Brewster's angle and the intensity of reflected light

[PainterGuy]

Hi,

Would it be correct to say that at Brewster's angle, all the incident light which has its electric field parallel to the plane of incidence gets refracted, and the rest of light whose electric field is perpendicular to the plane of incidence gets reflected? For example, if the light whose electric field is parallel to the plane of incidence contributes 70% towards the intensity of light then at Brewster angle, the reflected light has intensity of only 30%. Could you please help me with it?

 

[.Scott]

The "all the incident light which has its electric field parallel to the plane of incidence gets refracted" is correct.
The "and the rest of light whose electric field is perpendicular to the plane of incidence gets reflected" is not correct.

Some of the light with an electric field perpendicular to the plane of incidence will also be refracted.

In the example you gave:
The 70% that is parallel will be fully refracted.
The 30% that is perpendicular will be split - with most of it being reflected.

 

[PainterGuy]

The "and the rest of light whose electric field is perpendicular to the plane of incidence gets reflected" is not correct.

Thank you!

Wouldn't it be correct in an ideal case?

 

[.Scott]

No.
Under "ideal" conditions, the proportion of light reflected and refracted can be computer by using the Fresnel Equations.

 

[PainterGuy]

Thank you.

I think that it could be rephrased as follows.

At Brewster's angle, all the incident light which has its electric field parallel to the plane of incidence gets refracted, and all the reflected light has its electric field perpendicular to the plane of incidence. The refracted light would consist of both 'types' of light, but at Brewster's angle the portion of light whose electric field is perpendicular to the plane of incidence is at its minimum in the total of refracted light.