Show that there is no Brewster’s angle for s polarization.
To show that there is no Brewster's angle for s polarization, we need to first understand the concept of Brewster's angle and how it is related to polarization.
Brewster's angle is defined as the angle of incidence at which light undergoes only perpendicular polarization, also known as "p polarization," when reflecting off a dielectric surface. In other words, the reflected light at Brewster's angle is completely polarized in the plane of incidence (parallel to the surface).
Now, let's consider s polarization, which is also known as "TE polarization." In s polarization, the electric field of the incident light is parallel to the plane of incidence. The reflected light in s polarization remains in the same plane of incidence.
To determine if there is a Brewster's angle for s polarization, we need to find the angle at which the reflected light is completely polarized in the plane of incidence, as is the case for p polarization.
Let's assume that there is a Brewster's angle (let's call it θ_Bs) for s polarization. At this angle, the reflected light would be completely polarized in the plane of incidence. However, this assumption contradicts the fundamental nature of s polarization, which means the reflected light will always remain in the same plane of incidence, regardless of the angle of incidence.
Therefore, we can conclude that there is no Brewster's angle for s polarization. The reflected light in s polarization will always remain within the plane of incidence, irrespective of the angle of incidence.