91Ó°ÊÓ

Write the expressions forthe index of refraction.

$n=1+\frac{N{q}^2}{2m{\mathrm{Î\mu }}_0}\frac{({\mathrm{Ó\neg }}_0^2-{\mathrm{Ó\neg }}^2)}{[{({\mathrm{Ó\neg }}_0^2-{\mathrm{Ó\neg }}^2)}^2+{\mathrm{γ}}^2{\mathrm{Ó\neg }}^2]}$ …… (1)

Here, N is the number of molecules per unit volume, q is the charge, m is the mass,${\mathrm{Î\mu }}_0$ is the permittivity of free space,${\mathrm{Ó\neg }}_0$ is the resonance frequency and$\mathrm{γ}$ is the Lorentz contraction.

Let the denominator part of equation (1) is equal to $D$.

Differentiate the equation (1) with respect to $\mathrm{Ó\neg }$.

$\frac{dn}{d\mathrm{Ó\neg }}=\frac{N{q}^2}{2m{\mathrm{Î\mu }}_0}\{\frac{−2\mathrm{Ó\neg }}{D}−\frac{({\mathrm{Ó\neg }}_0^2−{\mathrm{Ó\neg }}^2)}{D^2}[2({\mathrm{Ó\neg }}_0^2−{\mathrm{Ó\neg }}^2)(−2\mathrm{Ó\neg })+2\mathrm{Ó\neg }{\mathrm{γ}}^2]\}$

Substitute $\frac{dn}{d\mathrm{Ó\neg }}=0$in the equation.

$\begin{array}{l}\frac{N{q}^2}{2m{\mathrm{Î\mu }}_0}\{\frac{−2\mathrm{Ó\neg }}{D}−\frac{({\mathrm{Ó\neg }}_0^2−{\mathrm{Ó\neg }}^2)}{D^2}[2({\mathrm{Ó\neg }}_0^2−{\mathrm{Ó\neg }}^2)(−2\mathrm{Ó\neg })+2\mathrm{Ó\neg }{\mathrm{γ}}^2]\}=0\\ \frac{−2\mathrm{Ó\neg }}{D}−\frac{({\mathrm{Ó\neg }}_0^2−{\mathrm{Ó\neg }}^2)}{D^2}[2({\mathrm{Ó\neg }}_0^2−{\mathrm{Ó\neg }}^2)(−2\mathrm{Ó\neg })+2\mathrm{Ó\neg }{\mathrm{γ}}^2]=0\\ 2\mathrm{Ó\neg }D=2\mathrm{Ó\neg }({\mathrm{Ó\neg }}_0^2−{\mathrm{Ó\neg }}^2)[2({\mathrm{Ó\neg }}_0^2−{\mathrm{Ó\neg }}^2)−{\mathrm{γ}}^2]\\ {({\mathrm{Ó\neg }}_0^2−{\mathrm{Ó\neg }}^2)}^2+{\mathrm{γ}}^2{\mathrm{Ó\neg }}^2=2{({\mathrm{Ó\neg }}_0^2−{\mathrm{Ó\neg }}^2)}^2−{\mathrm{γ}}^2({\mathrm{Ó\neg }}_0^2−{\mathrm{Ó\neg }}^2)\end{array}$

On further solving, the above equation becomes,

$\begin{array}{l}{({\mathrm{Ó\neg }}_0^2−{\mathrm{Ó\neg }}^2)}^2={\mathrm{γ}}^2({\mathrm{Ó\neg }}^2+{\mathrm{Ó\neg }}_0^2−{\mathrm{Ó\neg }}^2)\\ {({\mathrm{Ó\neg }}_0^2−{\mathrm{Ó\neg }}^2)}^2={\mathrm{γ}}^2{\mathrm{Ó\neg }}_0^2\\ {\mathrm{Ó\neg }}_0^2−{\mathrm{Ó\neg }}^2=Â\pm \mathrm{γ}{\mathrm{Ó\neg }}_0\\ {\mathrm{Ó\neg }}^2={\mathrm{Ó\neg }}_0^2∓\mathrm{γ}{\mathrm{Ó\neg }}_0\end{array}$

Again on further solving, the above equation becomes,

$\begin{array}{c}\mathrm{Ó\neg }={\mathrm{Ó\neg }}_0\sqrt{1∓\frac{\mathrm{γ}}{{\mathrm{Ó\neg }}_0}}\\ ={\mathrm{Ó\neg }}_0(1∓\frac{\mathrm{γ}}{2{\mathrm{Ó\neg }}_0})\\ ={\mathrm{Ó\neg }}_0∓\frac{\mathrm{γ}}{2}\end{array}$

Hence, the initial and final width of an anomalous region will be,

$\begin{array}{c}{\mathrm{Ó\neg }}_1={\mathrm{Ó\neg }}_0−\frac{\mathrm{γ}}{2}\\ {\mathrm{Ó\neg }}_2={\mathrm{Ó\neg }}_0+\frac{\mathrm{γ}}{2}\end{array}$

Calculate the width of the anomalous region.

$\begin{array}{c}\mathrm{Δ}\mathrm{Ó\neg }={\mathrm{Ó\neg }}_2−{\mathrm{Ó\neg }}_1\\ \mathrm{Δ}\mathrm{Ó\neg }={\mathrm{Ó\neg }}_0+\frac{\mathrm{γ}}{2}−{\mathrm{Ó\neg }}_0+\frac{\mathrm{γ}}{2}\\ \mathrm{Δ}\mathrm{Ó\neg }=\mathrm{γ}\end{array}$

Write the equation for the absorption coefficient.

$\mathrm{Î\pm }=\frac{N{q}^2{\mathrm{Ó\neg }}^2}{m{\mathrm{Î\mu }}_{0}c}\frac{\mathrm{γ}}{{({\mathrm{Ó\neg }}_0^2−{\mathrm{Ó\neg }}^2)}^2+{\mathrm{γ}}^2{\mathrm{Ó\neg }}^2}$ ……. (2)

Here $\mathrm{Ó\neg }={\mathrm{Ó\neg }}_0$,

Substitute $\mathrm{Ó\neg }={\mathrm{Ó\neg }}_0$in equation (2).

$\begin{array}{c}{\mathrm{Î\pm }}_{\max }=\frac{N{q}^2{\mathrm{Ó\neg }}_0^2}{m{\mathrm{Î\mu }}_{0}c}\frac{\mathrm{γ}}{{({\mathrm{Ó\neg }}_0^2−{\mathrm{Ó\neg }}_0^2)}^2+{\mathrm{γ}}^2{\mathrm{Ó\neg }}_0^2}\\ {\mathrm{Î\pm }}_{\max }=\frac{N{q}^2}{m{\mathrm{Î\mu }}_{0}c\mathrm{γ}}\end{array}$

At${\mathrm{Ó\neg }}_1$and ${\mathrm{Ó\neg }}_2$,${\mathrm{Ó\neg }}^2={\mathrm{Ó\neg }}_0^2∓\mathrm{γ}{\mathrm{Ó\neg }}_0$, so, the equation (2) becomes,

$\begin{array}{c}\mathrm{Î\pm }=\frac{N{q}^2{\mathrm{Ó\neg }}^2}{m{\mathrm{Î\mu }}_{0}c}\frac{\mathrm{γ}}{{\mathrm{γ}}^2{\mathrm{Ó\neg }}_0^2+{\mathrm{γ}}^2{\mathrm{Ó\neg }}^2}\\ \mathrm{Î\pm }={\mathrm{Î\pm }}_{\max }\left(\frac{{\mathrm{Ó\neg }}^2}{{\mathrm{Ó\neg }}_0^2+{\mathrm{Ó\neg }}^2}\right)\end{array}$ …… (3)

Here, ${\mathrm{Ó\neg }}^2={\mathrm{Ó\neg }}_0^2∓\mathrm{γ}{\mathrm{Ó\neg }}_0$.

Calculatethe value of $\left(\frac{{\mathrm{Ó\neg }}^2}{{\mathrm{Ó\neg }}_0^2+{\mathrm{Ó\neg }}^2}\right)$from equation (3).

$\begin{array}{c}\left(\frac{{\mathrm{Ó\neg }}^2}{{\mathrm{Ó\neg }}_0^2+{\mathrm{Ó\neg }}^2}\right)=\frac{{\mathrm{Ó\neg }}_0^2∓\mathrm{γ}{\mathrm{Ó\neg }}_0}{2{\mathrm{Ó\neg }}_0^2∓\mathrm{γ}{\mathrm{Ó\neg }}_0}\\ =\frac{1}{2}\frac{(1∓\mathrm{γ}/{\mathrm{Ó\neg }}_0)}{(1∓\mathrm{γ}/2{\mathrm{Ó\neg }}_0)}\\ â‰\dots \frac{1}{2}(1∓\frac{\mathrm{γ}}{{\mathrm{Ó\neg }}_0})(1Â\pm \frac{\mathrm{γ}}{2{\mathrm{Ó\neg }}_0})\\ â‰\dots \frac{1}{2}(1∓\frac{\mathrm{γ}}{2{\mathrm{Ó\neg }}_0})\end{array}$

Simplify further.

$\left(\frac{{\mathrm{Ó\neg }}^2}{{\mathrm{Ó\neg }}_0^2+{\mathrm{Ó\neg }}^2}\right)â‰\dots \frac{1}{2}$

Substitute$\left(\frac{{\mathrm{Ó\neg }}^2}{{\mathrm{Ó\neg }}_0^2+{\mathrm{Ó\neg }}^2}\right)â‰\dots \frac{1}{2}$ in equation (3).

$\mathrm{Î\pm }=\frac{1}{2}{\mathrm{Î\pm }}_{\max }$

Therefore, the width of the anomalous region is .it is proved that the index of refraction assumes its maximum and minimum values at points where the absorption coefficient is at half-maximum.