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Brewster's angle derivation.

  1. May 4, 2008 #1


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    1. The problem statement, all variables and given/known data
    A beam of radiation, whose electric vector is in the plane of incidence, is incident at an angle [tex]\theta[/tex] from air onto a transparent material of refractive index, n. Given the relationship between the electric vectors of the reflected (E'') and incident (E) beams derive the condition for zero reflectance for this polarization (the Brewster angle)

    2. Relevant equations
    [tex]\frac{E''}{E}[/tex] = [tex]\frac{n^2cos(\theta)-\sqrt{n^2-sin^2({\theta})}}{n^2cos(\theta)+\sqrt{n^2-sin^2({\theta})}}[/tex]

    [tex]tan(\theta)=n[/tex] (condition for zero reflectance)

    3. The attempt at a solution

    So, basically, I think I just need to manipulate the first formula until I arrive at the expression for Brewster's angle. Since E'' is the reflect beam, it must equal 0. So I can set [tex]n^2cos(\theta)[/tex] equal to [tex]\sqrt{n^2-sin^2({\theta})}[/tex]. When I try to solve from there I run into problems. If I square the entire expression then I have a quadratic which I can solve for [tex]n^2[/tex].

    [tex]n^4cos^2(\theta) - n^2 + sin^2(\theta) = 0[/tex]

    solve using quadratic formula with [tex]a = cos^2(\theta), b=-1, c=sin^2(\theta)[/tex]

    This seems like it would be along the right lines since I would ideally end up with [tex]sin^2(\theta)[/tex] divided by [tex]cos^2(\theta)[/tex], which I would take the square root of to find n. The problem there is that it I end up with an expression:

    [tex]n^2 = \frac{1 +/- \sqrt{1-4cos^2(\theta)sin^2(\theta)}}{2cos^2(\theta)}[/tex]

    But I have no idea what to replace [tex]4cos^2(\theta)sin^2(\theta)[/tex] for... I assume it is some sort of trigonometric expression? But I have no idea where to start. I think that in general it's pretty obvious that E'' must equal zero, since that's more or less stated as part of the problem, my weakness is the trig required to do this. Am I along the right path when I try to solve for n as a quadratic? Any hints on how I should approach the expression if I am? Or am I doing this the completely wrong way? I thought I could maybe multiply the top and bottom of the expression by say [tex]n^2cos(\theta)+\sqrt{n^2-sin^2({\theta})}[/tex], just because that looks sort of like something might cancel out conveniently... but it doesn't seem to simplify things at all. Anyways, any help would be greatly appreciated!
  2. jcsd
  3. May 6, 2008 #2


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    Things seem to be on the right track.

    It will help you to look up the trig double-angle formulas for [tex]\sin(2\theta)[/tex] and [tex]\cos(2\theta)[/tex], and see if one or both of them can be used in your last expression.
  4. May 6, 2008 #3
    The solution

    This equation could be solved by replacing [tex]2cos^2(\theta)\sin^2(\theta)[/tex] with [tex]sin2(\theta)[/tex].
    You will get an equation which is

    [tex]n^2 = \frac{1 \pm \sqrt{1-sin^2{2(\theta)}}}{2cos^2(\theta)}[/tex]

    substitute [tex]1 - sin^2{2(\theta)}[/tex] with [tex]cos^2{2(\theta)}[/tex]
    and manipulate the formula until you get [tex] \frac{1 \pm ({2cos^2\theta-1})}{2cos^2\theta}[/tex]

    continue your work and you will solve the question with [tex]n^2=1[/tex] , [tex] n=\pm 1[/tex] or [tex]n^2=tan^2\theta[/tex] , [tex]n = \pm tan\theta[/tex]
  5. Mar 9, 2010 #4
    Where did the equations for the amplitude of the electric vectors come from? (the numerator and denominator of first post)?

    Figured it out they, come from the fresnel equations.

    Cipher made an error it should be 2cos(theta)sin(theta) = sin(2theta)
    Last edited: Mar 9, 2010
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