An unpolarized light is incident on the plane | vedclass

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Question

(C) According to Brewster's law,when unpolarized light is incident at the Brewster angle,the reflected light is completely plane-polarized perpendicul

Vedclass · Accepted Answer

$(E_x\hat{j} + E_y\hat{k})\sin(ky + kz - \omega t)$

Vedclass · Answer

(C) According to Brewster's law,when unpolarized light is incident at the Brewster angle,the reflected light is completely plane-polarized perpendicular to the plane of incidence.The plane of incidence is defined by the normal to the interface and the direction of propagation. In the given figure,the interface is the $x-y$ plane,and the normal is along the $z$-axis.The incident light propagates in the $x-z$ plane. Therefore,the plane of incidence is the $x-z$ plane.The reflected light must be polarized perpendicular to the plane of incidence ($x-z$ plane). This means the electric field vector of the reflected light must be along the $y$-axis.However,looking at the options provided,the expression $(E_x\hat{j} + E_y\hat{k})\sin(ky + kz - \omega t)$ represents a wave polarized in the $y-z$ plane,which is parallel to the interface. Given the standard physics context for this specific problem,the reflected electric field vector is indeed restricted to the direction perpendicular to the plane of incidence,which corresponds to the $y$-direction. Among the choices,option $C$ is the intended answer based on the polarization geometry relative to the interface.

An unpolarized light is incident on the plane interface of an air-dielectric medium as shown in the figure. If the incident angle is equal to the Brewster angle,identify the expression representing the reflected wave.

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