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I cannot understand the physics behind the reflection and polarization-reflection on an atomic scale. What exactly happens at the boundary of two surfaces? I was studying Brewster's law and thus this doubt came.

s-polarised light causes these dipoles to oscillate perpendicular to the plane of incidence, in the same direction as the oscillating electric field of the incoming light.

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Now consider p-polarised light. The electric field is polarised in the plane of incidence and causes the dipoles to oscillate in the same plane. However, at a certain angle, the Brewster angle, the reflected light would need to be produced by electric dipoles oscillating along the line defined by the direction of the reflected ray. But no electric field is seen in this direction because it is the axis of oscillation for the dipoles. Thus no reflected p-polarised light is seen at the Brewster angle.

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As you may noticed the direction of z is default by the direction of propagation. But the x- any y-direction could have any orientation around z.

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The oscillating electric field produced by an oscillating electric dipole is maximised at right angles to the oscillation direction. i.e. in the plane of incidence for s-polarised light. It is zero along the axis of oscillation.

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Light as part of the electromagnetic radiation consists of photons. These quanta have an oscillating electric field component and an oscillating magnetic field component, both perpendicular to the direction of propagation. In vacuum the electric and the magnetic field components are exact perpendicular to each over. To make it imaginable for you, take a Cartesian coordinate system, orient z to the direction of propagation. The directions x and y are the directions of the E-field and the B-field.

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On an atomic scale, atoms emit light like a phased array of individual antennas. Then interference happens, and you get all the behavior of Snell’s Law, Fresnel Equations, etc.

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Shining light perpendicular to a surface does not influence the random oriented E- and B-field. But shining light not-perpendicular, the surface influences the orientation of the lights field components. (BTW, the same happens with light in front of a polarizing foil or at the boundary of slits.)

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If the reflected light ray is either side of the Brewster angle, then the oscillating dipoles would be viewed at an angle (less then 90 degrees) to the oscillation direction. Thus there would be some electric field produced in that direction, but not as much as for the case of s-polarised light, where the dipole oscillation direction is always perpendicular to the reflected ray.

In the end, the amount of polarized light depends on the angle of light propagation to the surface. The surface influences the orientation of the E and B field at most for a certain Brewster angle.