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As seen from the diagram, the reflected ray is said to be polarised when the light is incident to the interface making an angle called Brewster's angle with the normal. This polarisation only occurs when the angle between the reflected ray and the refracted ray is 90 degrees.

I want to know why is it that the reflected ray is said to be polarised only when the angle of between the reflected ray and refracted ray is 90 degrees? Is it only an experimental result or is there any theoretical explanation behind it?

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    $\begingroup$ For clarity, be sure you understand that you get complete polarization only at Brewster's angle, but you get partial polarization at other angles (and more when close to Brewster's angle than when far from it). This isn't something that switches on or off in one go, it is something that exhibits a continuum. $\endgroup$ – dmckee --- ex-moderator kitten Dec 26 '19 at 15:52
  • $\begingroup$ @my2cts Not in general it's not: physics.stackexchange.com/questions/294494/…. $\endgroup$ – dmckee --- ex-moderator kitten Dec 26 '19 at 17:35
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There is a physical reason.

Start with this question: what generates the reflected wave? What are the electromagnetic sources? The answer is that the source of the reflected wave is the polarization of the medium. The oscillating electric field of the refracted beam causes an oscillating polarization in the molecules of the medium. These oscillating dipoles radiate, and are the source of the reflected EM wave. The direction of the polarization is the direction of the electric field in the medium, and this is also the direction of the generated electric field.

If the reflected and refracted beams are perpendicular, then the generated electric field is in the same direction as the propagation of the reflected wave. But an EM wave is transverse. The electric field cannot be in the direction of propagation. Hence there is no p-polarized radiation in that direction.

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It arises directly from the boundary conditions that demand that the components of the E- and H-field of the electromagntic fields immediately either side of the interface are the same. These boundary conditions come from an applications of Maxwell's equations to closed loops that enclose the interface.

Application of thes boundary conditions lead to reflection (and transmission) coefficients for the electric field that are a function of the incidence angle and there are two different expressions for the cases that the incident wave's E-field is polarised in the plane of incidence or at right angles to the plane of incidence. These are known collectively as the Fresnel equations.

In your diagram, the incident wave has a polarisation in the plane of incidence and the reflection coefficient for this polarisation state is zero at the Brewster angle. The transmission coefficient is not zero however and the component of the transmitted E-field parallel to the interface will equal the E-field component of the incident wave that is parallel to the interface.

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