question about brewster's angle


by kor
Tags: angle, brewster
kor
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#1
Nov6-09, 10:42 AM
P: 12
except the maths. methods
can anyone explain why the reflected light is polarised while the refracted light is partially polarised in brewster angle ,please??
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Bob S
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#2
Nov6-09, 01:24 PM
P: 4,664
If the incident light is unpolarized, the light reflected at Brewster's angle will include only one polarization (100% polarized), and the refracted light will include all the light of the other polarization.
Bob S
clem
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#3
Nov6-09, 01:39 PM
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The polarization is determined by the Fresnel coefficients for reflection and refraction. The Fresnel coefficient for polarization parallel to the plane of incidence vanishes, while the coefficient for refraction does not. This is most easily seen from the algebra.

kor
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#4
Nov6-09, 07:05 PM
P: 12

question about brewster's angle


Quote Quote by Bob S View Post
If the incident light is unpolarized, the light reflected at Brewster's angle will include only one polarization (100% polarized), and the refracted light will include all the light of the other polarization.
Bob S
YUP That's brewster's law
but what's the principle behind it ???

Quote Quote by clem View Post
The polarization is determined by the Fresnel coefficients for reflection and refraction. The Fresnel coefficient for polarization parallel to the plane of incidence vanishes, while the coefficient for refraction does not. This is most easily seen from the algebra.
What's the fresnel coefficients ???
Bob S
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#5
Nov6-09, 08:19 PM
P: 4,664
Quote Quote by kor View Post
YUP That's brewster's law
but what's the principle behind it ???
What's the fresnel coefficients ???
Fresnel's equations for the transverse E and H components of light match the incident, reflected and refracted tangential components of E and H at the boundary.
Bob S
clem
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#6
Nov7-09, 08:30 AM
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Quote Quote by kor View Post
except the maths. methods
That's your problem. Buy a textbook.
jtbell
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#7
Nov7-09, 11:37 AM
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Quote Quote by kor View Post
What's the fresnel coefficients ???
http://en.wikipedia.org/wiki/Fresnel_equations

They can be derived from Maxwell's equations by applying the boundary conditions for E and B fields to an electromagnetic wave at a reflecting/refracting surface.
kor
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#8
Nov7-09, 11:25 PM
P: 12
Quote Quote by jtbell View Post
http://en.wikipedia.org/wiki/Fresnel_equations

They can be derived from Maxwell's equations by applying the boundary conditions for E and B fields to an electromagnetic wave at a reflecting/refracting surface.
[OFF TOPIC] What maths's level is required the above equation???
furthermore, at brewster's angle , what is the source of the reflected light ???
Bob S
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#9
Nov8-09, 03:56 PM
P: 4,664
Quote Quote by kor View Post
[OFF TOPIC] What maths's level is required the above equation???
furthermore, at brewster's angle , what is the source of the reflected light ???
kor-
You have been given several good references. Now read them. When you have, I would like you to solve the following real-world situation. You are fly-fishing in a high-elevation trout lake on a beautiful clear-blue (hint) morning. You are facing south, roughly 90 degrees away from the sun, at your left shoulder. Why can you see the fish clearly? Why is the reflection of the sky in the water pitch black? Read
http://farside.ph.utexas.edu/teachin...es/node97.html
Bob S
kor
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#10
Nov9-09, 12:57 AM
P: 12
Quote Quote by Bob S View Post
kor-
You have been given several good references. Now read them. When you have, I would like you to solve the following real-world situation. You are fly-fishing in a high-elevation trout lake on a beautiful clear-blue (hint) morning. You are facing south, roughly 90 degrees away from the sun, at your left shoulder. Why can you see the fish clearly? Why is the reflection of the sky in the water pitch black? Read
http://farside.ph.utexas.edu/teachin...es/node97.html
Bob S
although I still dont understand it (it needs time ), thank you for your help
Bob S
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#11
Nov9-09, 11:06 AM
P: 4,664
Quote Quote by kor View Post
although I still dont understand it (it needs time ), thank you for your help
In Rayleigh scattering of sunlight in clear unpolluted air, the light scattered at 90 degrees is completely polarized such that the E vector is perpendicular to the plane of scattering. In the url I posted previously

http://farside.ph.utexas.edu/teachin...es/node97.html

read the paragraph

"Light from the Sun is unpolarized. However, when it is scattered it becomes polarized, because light is scattered preferentially in some directions rather than others. Consider a light-ray from the Sun which grazes the Earth's atmosphere. The light-ray contains light which is polarized such that the electric field is vertical to the ground, and light which is polarized such that the electric field is horizontal to the ground (and perpendicular to the path of the light-ray), in equal amounts. However, due to the factor in the dipole emission formula (1119) (where, in this case, is the angle between the direction of the wave electric field and the direction of scattering), very little light is scattered downward from the vertically polarized light compared to the horizontally polarized light. Moreover, the light scattered from the horizontally polarization is such that its electric field is preferentially perpendicular, rather than parallel, to the direction of propagation of the solar light-ray (i.e., the direction to the Sun). Consequently, the blue light from the sky is preferentially polarized in a direction perpendicular to the direction to the Sun. "
On a clear blue-sky day, get a polarizing filter and look at the sky at 90 degrees to the sun (where a single photon scattering would be 90-degree scattering), and observe the polarization of blue light.
Bob S


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