Polarisation question from JEE Main 2026 (4 Apr, Shift 1)

Answer: (a)

1. The surface is the $xy$-plane with air on the $-z$ side, so the reflected wave must travel back into air: its wave vector has a $-z$ component.
2. Only (a) and (d) have a $-kz$ term; (b) runs along the surface and (c) goes into the medium.
3. At Brewster angle the reflected light is plane polarised perpendicular to the plane of incidence; (a) with $\vec k \propto \hat i - \hat k$ has plane of incidence $xz$ and field along $\hat j$ (with the $\hat i$ term as written in the option).
4. So the answer is (a) $(E_x\hat i + E_y\hat j)\sin(kx - kz - \omega t)$.