An unpolarized light is incident on the plane interface of air-dielectric medium shown in figure. If the incident angle is equal to Brewster angle, identify the expression representing reflected wave.
Step 1: Concept Used [4pt]
At Brewster angle, the reflected light becomes completely plane-polarized.
This means the electric field vector of reflected light is perpendicular to the plane of incidence.
Step 2: General Equation of EM Wave [4pt]
The electric field of a plane electromagnetic wave is written as:
\[
\vec{E}=\vec{E_0}\sin(\vec{k}\cdot\vec{r}-\omega t)
\]
where:
\[
\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}
\]
Step 3: Direction of Reflected Wave [4pt]
For the standard figure, the plane of incidence is the \(xz\)-plane.
Hence, for reflected light:
\[
\vec{k}=k_x\hat{i}-k_z\hat{k}
\]
Now,
\[
\vec{k}\cdot\vec{r}
=
(k_x\hat{i}-k_z\hat{k})\cdot(x\hat{i}+y\hat{j}+z\hat{k})
\]
\[
= k_xx-k_zz
\]
Therefore, phase term becomes:
\[
k_xx-k_zz-\omega t
\]
or simply,
\[
kx-kz-\omega t
\]
Step 4: Compare with Options [4pt]
The option matching this phase term is:
\[
(E_x \hat{i}+E_y \hat{j})\sin(kx-kz-\omega t)
\]
Final Answer:
\[
\boxed{(E_x \hat{i}+E_y \hat{j})\sin(kx-kz-\omega t)}
\]
