Expression for Reflected Wave at Brewster Angle

To identify the correct expression representing the reflected wave, we analyze the propagation vector and the polarization state of the wave:

1. Determination of Wave Vector and Phase Factor

From the given diagram:

• The plane interface separating air and the dielectric medium lies in the $xy$-plane ($z = 0$).
• Air lies in the region $z < 0$, while the dielectric medium lies in $z > 0$.
• The plane of incidence is defined by the incident wave vector and the normal to the boundary ($z$-axis), which is the $xz$-plane.
• The incident light travels in air towards the interface, having positive velocity components along both the $+x$ and $+z$ directions. Thus, its wave vector is: $\vec{k}_{\text{inc}} = k_x \hat{i} + k_z \hat{k}$

Upon reflection at the plane interface $z = 0$:

• The parallel component of the wave vector ($x$-component) remains unchanged: $k_{rx} = k_x$.
• The normal component ($z$-component) reverses its direction as the wave reflects back into air ($z < 0$): $k_{rz} = -k_z$.
• Therefore, the reflected wave vector is: $\vec{k}_{\text{ref}} = k_x \hat{i} - k_z \hat{k}$

Assuming $k_x = k_z = k$, the spatial phase part of the wave equation $\vec{k}_{\text{ref}} \cdot \vec{r} - \omega t$ becomes: $\vec{k}_{\text{ref}} \cdot \vec{r} - \omega t = (k\hat{i} - k\hat{k}) \cdot (x\hat{i} + y\hat{j} + z\hat{k}) - \omega t = kx - kz - \omega t$

2. Polarization at Brewster's Angle

• When unpolarized light is incident at Brewster's angle, the reflected wave becomes completely linearly polarized with its electric field vector $\vec{E}$ oscillating perpendicular to the plane of incidence (s-polarization).

• Since the plane of incidence is the $xz$-plane, the direction perpendicular to it is along the $y$-axis ($\hat{j}$).

• Furthermore, for a transverse electromagnetic wave, the electric field vector $\vec{E}$ must be perpendicular to the direction of propagation ($\vec{E} \cdot \vec{k}_{\text{ref}} = 0$).

  For $\vec{E} = E_x \hat{i} + E_y \hat{j}$ and $\vec{k}_{\text{ref}} = k\hat{i} - k\hat{k}$: $\vec{E} \cdot \vec{k}_{\text{ref}} = (E_x \hat{i} + E_y \hat{j}) \cdot (k\hat{i} - k\hat{k}) = E_x k = 0 \implies E_x = 0$

Thus, the electric field vector is purely along the $y$-direction ($\vec{E} = E_y \hat{j}$), perfectly consistent with s-polarization at Brewster's angle.

Conclusion

Combining the polarization state and the phase term, the expression for the reflected wave is: $(E_x\hat{i} + E_y\hat{j})\sin(kx - kz - \omega t)$

Correct Answer: A