A linearly polarized electromagnetic wave given as E→=Eocos(kz−wt) i^ 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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a
Er→=−Eoi^cos(kz−wt)
b
Er→=Eoi^cos(kz+wt)
c
Er→=−Eoi^cos(kz+wt)
d
Er→=Eoi^cos(kz-wt)
answer is B.
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Detailed Solution
When a wave is reflected from a denser medium, the reflected wave is without change in type of wave but with a change in phase by 1800 or π radian. Therefore, additional phase of π is introduced in the incident wave . For the reflected wave direction is reversed means reflected wave travels in -i^ . As a result in the phase +kz is replaced by -kz . The incident electromagnetic wave isE→=Eocos(kz−wt)i^ The reflected em wave will be ,Er→=Eo(−i^)cos[-k(z)-wt+π] =−Eoi^cos[−(kz+wt)+π] =Eoi^cos[−(kz+wt)][cos(θ+π)=−cosθ] =Eoi^cos(kz+wt) [cos(−θ)=cosθ]