# How to solve this reflectivity equations?

1. Mar 30, 2010

### getcarter

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. Mar 30, 2010

### LCKurtz

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

$$\frac{(n-1)-ik}{(n+1)-ik}\cdot \frac{(n+1)+ik}{(n+1)+ik}$$

Disclaimer: I haven't actually worked it out, but that will separate the real and imaginary parts to get you started.

3. Mar 30, 2010

### getcarter

but there is another imaginary part on cosa + isina ?

4. Mar 30, 2010

### Staff: Mentor

The left side can be separated into sqrt(R)cos(a) + i*sqrt(R)sin(a).

5. Mar 31, 2010

### getcarter

can i say that imaginary parts of RHS is equal to imaginary part of LHS ?

6. Mar 31, 2010

### Staff: Mentor

Yes, and the same for the real parts.

7. Mar 31, 2010

### getcarter

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. Mar 31, 2010

### Staff: Mentor

Show us the work you have done.