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Homework Statement
The dispersive power of glass is defined as the ratio [itex]\frac{n_{F} - n_{C}}{n_{D} - 1}[/itex], where C, D, and F refer to the Fraunhofer wavelengths, [itex]λ_{C} = 6563 \stackrel{o}{A}[/itex], [itex]λ_{D} = 5890 \stackrel{o}{A}[/itex], and [itex]λ_{F} = 4861 \stackrel{o}{A}[/itex]. Find the approximate group velocity in glasss whose dispersive power is [itex]frac{1}{30}[/itex] and for which [itex]n_{D} = 1.50[/itex].
Homework Equations
The Attempt at a Solution
I start off with the given information
[itex]\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}[/itex]
I simplify
[itex]n_{F} - n_{C} = \frac{1}{60} = Δn[/itex]
I know that
[itex]Δλ = λ_{F} - λ_{C} = 4861 \stackrel{o}{A} - 6563 \stackrel{o}{A} = -1702 \stackrel{o}{A}[/itex]
I use the formula for group velocity
[itex]v_{g} = v_{p}(1 + \frac{λ}{n}\frac{dn(λ)}{dλ})[/itex]
I use the approximation that
[itex]\frac{dn(λ)}{dλ}) ≈ \frac{Δn}{Δλ} = \frac{1}{60(-1702 \stackrel{o}{A})}[/itex]
[itex]v_{g} = v_{p}(1 - \frac{5890 \stackrel{o}{A}}{1.5}\frac{1}{60(1702 \stackrel{o}{A})})[/itex]
simplify and round to three decimal places
[itex]v_{g} = v_{p}(1 - 3.845x10^{-2})[/itex]
From here I'm not really sure what to do. Someone told me that I should use [itex]v_{p} = \frac{c}{n}[/itex]. However I'm not sure how this is correct as [itex]v_{p} = \frac{ω_{p}}{k_{p}}[/itex].
Thanks for any help.