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Brewster Angle

  1. Jan 6, 2014 #1
    1. The problem statement, all variables and given/known data

    I have solved the first two parts, I'm having trouble with the final part.

    Given the following relations, show the following relation of brewster angle:

    [tex] tan θ = \frac {n_2}{n_1} [/tex]

    bh10rd.png

    2. Relevant equations



    3. The attempt at a solution

    Using:

    [tex] \frac {n_2}{n1} = \frac {sin θ_1}{sin θ_3} [/tex]

    Starting from:

    [tex] sin 2θ_3 = sin 2θ_1 [/tex]
    [tex] sin θ_3 cos θ_3 = sin θ_1 cos θ_1 [/tex]
    [tex] \frac {sin θ_1}{sin θ_3} = \frac {cos θ_3}{cos θ_1} [/tex]
    [tex] \frac {n_2}{n_1} = \frac {\sqrt {1 - sin^{2}θ_3}}{cos θ_1} [/tex]
    [tex] \frac {n_2}{n_1} = \sqrt { \frac {1}{sin^{2}θ_1} - ( \frac {n_1}{n_2} )^{2} } tan θ_1 [/tex]

    How do i show the square root term at the bottom = 1?
     
    Last edited: Jan 6, 2014
  2. jcsd
  3. Jan 6, 2014 #2

    TSny

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    Since ##\theta_1## and ##\theta_3## lie between 0 and ##\frac{\pi}{2}##, ##2\theta_1## and ##2\theta_3## lie between 0 and ##\pi##.

    One way to solve ##\sin 2θ_3 = \sin 2θ_1## is to have ##\theta_1 = \theta_3##.

    But there is also another relation between ##\theta_1## and ##\theta_3## that will satisfy ##\sin 2θ_3 = \sin 2θ_1##.
     
  4. Jan 6, 2014 #3
    [tex]θ_1 = θ_3 + 2\pi[/tex]

    Not sure if this helps at all..
     
  5. Jan 6, 2014 #4

    TSny

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    We need to keep ##\theta_1## and ##\theta_3## less than ##\frac{\pi}{2}##, so ##2\theta_1## and ##2\theta_3## must lie between 0 and ##\pi##.

    Sketch a graph of the sine function between 0 and ##\pi##. Draw a horizontal line that intersects the graph at two different angles. How are the two angles related?
     
  6. Jan 6, 2014 #5
    [tex] θ_1 + θ_3 = \frac {\pi}{2} [/tex]
     
  7. Jan 6, 2014 #6

    TSny

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    OK. Use this in Snell's law.
     
  8. Jan 6, 2014 #7
    Ha ha, the answer just pops right out!
     
  9. Jan 6, 2014 #8

    TSny

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    :bugeye:

    Good.
     
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