by kor
Tags: angle, brewster
 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??
 P: 4,663 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
 Sci Advisor P: 1,258 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.
P: 12

 Quote by Bob S 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 by clem 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 ???
P: 4,663
 Quote by kor 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
P: 1,258
 Quote by kor except the maths. methods
Mentor
P: 11,739
 Quote by kor 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.
P: 12
 Quote by jtbell 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 ???
P: 4,663
 Quote by kor [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
P: 12
 Quote by Bob S 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
P: 4,663
 Quote by kor 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