# Understanding Brewster's Angle

• fred3142
In summary: The reflected beam can be considered as radiation of the exited dipoles in the material. It can be shown that a dipole does not radiate in its own direction. The dipoles in the material are perpendicular to the refracted beam. At the Brewster Angle, the refracted and reflected beams are perpendicular, so the dipoles are oriented parallel to the reflected beam in case of in-plane polarization (when the electric field is parallel to the plane of incidence) . At perpendicular polarization, the polarization is out-of plane, perpendicular to the reflected ray. -For the one polarization, the axis of the induced dipoles lines up with the reflected ray at Brewster's angle, thus no reflected ray can be generated
fred3142
Hi,

I'm struggling to conceptually understand why Brewster's angle occurs. I know Fresnel's equations and can see that at some angle, the reflected parallel component goes to zero. What I don't understand is why this is conceptually happening. I understand that the material is made up of 'dipoles' and free charges which can absorb and deflect the wave, but I cannot see why the perpendicular component is not treated the same way as the parallel component. I don't see how they appear any differently to the material. If someone could provide some insight into this, that would be great.

Thank you.

The Fresnel formulas come from Maxwell's equations and continuity requirements at the boundary between the two materials. The boundary conditions are different for the field components in tangential and perpendicular direction of the medium. See, e.g., Jackson Classical Electrodynamics.

Thank you for the reply. I understand the derivation of these equations (at least I think I do), and I understand that they are based upon the boundary conditions (tangential fields equal on either side of the boundary and perpendicular components not necessarily equal, and a charge distribution forms on the boundary to oppose it).

My question is more a conceptual one. I don't 'get' it. I see the maths, the maths works nicely. I get the equations and everything, but I don't understand conceptually what is going on. I don't see how changing the angle of the field will affect anything if initially, the incoming field is unpolarised. As the angle increase, why are the parallel and perpendicular components interacting differently with the boundary?

The reflected beam can be considered as radiation of the exited dipoles in the material. It can be shown that a dipole does not radiate in its own direction. The dipoles in the material are perpendicular to the refracted beam. At the Brewster Angle, the refracted and reflected beams are perpendicular, so the dipoles are oriented parallel to the reflected beam in case of in-plane polarization (when the electric field is parallel to the plane of incidence) . At perpendicular polarization, the polarization is out-of plane, perpendicular to the reflected ray.

ehild

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An oscillating dipole cannot radiate along its axis (you can convince yourself of this by drawing a charge bobbing up and down with the electric field lines attached at different points in time). For the one polarization, the axis of the induced dipoles lines up with the reflected ray at Brewster's angle, thus no reflected ray can be generated. For the other polarization, the axis of the dipoles is parallel to the material's surface and never lines up with reflected ray direction, even at Brewster's angle. The axis of the induced dipoles in the material is lined up with the electric field vector of the transmitted wave.

Take a look at my lecture notes, page 17: http://faculty.uml.edu/cbaird/95.658%282013%29/Lecture1.pdf

1 person

## What is Brewster's Angle and how is it defined?

Brewster's Angle is an important concept in optics, specifically in the study of reflection and refraction of light. It is defined as the angle at which light is incident on the surface of a material such that the reflected light is completely polarized parallel to the surface.

## What is the mathematical formula for Brewster's Angle?

The mathematical formula for Brewster's Angle is tanθ = n2/n1, where θ represents the angle of incidence, n1 is the refractive index of the incident medium, and n2 is the refractive index of the medium the light is entering.

## What is the physical significance of Brewster's Angle?

Brewster's Angle is significant because it allows us to determine the refractive index of a material by measuring the angle of incidence at which polarization occurs. It also helps us understand the behavior of light at interfaces between different materials.

## What are some real-world applications of Brewster's Angle?

Brewster's Angle has several practical applications, such as in the design of polarizing lenses for sunglasses, liquid crystal displays, and anti-glare coatings for windows and screens. It is also used in the manufacturing of optical fibers and in studying the properties of minerals and crystals.

## How does Brewster's Angle relate to Snell's Law?

Brewster's Angle is closely related to Snell's Law, which describes the relationship between the angle of incidence and the angle of refraction when light passes through a boundary between two materials. When light is incident at Brewster's Angle, the angle of refraction is 90 degrees, resulting in complete polarization of the reflected light.

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