91Ó°ÊÓ

Write the expressions for the index of refraction.

$\mathit{n}\mathbf{=}\mathbf{1}\mathbf{+}\frac{\mathbf{N}{\mathbf{q}}^{\mathbf{2}}}{\mathbf{2}\mathbf{m}{\mathbf{Î\mu }}_{\mathbf{0}}}\frac{\left({\mathrm{Ó\neg }}_0^2-{\mathrm{Ó\neg }}^2\right)}{\left[{\left({\mathrm{Ó\neg }}_0^2-{\mathrm{Ó\neg }}^2\right)}^2+v^2{\mathrm{Ó\neg }}^2\right]}$…… (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 .

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

$\frac{dn}{d\mathrm{Ó\neg }}=\frac{N{q}^3}{2m{\mathrm{Î\mu }}_0}\left\{\frac{-2\mathrm{Ó\neg }}{D}-\frac{\left({\mathrm{Ó\neg }}_0^2-{\mathrm{Ó\neg }}^2\right)}{D^2}\left[2\left({\mathrm{Ó\neg }}_0^2-{\mathrm{Ó\neg }}^2\right)(-2\mathrm{Ó\neg })+2\mathrm{Ó\neg }{\mathrm{γ}}^2\right]\right\}$

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

localid="1657517748616" $$\begin{gathered} \frac{N{q}^2}{2m{\mathrm{Î\mu }}_0}\left\{\frac{-2\mathrm{Ó\neg }}{D}-\frac{\left({\mathrm{Ó\neg }}_0^2-{\mathrm{Ó\neg }}^2\right)}{D^2}\left[2\left({\mathrm{Ó\neg }}_0^2-{\mathrm{Ó\neg }}^2\right)(-2\mathrm{Ó\neg })+2\mathrm{Ó\neg }{\mathrm{γ}}^2\right]\right\}=0 \\ \frac{-2\mathrm{Ó\neg }}{D}-\frac{\left({\mathrm{Ó\neg }}_0^2-{\mathrm{Ó\neg }}^2\right)}{D^2}\left[2\left({\mathrm{Ó\neg }}_0^2-{\mathrm{Ó\neg }}^2\right)(-2\mathrm{Ó\neg })+2\mathrm{Ó\neg }{\mathrm{γ}}^2\right]=0 \\ 2\mathrm{Ó\neg }D=2\mathrm{Ó\neg }\left({\mathrm{Ó\neg }}_0^2-{\mathrm{Ó\neg }}^2\right)\left[2\left({\mathrm{Ó\neg }}_0^2-{\mathrm{Ó\neg }}^2\right)-{\mathrm{γ}}^2\right] \\ {\left({\mathrm{Ó\neg }}_0^2-{\mathrm{Ó\neg }}^2\right)}^2+{\mathrm{γ}}^2{\mathrm{Ó\neg }}^2=2{\left({\mathrm{Ó\neg }}_0^2-{\mathrm{Ó\neg }}^2\right)}^2-{\mathrm{γ}}^2\left({\mathrm{Ó\neg }}_0^2-{\mathrm{Ó\neg }}^2\right) \end{gathered}$$

On further solving, the above equation becomes,

$$\begin{gathered} {\left({\mathrm{Ó\neg }}_0^2-{\mathrm{Ó\neg }}^2\right)}^2={\mathrm{γ}}^2\left({\mathrm{Ó\neg }}^2+{\mathrm{Ó\neg }}_0^2-{\mathrm{Ó\neg }}^2\right) \\ {\left({\mathrm{Ó\neg }}_0^2-{\mathrm{Ó\neg }}^2\right)}^2={\mathrm{γ}}^2{\mathrm{Ó\neg }}_0^2 \\ {\mathrm{Ó\neg }}_0^2-{\mathrm{Ó\neg }}^2=Â\pm \mathrm{γ}{\mathrm{Ó\neg }}_0 \\ {\mathrm{Ó\neg }}^2={\mathrm{Ó\neg }}_0^2∓\mathrm{γ}{\mathrm{Ó\neg }}_0 \end{gathered}$$

On further solving, the above equation becomes,

$$\begin{gathered} \mathrm{Ó\neg }={\mathrm{Ó\neg }}_0\sqrt{1∓\frac{\mathrm{γ}}{{\mathrm{Ó\neg }}_0}} \\ ={\mathrm{Ó\neg }}_0\left(1∓\frac{\mathrm{γ}}{2{\mathrm{Ó\neg }}_0}\right) \\ ={\mathrm{Ó\neg }}_0∓\frac{\mathrm{γ}}{2} \end{gathered}$$

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

$$\begin{gathered} {\mathrm{Ó\neg }}_1={\mathrm{Ó\neg }}_0-\frac{\mathrm{γ}}{2} \\ {\mathrm{Ó\neg }}_2={\mathrm{Ó\neg }}_0+\frac{\mathrm{γ}}{2} \end{gathered}$$

Calculate the width of the anomalous region.

$$\begin{gathered} \mathrm{Δ}\mathrm{Ó\neg }={\mathrm{Ó\neg }}_2-{\mathrm{Ó\neg }}_1 \\ \mathrm{Δ}\mathrm{Ó\neg }={\mathrm{Ó\neg }}_0+\frac{\mathrm{γ}}{2}-{\mathrm{Ó\neg }}_0+\frac{\mathrm{γ}}{2}1 \\ \mathrm{Δ}\mathrm{Ó\neg }=\mathrm{γ} \end{gathered}$$

Write the equation for the absorption coefficient.

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

Here, localid="1657518190348" $\mathrm{Ó\neg }={\mathrm{Ó\neg }}_0$

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

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

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,

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

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

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

$\left(\frac{{\mathrm{Ó\neg }}^2}{{\mathrm{Ó\neg }}_0^2+{\mathrm{Ó\neg }}^2}\right)=\frac{{\mathrm{Ó\neg }}_0^2∓\mathrm{γ}{\mathrm{Ó\neg }}_0}{2{\mathrm{Ó\neg }}_0^2∓\mathrm{γ}{\mathrm{Ó\neg }}_0}$

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

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

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$\mathrm{γ}$.it is proved that the index of refraction assumes its maximum and minimum values at points where the absorption coefficient is at half-maximum.