Question

1. Consider a dispersive material whose (real) refractive index $n = n(\lambda)$ depends on the wavelength. Starting from the general definition of the group velocity: $v_g = \frac{\partial \omega}{\partial k}$ derive the alternative expression: $v_g = v_p \left[ 1 + \frac{\lambda}{n} \left( \frac{dn}{d\lambda} \right) \right]$ where we denoted the phase velocity by $v_p$.

          1. Consider a dispersive material whose (real) refractive index $n = n(\lambda)$ depends on the
wavelength. Starting from the general definition of the group velocity:
$v_g = \frac{\partial \omega}{\partial k}$
derive the alternative expression:
$v_g = v_p \left[ 1 + \frac{\lambda}{n} \left( \frac{dn}{d\lambda} \right) \right]$
where we denoted the phase velocity by $v_p$.

$1. Consider a dispersive material whose (real) refractive index n = n(λ) depends on the wavelength. Starting from the general definition of the group velocity: vg = (∂ω)/(∂ k) derive the alternative expression: vg = vp [ 1 + (λ)/(n)( (dn)/(dλ)) ] where we denoted the phase velocity by vp.$

Added by Vanessa D.

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