Example 6.3 gives the refractive index for high-frequency electromagnetic radiation passing through Earth’s ionosphere. The constant $\mathbf{b}$, related to the so-called plasma frequency, varies with atmospheric conditions, but a typical value is$\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{15}}{\mathbf{rad}}^{\mathbf{2}}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$ . Given a GPS pulse of frequency$\mathbf{1.5}\mathbf{GHz}$ traveling through $\mathbf{8}\mathbf{}\mathbf{km}$of ionosphere, by how much, in meters, would the wave group and a particular wave crest be ahead of or behind (as the case may be) a pulse of light passing through the same distance of vacuum?

The constant in the refractive index equation is$\mathrm{b}=8\times {10}^{15}{\mathrm{rad}}^2/{\mathrm{s}}^2$ .

The frequency of the pulse is$f=1.5\times {10}^{9}Hz$

The distance covered by the pulse is $8\mathrm{km}$.

A particle moving can be associated as moving group of plane waves. This is regarded as sum of planes which is called as wave group and the velocity of this wave group is called group velocity.

The relation between group velocity and angular frequency is expressed as below,

${\mathrm{v}}_{\mathrm{g}}=\mathrm{c}\sqrt{1-\frac{\mathrm{b}}{{\mathrm{\omega }}^2}}$

Here, $\mathrm{c}$is the speed of light and $\omega$is angular frequency. Inserting the given data in the above expression

$\begin{array}{c}{\mathrm{v}}_{\mathrm{g}}=\mathrm{c}\sqrt{1-\frac{8\times {10}^{15}{\mathrm{rad}}^2/{\mathrm{s}}^2}{{\left(2\mathrm{\pi f}\right)}^2}}\\ =\mathrm{c}\sqrt{1-\frac{8\times {10}^{15}{\mathrm{rad}}^2/{\mathrm{s}}^2}{{\left(2\mathrm{\pi }(1.5\times {10}^9\mathrm{cycles}/\mathrm{s})\right)}^2}}\\ =(3\times {10}^8\mathrm{m}/\mathrm{s})\sqrt{1-\frac{8\times {10}^{15}{\mathrm{rad}}^2/{\mathrm{s}}^2}{{(3\mathrm{\pi }\times {10}^9\mathrm{rad}/\mathrm{s})}^2}}\\ =2.999865\times {10}^8\mathrm{m}/\mathrm{s}\end{array}$

The time taken by wave group to cover$8\mathrm{km}$is

$\begin{array}{c}{\mathrm{t}}_{\mathrm{g}}=\frac{(8\times {10}^3\mathrm{m})}{2.999865\times {10}^8\mathrm{m}/\mathrm{s}}\\ =2.66679\times {10}^{-5}\mathrm{s}\end{array}$

And the time taken by the light pulse in vacuum to travel the same distance is

$\begin{array}{c}{\mathrm{t}}_{\mathrm{c}}=\frac{8\times {10}^3\mathrm{m}}{3\times {10}^8\mathrm{m}/\mathrm{s}}\\ =2.66667\times {10}^{-5}\mathrm{s}\end{array}$

The time difference is

$\begin{array}{c}\mathrm{\Delta t}=2.66679\times {10}^{-5}\mathrm{s}-2.66667\times {10}^{-5}\mathrm{s}\\ =0.00012\times {10}^{-5}\mathrm{s}\end{array}$

It takes more time for wave group to reach $8\mathrm{km}$than light pulse. Hence, the wave group will be behind the light pulse by an amount

$\begin{array}{c}\mathrm{d}={\mathrm{v}}_{\mathrm{g}}\mathrm{\Delta t}\\ =(2.999865\times {10}^8\mathrm{m}/\mathrm{s})(0.00012\times {10}^{-5}\mathrm{s})\\ =0.36\mathrm{m}\\ =36\mathrm{cm}\end{array}$

The phase velocity is related to angular frequency by the relation

${\mathrm{v}}_{\mathrm{P}}=\frac{\mathrm{c}}{\sqrt{1-\frac{\mathrm{b}}{{\mathrm{\omega }}^2}}}$

Here, $\mathrm{c}$is the speed of light and $\mathrm{\omega }$is angular frequency. Inserting the given data in the above expression

$\begin{array}{c}{\mathrm{v}}_{\mathrm{P}}=\frac{3\times {10}^8\mathrm{m}/\mathrm{s}}{\sqrt{1-\frac{8\times {10}^{15}{\mathrm{rad}}^2/{\mathrm{s}}^2}{{\left(2\mathrm{\pi f}\right)}^2}}}\\ =\frac{3\times {10}^8\mathrm{m}/\mathrm{s}}{\sqrt{1-\frac{8\times {10}^{15}{\mathrm{rad}}^2/{\mathrm{s}}^2}{{\left(2\mathrm{\pi }(1.5\times {10}^9\mathrm{cycles}/\mathrm{s})\right)}^2}}}\\ =\frac{3\times {10}^8\mathrm{m}/\mathrm{s}}{\sqrt{1-\frac{8\times {10}^{15}{\mathrm{rad}}^2/{\mathrm{s}}^2}{{(3\mathrm{\pi }\times {10}^9\mathrm{rad}/\mathrm{s})}^2}}}\\ =3.000135\times {10}^8\mathrm{m}/\mathrm{s}\end{array}$

The time taken by crest of the plane wave to cover $8km$is

$\begin{array}{c}{\mathrm{t}}_{\mathrm{p}}=\frac{(8\times {10}^3\mathrm{m})}{3.000135\times {10}^8\mathrm{m}/\mathrm{s}}\\ =2.66655\times {10}^{-5}\mathrm{s}\end{array}$

And the time taken by the light pulse in vacuum to travel the same distance is

$\begin{array}{c}{\mathrm{t}}_{\mathrm{c}}=\frac{8\times {10}^3\mathrm{m}}{3\times {10}^8\mathrm{m}/\mathrm{s}}\\ =2.66667\times {10}^{-5}\mathrm{s}\end{array}$

The time difference is

$\begin{array}{c}\mathrm{\Delta t}=2.66667\times {10}^{-5}\mathrm{s}-2.66655\times {10}^{-5}\mathrm{s}\\ =0.00012\times {10}^{-5}\mathrm{s}\end{array}$

It takes less time for crest of the plane wave to reach$8\mathrm{km}$ than light pulse. Hence, the crest will be ahead of the light pulse by an amount