Product of phase velocity and group velocity is equal to square of speed of light
Prove that $\rightarrow$
Proof →
We know that
$V_{p}=\nu \lambda \qquad(1)$
And de Broglie wavelength-
$\lambda =\frac{h }{mv}\qquad(2)$
According to Einstein's mass-energy relation-
$E=mc^{2}$
$h\nu=mc^{2}$
$\nu=\frac{mc^{2}}{h}\qquad(3)$
Now put the value of $\lambda $ and $\nu $ in equation$(1)$
$V_{p}= [\frac{mc^{2}}{h}] [\frac{h}{mv}]$
$V_{p}=\frac{C^{2}}{v}$
Since group velocity is equal to particle velocity i.e. $V_{g}=v$. So above equation can be written as
$V_{p}=\frac{C^{2}}{V_{g}}$
$V_{p}.V_{g}=C^{2}$
Note →
➢ $V_{g}=V_{p}$ for a non-dispersive medium ( in a non-dispersive medium all the waves travel with
phase velocity).
➢ $V_{g}< V_{p}$ for normal dispersive medium
➢ $V_{g}> V_{p}$ for anomalous dispersive media.
Dispersive medium →
The medium in which the phase velocity varies with wavelength or frequency is called a dispersive medium. In such a medium, waves of different wavelengths travel with different phase velocities.
Non-dispersive medium →
The medium in which the phase velocity does not vary with wavelength or frequency is called a Non-dispersive medium.
Dispersive waves →
Those waves in the medium for which phase velocity varies with wavelength or frequency are called dispersive waves.
Non-dispersive waves →
Those waves in which phase velocity does not vary with wavelength are called non-dispersive waves. So phase velocity independent of wavelength.
The Product of phase velocity and group velocity is equal to the square of the speed of light i.e. $\left( V_{p}.V_{g}=c^{2} \right)$