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Polarization by reflection. At what angle above the horizontal is the sun if sunlight reflected from the surface of calm lake is completely polarized? (b) If the su...

Question

Polarization by reflection. At what angle above the horizontal is the sun if sunlight reflected from the surface of calm lake is completely polarized? (b) If the sun is due south; along what direction does the electric-field vector oscillate in the reflected light?

Polarization by reflection. At what angle above the horizontal is the sun if sunlight reflected from the surface of calm lake is completely polarized? (b) If the sun is due south; along what direction does the electric-field vector oscillate in the reflected light?



Answers

(a) Sunlight reflected from the still surface of a lake is totally polarized when the incident light is at what angle with respect to the horizontal? (b) In what direction is the reflected light polarized? (c) Is any light incident at this angle transmitted into the water? If so, at what angle below the horizontal does the transmitted light travel?

Okay. The angle above the horizontal when the light reflected is completely polarized can be written using the relation we know. Dane. The angle off polarization is equal to reflect index off the water which is 1.33 or the refracting that's off air in in which the light is reflected and this gives us the polarization angle here will be the position will angle will be 53.1 and this is equal to reflect it angle so we get right. Also, this is a call to reflect it. Angle, angle now the angle above the horizontal. We can find it by subtracting the reflected angle which is a 53.1. So 53.1. This gives us 36 36.9. This is there angle above the horizontal, but be the plan off the electric field factor, uh, in the reflected light will be holds gentle. So this will be what I'm told. Horizontal, gentle. End off the problem. Thank you for watching

Everyone. This is the problem based on poor tradition. Here we have toe Wanda angle off incident with her gentle so that like incident on the water surface. God's reflected and it is to be completely polarized. As you know, According Toa Mr Stop Wow! And global radiation is then in worse off and to upon anyone and two is effective index off water that is 1.33 and one is refractive index off year. So all writing angle you will get 33.6 degrees. So angle made white incident light with horrid gentle would be 19 minus 53.6 So it is equal to 36.94 degree that sort on the direction off and a big field actor is off. Our right is perpendicular to plane off and Children plight. Okay, which is the original plane for? That's so thanks for watching it

Okay, so the answer part A. We need to consider Brewster's angle, which is equal to the inverse tan of the index of refraction for the ice over that of the air. We can substitute in those values and we get 1.31 for ice over one for error, and that gives us an angle of 52.64 degrees. And then the angle is complementary here. So we take 90 minus 52.64 It gives us an angle of 37.36 degrees. And for part B, uh, the reflected light is parallel to the reflecting surface and perpendicular to the incidents plane so parallel to the reflecting surface and perpendicular to the Incidents plane. And that is because we're looking at total polarization and then to answer part. See, the angle that it makes with the horizontal would actually be equal to Brewster's angle. So that angle is 52.64 degrees

We can say that the incident and the reflected rays are traveled in the air so we can use boosters Law. And we have that tangent of fate. Assad Bey would be equal to end. Therefore, Brewster's angle would be equal to are tan of the refractive index. This would be equal to our 10 of the refractive index of water 1.33 and this would be equal to 53.1 degrees. Therefore, we can say the angle of polarizing were angle of polarization. Rather, for water equals 53 0.1 degrees and four part, This would be our answer for part A for part B. Ah, we know that this, uh, this polarizing angle does does depend on the wavelength of the light. So this, um and this is because the refractive index for water depends on the wavelength of the light through Father. So the answer to part B is simply yes. It does depend on the wavelength of the light. That is the end of the solution. Thank you.


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