What is Brewster's Angle for Sunlight Reflecting on Water?

squirrelly
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Homework Statement



As the sun rises over a still pond, an angle will be reached where its image seen on the water's surface(n=1.33) will be completely polarized in a plan parallel to the surface.

i) Compute the appropriate incident angle(that is the polarization or Brewster's angle)

ii)At what angle(from the normal) will the transmitted beam propagate through the water(that is, what is the angle of refraction)

iii)What is the angle between the reflected and refracted beams.



The Attempt at a Solution


i) n2cos(θo)-n1√[(1-(n21/n22)sin2o)]

(n22/n21)=tan2o)+1-(n21/n22)tan2o)

tan2o)={[(n22/n21-1]/[1-(n21/n22)]}=(n22/n21)

tan(θo)=(n2/n1)=(1.33/1)

θo=tan-1(1.33)=53°

Homework Statement




and that's as far as I could get. :/ I am pretty sure part(i) is right.
 
on Phys.org
squirrelly said:

Homework Statement



As the sun rises over a still pond, an angle will be reached where its image seen on the water's surface(n=1.33) will be completely polarized in a plan parallel to the surface.

i) Compute the appropriate incident angle(that is the polarization or Brewster's angle)

ii)At what angle(from the normal) will the transmitted beam propagate through the water(that is, what is the angle of refraction)

iii)What is the angle between the reflected and refracted beams.

The Attempt at a Solution


i) n2cos(θo)-n1√[(1-(n21/n22)sin2o)]

(n22/n21)=tan2o)+1-(n21/n22)tan2o)

tan2o)={[(n22/n21-1]/[1-(n21/n22)]}=(n22/n21)

tan(θo)=(n2/n1)=(1.33/1)

θo=tan-1(1.33)=53°

Homework Statement




and that's as far as I could get. :/ I'm pretty sure part(i) is right.
Hello squirrelly. Welcome to PF !

ii) Use Snell's Law to find the angle of refraction.

iii) Draw a sketch. Use geometry.
 

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