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Refraction in a prism, Snell's law: Please help!

  1. Mar 29, 2012 #1
    1. The problem statement, all variables and given/known data
    A horizontal beam of light enters a 45-90-45 prism at the center of it's long side, as shown below. The emerging ray moves in a direction that is 34˚ below the horizontal. What is the index of refraction for the prism?

    Walker4e.ch26.Pr068.JPG


    2. Relevant equations
    n[itex]_{1}[/itex]sinø[itex]_{1}[/itex] = n[itex]_{2}[/itex]sinø[itex]_{2}[/itex]


    3. The attempt at a solution
    n1 x sin(i) = n2 x sin(r1): (1st refrection)

    and n2 x sin(r2) = n1 x sin(34˚): (2nd refraction)

    ^^Here n1 is the refractive index of air, n2 is the refractive index of prism, r1 and r2 are the angles of reflection at the two surfaces, and i is the incidence angle.

    r1 + r2 = 45˚ or r2 = 45˚ - r1.
    Substituting the above value of r2 in the equation(2),we get
    n2 x sin(45 - r1) = 1 x sin(34˚)

    ...and then I'm stuck!

    To simplify what I've figured out:
    first refraction: sin45 = n sin a

    second refraction: n sinb = sin34

    and using geometry: a + b = 45

    BUT HOW DO I FIND a and b?!?!?
     
  2. jcsd
  3. Mar 29, 2012 #2

    tiny-tim

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    Hi ObviousManiac! :smile:
    Learn your trigonometric identities

    sin(45° - a) = … ? :wink:
     
  4. Mar 29, 2012 #3
    okay so sin(45˚ - a) = sin(45)cos(a) - cos(45)sin(a)

    but I'm not sure where to take it from there...
    I tried setting up something like:

    sin(45)/sin(34) = sin(a)/sin(b)
    sin(45)/sin(34) = sin(a)/sin(45-a)
    sin(45)/sin(34) = sin(a)/[sin(45)cos(a) - cos(45)sin(a)]

    but I don't even know if any of that ^^ is right, I could be going in a totally wrong direction.
     
  5. Mar 29, 2012 #4

    tiny-tim

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    Hi ObviousManiac! :smile:
    Substitute from the first equation into the second. :wink:
     
  6. Mar 29, 2012 #5
    Sin(45-a) = nsin(a)cos(a) - cos(45)sin(a)

    ...doesn't this just introduce a new variable? (n?)
     
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