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How to solve this reflectivity equations?

  • Thread starter getcarter
  • Start date
  • #1
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Homework Statement



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.

Homework Equations





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.
 

Answers and Replies

  • #2
LCKurtz
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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.
 
  • #3
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but there is another imaginary part on cosa + isina ?
 
  • #4
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but there is another imaginary part on cosa + isina ?
The left side can be separated into sqrt(R)cos(a) + i*sqrt(R)sin(a).
 
  • #5
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can i say that imaginary parts of RHS is equal to imaginary part of LHS ?
 
  • #6
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Yes, and the same for the real parts.
 
  • #7
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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.
 
  • #8
33,158
4,841
Show us the work you have done.
 

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