Dispersive power of glass

1. Nov 24, 2013

GreenPrint

1. The problem statement, all variables and given/known data

The dispersive power of glass is defined as the ratio $\frac{n_{F} - n_{C}}{n_{D} - 1}$, where C, D, and F refer to the Fraunhofer wavelengths, $λ_{C} = 6563 \stackrel{o}{A}$, $λ_{D} = 5890 \stackrel{o}{A}$, and $λ_{F} = 4861 \stackrel{o}{A}$. Find the approximate group velocity in glasss whose dispersive power is $frac{1}{30}$ and for which $n_{D} = 1.50$.

2. Relevant equations

3. The attempt at a solution
I start off with the given information
$\frac{n_{F} - n_{C}}{n_{D} - 1} = \frac{n_{F} - n_{C}}{1.50 - 1} = \frac{1}{30} = \frac{n_{F} - n_{C}}{.5}$
I simplify
$n_{F} - n_{C} = \frac{1}{60} = Δn$
I know that
$Δλ = λ_{F} - λ_{C} = 4861 \stackrel{o}{A} - 6563 \stackrel{o}{A} = -1702 \stackrel{o}{A}$
I use the formula for group velocity
$v_{g} = v_{p}(1 + \frac{λ}{n}\frac{dn(λ)}{dλ})$
I use the approximation that
$\frac{dn(λ)}{dλ}) ≈ \frac{Δn}{Δλ} = \frac{1}{60(-1702 \stackrel{o}{A})}$
$v_{g} = v_{p}(1 - \frac{5890 \stackrel{o}{A}}{1.5}\frac{1}{60(1702 \stackrel{o}{A})})$
simplify and round to three decimal places
$v_{g} = v_{p}(1 - 3.845x10^{-2})$

From here I'm not really sure what to do. Someone told me that I should use $v_{p} = \frac{c}{n}$. However I'm not sure how this is correct as $v_{p} = \frac{ω_{p}}{k_{p}}$.

Thanks for any help.

2. Nov 24, 2013

never mind