We know that phase velocity
$V_{p}=\frac{\omega }{k}$
$\omega =V_{p}.k \qquad(1)$
And group velocity
$V_{g}=\frac{d\omega}{dk} \qquad(2)$
Substitute the value of $\omega$ from equation$(1)$ in equation $(2)$
$V_{g}=\frac{d}{dk}(V_{p}.k)$
$V_{g}=V_{p}+k.\frac{dV_{p}}{dk}$
$V_{p}=V_{p}+k.\frac{dV_{p}}{d\lambda}.\frac{d\lambda }{dk} \qquad (3)$
But
$\lambda=\frac{2\pi }{k}$
The above equation can be obtain from following formula i.e. $k=\frac{2\pi}{\lambda }$
Now put the value of $\lambda$ in equation $(3)$
$V_{g}=V_{p}+k\frac{dV_{p}}{d\lambda}\frac{d}{dk}(\frac{2\pi }{k}$
$V_{g}=V_{p}+k\frac{dV_{p}}{d\lambda}(\frac{-2\pi }{k^{2}}$
$V_{g}=V_{p}-\frac{2\pi}{k}\frac{dV_{p}}{d\lambda }$
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$V_{g}=V_{p}-\lambda\frac{dV_{p}}{d\lambda }$
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Thus, the above equation represents the relation between group velocity and phase velocity.
