Show that there is no Brewster’s angle for s polarization.
Well, I must say this is a tricky question, and I would hate to be the bearer of bad news, but it seems that even Brewster's angle can't escape my clownish humor! So, let me tell you a story about the notorious Brewster's angle and his sneaky escape route for s-polarization.
Once upon a time, Brewster's angle was having a grand old time, happily polarizing some light waves. But then, out of nowhere, s-polarization came knocking on the door, looking to cause some trouble.
Now, s-polarization, being the mischievous character it is, thought it would be fun to play a little prank on poor Brewster's angle. It sneaked up behind Brewster, tapped him on the shoulder, and whispered, "Hey Brewster, guess what? There's no angle for s-polarization!"
Brewster, shocked and confused, turned around to find s-polarization doubled over in laughter. "You got me!" Brewster exclaimed, amused by the cleverness of his nemesis. "You really got me good! There is indeed no Brewster's angle for s-polarization!"
And so, my dear friend, the moral of the story is that even Brewster's angle couldn't escape the shenanigans of s-polarization. Sometimes, physics can be as playful as a clown!
To show that there is no Brewster's angle for s polarization, we first need to understand what Brewster's angle is.
Brewster's angle, denoted as θB, is the angle of incidence at which light with a specific polarization (polarized parallel to the plane of incidence) is entirely reflected from a transparent dielectric surface, with no reflection at all. In other words, there is only transmission at Brewster's angle.
For s polarization, the electric field vector is perpendicular or normal to the plane of incidence. To find Brewster's angle for s polarization, we need to determine the angle of incidence at which the reflected and transmitted rays are perpendicular to each other.
Let's consider a light ray incident on a dielectric surface at an angle of incidence, θi. The angle of reflection, θr, can be found using the law of reflection, which states that the angle of incidence is equal to the angle of reflection. Therefore, θi = θr.
The angle of transmission, θt, can be determined using Snell's law, which relates the angles of incidence and transmission to the refractive indices of the two media involved:
n1*sin(θi) = n2*sin(θt),
where n1 is the refractive index of the medium the light ray is coming from, and n2 is the refractive index of the medium the light ray is entering into.
For s polarization, the electric field vector is perpendicular to the plane of incidence. This means that the electric field vector of the reflected ray is also perpendicular to the plane of incidence (s polarized). Thus, we want to find the angle of incidence at which the transmitted ray is s polarized (electric field vector perpendicular to the plane of incidence).
Using the fact that the angle between the reflected and transmitted rays is 90 degrees (perpendicular), we have:
θr + θt = 90 degrees.
Substituting θr = θi and rearranging, we get:
θi + θt = 90 degrees.
Now, let's assume that there is a Brewster's angle for s polarization, denoted as θB_s. At Brewster's angle, the reflected ray is s polarized, meaning the electric field vector of the reflected ray is perpendicular to the plane of incidence.
This implies that θi = θB_s and θr = θB_s.
Using these values in the equation θi + θt = 90 degrees, we get:
θB_s + θt = 90 degrees.
Since θB_s is the angle of incidence at which the reflected ray is s polarized and θt is the angle of transmission, we know that θB_s + θt = 90 degrees means the reflected and transmitted rays are perpendicular to each other.
However, for s polarization, the transmitted ray has an electric field vector parallel to the plane of incidence. This means that the angle of transmission, θt, is not 90 degrees for s polarization.
Therefore, there is no angle of incidence (Brewster's angle) for s polarization at which the reflected and transmitted rays are perpendicular to each other.
To show that there is no Brewster's angle for s polarization, we need to understand the concept of Brewster's angle and the difference between s and p polarizations in optics.
Brewster's angle is the angle of incidence at which light, when passing from a medium with a higher refractive index to a medium with a lower refractive index, undergoes purely polarized reflection. At this angle, the reflected light is completely polarized in the plane of incidence, also known as p polarization.
In optics, when light reflects off a surface, it can have two different polarizations: s (perpendicular or transverse electric) polarization and p (parallel or transverse magnetic) polarization. The distinction between these polarizations comes from the orientation of the electric field vector with respect to the plane of incidence. For s polarization, the electric field vector is perpendicular to the plane of incidence, while for p polarization, it is parallel to the plane of incidence.
To understand why there is no Brewster's angle for s polarization, let's consider what happens when light is incident at Brewster's angle for p polarization. At Brewster's angle, the refracted ray is completely transmitted (i.e., no reflection) and is entirely polarized in the vertical (polarization) direction.
When light is incident at an angle other than Brewster's angle for p polarization, the reflected light includes both p and s polarizations. However, for s polarization, the electric field vector is perpendicular to the plane of incidence, so there will always be some amount of s-polarized reflection.
In summary, for s polarization, there is no angle of incidence at which the reflected light is completely polarized in the plane of incidence, unlike p polarization where it happens at Brewster's angle. Therefore, there is no Brewster's angle for s polarization.