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How to solve this reflectivity equations?

  1. Mar 30, 2010 #1
    1. The problem statement, all variables and given/known data

    sqrt(R)*(cos a + i sin a) = (n - ik - 1) / (n - ik + 1)

    what is n in terms of others.
    "i" is imaginary part.
    By using this equation, can you seperate imaginary and real part for finding

    R= (1-R) / [1+R-2*sqrt(R) * cos a] and

    K = 2 * sqrt(R) * sin a / [1 + R - 2 * sqrt (R) * cos a ]


    thank you.

    2. Relevant equations



    3. The attempt at a solution
    i have found n iK + [ (1+sqr(R) * (Cosa+isina) ]/[(1-sqr(R) * (Cosa+isina)]
    but i couldn't separete imaginary and real parts.

    if i separeta n has not imaginary part. by using this info. we can gain K.
     
  2. jcsd
  3. Mar 30, 2010 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Why don't you begin by rationalizing the right side:

    [tex]\frac{(n-1)-ik}{(n+1)-ik}\cdot \frac{(n+1)+ik}{(n+1)+ik}[/tex]

    Disclaimer: I haven't actually worked it out, but that will separate the real and imaginary parts to get you started.
     
  4. Mar 30, 2010 #3
    but there is another imaginary part on cosa + isina ?
     
  5. Mar 30, 2010 #4

    Mark44

    Staff: Mentor

    The left side can be separated into sqrt(R)cos(a) + i*sqrt(R)sin(a).
     
  6. Mar 31, 2010 #5
    can i say that imaginary parts of RHS is equal to imaginary part of LHS ?
     
  7. Mar 31, 2010 #6

    Mark44

    Staff: Mentor

    Yes, and the same for the real parts.
     
  8. Mar 31, 2010 #7
    but the imaginary part has n and also k, but i need to find k as an equation that i have read above.

    but i could't.

    if one of member can solve , i would be appreciate.
    i am very sorry for spending ur time.
     
  9. Mar 31, 2010 #8

    Mark44

    Staff: Mentor

    Show us the work you have done.
     
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