What Is the Value of n in the Given Reflectivity Equation?

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Homework Help Overview

The discussion revolves around an equation involving reflectivity, where the original poster seeks to express the variable n in terms of other parameters. The context includes complex numbers, specifically the imaginary unit i, and involves separating real and imaginary components of the equation.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss methods for rationalizing the equation to separate real and imaginary parts. There are questions about the presence of imaginary components in the terms involving cosine and sine, and whether the imaginary parts of both sides of the equation can be equated.

Discussion Status

The discussion is ongoing, with participants exploring various methods to separate the components of the equation. Some guidance has been offered regarding rationalization, but there is no consensus on the approach or resolution of the problem yet.

Contextual Notes

There is mention of specific equations for R and K that are relevant to the problem, but the original poster expresses difficulty in isolating k and understanding the implications of the imaginary parts in their calculations.

getcarter
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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 separate 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.
 
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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.
 
but there is another imaginary part on cosa + isina ?
 
getcarter said:
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).
 
can i say that imaginary parts of RHS is equal to imaginary part of LHS ?
 
Yes, and the same for the real parts.
 
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.
 
Show us the work you have done.
 

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