat{i}$ and $z = - z$ creating the additional phase of $π^{c}$ . We have given the equation of the incident wave, $E = E_{\circ} \hat{i} \cos (k z - ω t)$ .
Therefore, we can write the reflected wave equation as,
$E = E_{\circ} (- \hat{i}) \cos (k (- z) - ω t + π)$
$\Rightarrow E = - E_{\circ} \hat{i} \cos (- (k z + ω t) + π)$
$\Rightarrow E = - E_{\circ} \hat{i} \cos (π - (k z + ω t))$
We know the identity, $\cos (π - θ) = - \cos θ$
Using this identity in the above equation, we get,
$∴ E = E_{\circ} \hat{i} \cos (k z + ω t)$