1 Answer
Yes, light is incident at Brewster's angle when it strikes a surface, and this angle is special because it results in perfectly polarized light when reflected.
Brewster's angle is the angle of incidence at which light with a particular polarization (the plane of polarization) is perfectly transmitted through a transparent dielectric surface, with no reflection. At this angle, the reflected light is completely polarized perpendicular to the plane of incidence.
Mathematically, Brewster's angle (\( \theta_B \)) can be found using the formula:
\[
\tan(\theta_B) = \frac{n_2}{n_1}
\]
where:
At Brewster's angle, the angle between the reflected and refracted rays is 90°, meaning that the reflected ray is completely polarized.
This phenomenon is often seen in photography and optics when glare needs to be reduced, such as through polarized filters.
Brewster's angle is the angle of incidence at which light with a particular polarization (the plane of polarization) is perfectly transmitted through a transparent dielectric surface, with no reflection. At this angle, the reflected light is completely polarized perpendicular to the plane of incidence.
Mathematically, Brewster's angle (\( \theta_B \)) can be found using the formula:
\[
\tan(\theta_B) = \frac{n_2}{n_1}
\]
where:
- \( n_1 \) is the refractive index of the first medium (usually air, with \( n_1 \approx 1 \)),
- \( n_2 \) is the refractive index of the second medium (like glass or water).
At Brewster's angle, the angle between the reflected and refracted rays is 90°, meaning that the reflected ray is completely polarized.
This phenomenon is often seen in photography and optics when glare needs to be reduced, such as through polarized filters.