Sellmeier's equation to cauchy's equation

[ehreming]

My problem is to show that that where (lambda)>>(lambda_0) then Cauchy's Equation is an approximation of Sellmeier's.

Now I know generally how to solve it, but my trouble is in some of the expansions. The hint that is provided with the problem is "Write Sellmeier's Equation with only the first term in the sum; expand it by the binomial theorem; take the square root of n^2 and expand again."

it is the "expand by the binomial theorem" that gives me trouble. i looked up the binomial theorem and understand what it says but don't know how to apply it for the first expansion.

I'll try to attach some of my scratch work later, to clear up any confusion. thanks in advance for any help you can offer.

eric

 

[ehreming]

ok i found some direction to solving the problem but I'm still having trouble following the first expansion... see the attachment (Word) to see what I'm talking about.

thank
eric

 

[ehreming]

so I've stumped the Physics Forum??

 

[Physics Monkey]

Apparently so, although I think some of the trouble is that your doc appears unreadable. Let me give it a try though.

The Sellmeier dispersion formula to leading order is n^2 = 1 + A \frac{\lambda^2}{\lambda^2 - \lambda^2_0}. You can rewrite this as n^2 = 1 + A\frac{1}{1-\lambda^2_0/\lambda^2}. Since \lambda_0 >> \lambda, you want to expand \frac{1}{1-x} to leading order in x. The binomial theorem is nothing really but a taylor expansion, so as long as you can taylor expand you should be fine.

 

[ehreming]

sorry about the doc... i copied and pasted from a pdf so that might have something to do with it.

btw... you said \lambda_0 >> \lambda but it is actually the other way around... does that change anything?

lemme try that and i'll get back to ya. thanks a lot

 

[ehreming]

so you can set \lambda^2_0 / \lambda^2 = x

That changes everything... thanks i didn't think to do that. i really appreciate it.

 

[Physics Monkey]

btw... you said \lambda_0 >> \lambda but it is actually the other way around... does that change anything?

Woops! I gather that you've seen through my typo so all is well.