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 understand the concept of Brewster's angle and the polarization of light.
Brewster's angle, also known as the polarizing angle, is the specific angle of incidence at which light is polarized when it reflects off a dielectric surface. At Brewster's angle, the reflected light becomes completely polarized, in either the p-polarized (parallel to the plane of incidence) or s-polarized (perpendicular to the plane of incidence) direction.
For light to be p-polarized, the incident light must be at the Brewster's angle with respect to the surface. This angle satisfies the condition:
tan(θ_B) = n2 / n1
Where:
θ_B is the Brewster's angle,
n1 is the refractive index of the incident medium (usually air),
n2 is the refractive index of the medium the light is incident on (e.g., a dielectric material).
On the other hand, for light to be s-polarized, the angle of incidence must be 90 degrees (perpendicular) to the plane of incidence. At this angle, no reflection occurs, and the light is fully transmitted into the medium.
To determine if there is a Brewster's angle for s-polarization, we evaluate the equation with θ = 90 degrees:
tan(90) = n2 / n1
The tangent of 90 degrees is undefined (or infinite). Thus, there is no angle of incidence that satisfies this equation. Consequently, there is no Brewster's angle for s-polarization.
In summary, the absence of a Brewster's angle for s-polarization results from the fact that s-polarized light is already fully transmitted at normal incidence (θ = 0) and does not undergo any reflection.