# Derive the expression

• Apr 22nd 2013, 04:03 AM
bobred
Derive the expression
Derive the expression

$\frac{A+B}{A-B}=\frac{k_1}{k_2}\frac{C+D}{C-D}=\frac{k^2_1}{k^2_2}$

Using

$A+B=C+D$ and $k_{1}A- k_{1}B = k_{2}C- k_{2}D$

$C e^{i k_{2}L}+D e^{- ik_{2}L} = F e^{i k_{1}L}$ and $k_{2}C e^{ ik_{2}L}- k_{2}D e^{-i k_{2}L} = k_{1}F e^{i k_{1}L}$

$k_2 L=\pi/2$

I can get

$\frac{A+B}{A-B}=\frac{k_1}{k_2}\frac{C+D}{C-D}$

But can't see how to get

$\frac{k^2_1}{k^2_2}$, probably easy but can't see it.

James
• Apr 22nd 2013, 04:24 AM
chiro
Re: Derive the expression
Hey bobred.

Hint: Try dividing C+D from RHS and multiplying A-B from LHS.
• Apr 22nd 2013, 10:42 AM
bobred
Re: Derive the expression
Hi, I've done that but still drawing a blank, I keep going round in circles. Going to have a break.
• Apr 24th 2013, 02:12 AM
bobred
Re: Derive the expression
Sorry, is there another small hint?
• Apr 24th 2013, 05:18 PM
chiro
Re: Derive the expression
Can you try squaring the terms and simplifying to show that the result holds? (In other words, square both sides and collect together)
• Apr 25th 2013, 03:39 AM
bobred
Re: Derive the expression
Hi, got it, was staring me in the face, many thanks. Bob