Question

Find the width of the anomalous dispersion region for the case of a single resonance at frequency ${\mathit{\omega }}_{\mathbf{0}}$. Assume$\mathit{\gamma }\mathbf{<}\mathbf{<}{\mathit{\omega }}_{\mathbf{0}}$ . Show that the index of refraction assumes its maximum and minimum values at points where the absorption coefficient is at half-maximum.

Short answer

The width of the anomalous region is$\gamma$ . It is proved that the index of refraction assumes its maximum and minimum values at points where the absorption coefficient is at half-maximum.

Step-by-step solution

Expressions for the index of refraction:

Write the expressions for the index of refraction.

$\mathit{n}\mathbf{=}\mathbf{1}\mathbf{+}\frac{\mathbf{N}{\mathbf{q}}^{\mathbf{2}}}{\mathbf{2}\mathbf{m}{\mathbf{\epsilon }}_{\mathbf{0}}}\frac{\left({\omega }_0^2-{\omega }^2\right)}{\left[{\left({\omega }_0^2-{\omega }^2\right)}^2+v^2{\omega }^2\right]}$…… (1)

Here, N is the number of molecules per unit volume, q is the charge, m is the mass, ${\epsilon }_0$is the permittivity of free space,${\omega }_0$ is the resonance frequency and $\gamma$is the Lorentz contraction.

Determine the width of the anomalous region:

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

Differentiate the equation (1) with respect to $\omega .$

$\frac{dn}{d\omega }=\frac{N{q}^3}{2m{\epsilon }_0}\left\{\frac{-2\omega }{D}-\frac{\left({\omega }_0^2-{\omega }^2\right)}{D^2}\left[2\left({\omega }_0^2-{\omega }^2\right)(-2\omega )+2\omega {\gamma }^2\right]\right\}$

Substitute$\frac{dn}{d\omega }=0$in the equation.

localid="1657517748616" $$\begin{gathered} \frac{N{q}^2}{2m{\epsilon }_0}\left\{\frac{-2\omega }{D}-\frac{\left({\omega }_0^2-{\omega }^2\right)}{D^2}\left[2\left({\omega }_0^2-{\omega }^2\right)(-2\omega )+2\omega {\gamma }^2\right]\right\}=0 \\ \frac{-2\omega }{D}-\frac{\left({\omega }_0^2-{\omega }^2\right)}{D^2}\left[2\left({\omega }_0^2-{\omega }^2\right)(-2\omega )+2\omega {\gamma }^2\right]=0 \\ 2\omega D=2\omega \left({\omega }_0^2-{\omega }^2\right)\left[2\left({\omega }_0^2-{\omega }^2\right)-{\gamma }^2\right] \\ {\left({\omega }_0^2-{\omega }^2\right)}^2+{\gamma }^2{\omega }^2=2{\left({\omega }_0^2-{\omega }^2\right)}^2-{\gamma }^2\left({\omega }_0^2-{\omega }^2\right) \end{gathered}$$

On further solving, the above equation becomes,

$$\begin{gathered} {\left({\omega }_0^2-{\omega }^2\right)}^2={\gamma }^2\left({\omega }^2+{\omega }_0^2-{\omega }^2\right) \\ {\left({\omega }_0^2-{\omega }^2\right)}^2={\gamma }^2{\omega }_0^2 \\ {\omega }_0^2-{\omega }^2=\pm \gamma {\omega }_0 \\ {\omega }^2={\omega }_0^2\mp \gamma {\omega }_0 \end{gathered}$$

On further solving, the above equation becomes,

$$\begin{gathered} \omega ={\omega }_0\sqrt{1\mp \frac{\gamma }{{\omega }_0}} \\ ={\omega }_0\left(1\mp \frac{\gamma }{2{\omega }_0}\right) \\ ={\omega }_0\mp \frac{\gamma }{2} \end{gathered}$$

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

$$\begin{gathered} {\omega }_1={\omega }_0-\frac{\gamma }{2} \\ {\omega }_2={\omega }_0+\frac{\gamma }{2} \end{gathered}$$

Calculate the width of the anomalous region.

$$\begin{gathered} \Delta \omega ={\omega }_2-{\omega }_1 \\ \Delta \omega ={\omega }_0+\frac{\gamma }{2}-{\omega }_0+\frac{\gamma }{2}1 \\ \Delta \omega =\gamma \end{gathered}$$

Show that the index of refraction assumes its maximum and minimum values at points where the absorption coefficient is at half-maximum:

Write the equation for the absorption coefficient.

$\alpha =\frac{N{q}^2{\omega }^2}{m{\epsilon }_{0}c}\frac{\gamma }{{\left({\omega }_0^2-{\omega }^2\right)}^2+{\gamma }^2{\omega }^2}$ ……. (2)

Here, localid="1657518190348" $\omega ={\omega }_0$

Substitute $\omega ={\omega }_0$in equation (2).

$$\begin{gathered} {\alpha }_{\max }=\frac{N{q}^2{\omega }_0^2}{m{\epsilon }_{0}C}\frac{\gamma }{{\left({\omega }_0^2-{\omega }_0^2\right)}^2+{\gamma }^2{\omega }_0^2} \\ {\alpha }_{\max }=\frac{N{q}^2}{m{\epsilon }_{0}C\gamma } \end{gathered}$$

At ${\omega }_1$and ${\omega }_2$, ${\omega }^2={\omega }_0^2\mp \gamma {\omega }_0$, so, the equation (2) becomes,

$$\begin{gathered} \alpha =\frac{N{q}^2{\omega }^2}{m{\epsilon }_{0}c}\frac{\gamma }{{\gamma }^2{\omega }_0^2+{\gamma }^2{\omega }^2} \\ \alpha ={\alpha }_{\max }\left(\frac{{\omega }^2}{{\omega }_0^2+{\omega }^2}\right) \end{gathered}$$ …… (3)

Here, ${\omega }^2={\omega }_0^2\mp \gamma {\omega }_0$

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

$\left(\frac{{\omega }^2}{{\omega }_0^2+{\omega }^2}\right)=\frac{{\omega }_0^2\mp \gamma {\omega }_0}{2{\omega }_0^2\mp \gamma {\omega }_0}$

$$\begin{gathered} =\frac{1}{2}\frac{\left(1\mp \gamma /{\omega }_0\right)}{\left(1\mp \gamma /2{\omega }_0\right)} \\ \cong \frac{1}{2}\left(1\mp \frac{\gamma }{{\omega }_0}\right)\left(1\pm \frac{\gamma }{2{\omega }_0}\right) \\ \cong \frac{1}{2}\left(1\mp \frac{\gamma }{2{\omega }_0}\right) \end{gathered}$$

Simplify further.$\left(\frac{{\omega }^2}{{\omega }_0^2+{\omega }^2}\right)\cong \frac{1}{2}\mid$

Substitute $\left(\frac{{\omega }^2}{{\omega }_0^2+{\omega }^2}\right)\cong \frac{1}{2}$in equation (3).

$\alpha =\frac{1}{2}{\alpha }_{\max }$

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