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Jyoti123 (24)

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A question from "Jason complements on electromagnetic polarisation"  whose answer I want :-
A beam of natural light falls on the surface of  a dielectric at an angle .Then what is the degree of polarisation of  the refracted light ?
[Hint:Fresnel equations have to be used.]

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edison (4929)

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Polarization by Reflection

Since the reflection coefficient for light which has electric field parallel to the plane of incidence goes to zero at some angle between 0° and 90°, the reflected light at that angle is linearly polarized with its electric field vectors perpendicular to the plane of incidence. The angle at which this occurs is called the polarizing angle or the Brewster angle. At other angles the reflected light is partially polarized.
From Fresnel's equations it can be determined that the parallel reflection coefficient is zero when the incident and transmitted angles sum to 90°. The use of Snell's law gives an expression for the Brewster angle.
 

The Scientist does not study nature because it is useful; he studies it because he delights in it, & he delights in it because it is beautiful. If nature were not beautiful, it would not be worth knowing, life would not be worth living. Ofcourse I do not here speak of that beauty that strikes the senses, the beauty of qualities & appearances; not that I undervalue such beauty, far from it, but it has nothing to do with science; I mean that profounder beauty which comes from the harmoniuos order of the parts, & which a pure intelligence can grasp.
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edison (4929)

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Fresnel's Equations

Fresnel's equations describe the reflection and transmission of electromagnetic waves at an interface. That is, they give the reflection and transmission coefficients for waves parallel and perpendicular to the plane of incidence. For a dielectric medium where Snell's Law can be used to relate the incident and transmitted angles, Fresnel's Equations can be stated in terms of the angles of incidence and transmission.
For light from a medium of index n1
incident upon a medium of index n2
at an angle ?i °
the angle of transmission is ?t°
Fresnel's equations give the reflection coefficients:
 and
The transmission coefficients are
 and

Note that these coefficients are fractional amplitudes, and must be squared to get fractional intensities for reflection and transmission. The signs of the coefficients depend on the original choices of field directions.
You can choose values of parameters which will give transmission coefficients greater than 1, and that would appear to violate conservation of energy. (For example, try light incident from a medium of n1=1.5 upon a medium of n2=1.0 with an angle of incidence of 30°.) But the square of the transmission coefficient gives the transmitted energy flux per unit area (intensity), and the area of the transmitted beam is smaller in the refracted beam than in the incident beam if the index of refraction is less than that of the incident medium. When you take the intensity times the area for both the reflected and refracted beams, the total energy flux must equal that in the incident beam. For further details, see Jenkins and White.
Checking out conservation of energy in this situation leads to the relationship
which applies to both the parallel and perpendicular cases.

The Scientist does not study nature because it is useful; he studies it because he delights in it, & he delights in it because it is beautiful. If nature were not beautiful, it would not be worth knowing, life would not be worth living. Ofcourse I do not here speak of that beauty that strikes the senses, the beauty of qualities & appearances; not that I undervalue such beauty, far from it, but it has nothing to do with science; I mean that profounder beauty which comes from the harmoniuos order of the parts, & which a pure intelligence can grasp.
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Jyoti123 (24)

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Thank you edison.But I am asking the degree of polarisation not the reflection coefficient.Your explanation is superb , I got to know many things from this .

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