Binomial theorem question

1. Sep 18, 2005

Luminous Blob

I am trying to do a question from Eugene Hecht's Optics book, which goes something like this:

Given the following equations:

Cauchy's Equation:

$$n = C_1 + \frac{C_2}{\lambda^2} + \frac{C_3}{\lambda^4} + ...$$

Sellmeier's Equation:

$$n^2 = 1 + \sum_{j} \frac{A_j\lambda^2}{\lambda^2-\lambda_0_j^2}$$

where the $$A_j$$ terms are constants and each $$\lambda_0_j$$ is the vacuum wavelength associated with a natural frequency $$v_0_j$$, such that $$\lambda_0_jv_0_j = c$$.

Show that where $$\lambda >> \lambda_0_j$$, Cauchy's Equation is an approximation of Sellmeier's Equation.

Now it also gives a hint which is as follows:

Write the above expression with only the first term in the sum; expand it by the binomial theorem; take the square root of $$n^2$$ and expand again.

From the hint, I gather that it means to rewrite Sellmeier's Equation as:

$$n^2 = 1 + \frac{A\lambda^2}{\lambda^2 - \lambda_0^2}$$

From there though, I have no idea how to apply the binomial theorem to expand it. I just don't see how anything in that equation has the form $$(x+y)^n$$, except for where n = 1.

If anyone can explain to me how to apply the binomial theorem to the equation, or if I've misunderstood what the hint means, it would be much appreciated.

2. Sep 18, 2005

Galileo

You can use the binomial theorem to expand $(1+x)^{1/2}$ when x<<1.

3. Sep 18, 2005

Luminous Blob

So you mean first take the square root of both sides, then expand it using the binomial theorem , letting $$x = \frac{A\lambda^2}{\lambda^2 - \lambda_0^2}$$, rather than first applying the binomial theorem, then taking the square root of both sides and then expanding again like the hint suggests?

4. Sep 18, 2005

Dr Transport

Rewrite $$\frac{A_j\lambda^2}{\lambda^2-\lambda_0_j^2}$$ as

$$\frac{A_j}{\lambda^2}\frac{1}{1-\frac{\lambda_0_j^2}{\lambda^2}}$$ and expand the second part as

$$\frac{1}{1-x^2} \approx 1 - x^2 + x^4 - x^6 \ldots$$ where $$x = \frac{\lambda}{\lambda_0_j}$$

Last edited: Sep 18, 2005
5. Sep 18, 2005

Luminous Blob

Aah, I didn't think to do that. Thanks, that was a great help.