The fact that a laser's resonant cavity so effectively sharpens the wavelength can lead to the output of several closely spaced laser wavelengths, called longitudinal modes. Here we see how. Suppose the spontaneous emission serving as the seed for stimulated emission is of wavelength , but somewhat fuzzy, with a line width of roughly$\mathbf{0}\mathbf{.001}\mathbf{}\mathbf{\text{nm}}$ either side of the central value. The resonant cavity is exactly$\mathbf{60}\mathbf{\text{cm}}$ long. (a) How many wavelengths fit the standing-wave condition'? (b) If only u single wavelength were desired, would change the length of the cavity help? Explain.

A standing wave, also called a stationary wave, a combination of two waves moving in opposite directions, each having the same amplitude and frequency.

The standing wave condition is given by

$\mathbf{L}\mathbf{=}\frac{\mathbf{n\lambda }}{\mathbf{2}}$ ….. (1)

Where, is the length of the cavity, is the wavelength, and is number of wavelengths.

$\mathit{n}\mathbf{=}\frac{\mathbf{2}\mathbf{L}}{\mathbf{\lambda }}$ ….. (2)

${\mathit{\lambda }}_{\mathbf{1}}\mathbf{-}{\mathit{\lambda }}_{\mathbf{0}}\mathbf{>}\mathit{d}$ ….. (3)

Here, ${\mathit{\lambda }}_{\mathbf{0}}$ is the base wavelength of light, ${\mathit{\lambda }}_{\mathbf{1}}$ is the wavelength of a longitudinal mode, and is the half-line width of the emission line.

Shorter wavelengths are formed when line width is added to wavelength of stimulated emission

$\begin{array}{c}\mathrm{\lambda }=\left(633\text{nm}+0.001\text{nm}\right)\left(\frac{1\times {10}^{-9}\text{m}}{1\text{nm}}\right)\\ =633.001\times {10}^{-9}\text{m}\end{array}$

Longer wavelengths are formed when line width is subtracted from wavelength of stimulated emission

$\begin{array}{c}\mathrm{\lambda }\text{'}=\left(633\text{nm}-0.001\text{nm}\right)\left(\frac{1\times {10}^{-9}\text{m}}{1\text{nm}}\right)\\ =632.999\times {10}^{-9}\text{m}\end{array}$

Cavity length shall be converted to standard units:

$\begin{array}{c}\mathrm{L}=\left(60\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\\ =0.6\text{m}\end{array}$

To get the number of half wavelengths of a somewhat shorter wavelength would fit the condition; substitute $633.001\text{nm}$ for $\lambda$ and $0.6\text{m}$ for $\mathrm{L}$ in equation (2).

$\begin{array}{c}\mathrm{n}=\frac{2(0.6\text{m})}{\left(633.001\times {10}^{-9}\text{m}\right)}\\ =1,895,731.6\end{array}$

So, $1,895,732$ half wavelength somewhat shorter wavelength would fit the condition

To find the number of loner wave lengths, substitute $632.999\text{nm}\lambda$ for $\lambda \text{'}$and $0.6\text{m}$ for $\text{L}$ in equation (1).

$\begin{array}{c}\mathrm{n}=\frac{2(0.6\text{m})}{\left(632.999\times {10}^{-9}\text{m}\right)}\\ =18,95,737.6\end{array}$

So, $1,895,732$ half wavelength somewhat shorter wavelength would fit the condition

The difference between the numbers of waves in the first to second situation is,

$\begin{array}{c}\mathrm{\Delta n}=1,895,737.6-1,895,731.6\\ =6\end{array}$

To use that though, equation (1) will first be rewritten to solve for the wavelength:

$\begin{array}{l}\mathrm{L}=\frac{\mathrm{n\lambda }}{2}\\ \mathrm{\lambda }=\frac{2\mathrm{L}}{\mathrm{n}}\end{array}$

That can then be inserted in for the ${\diamond }_1$ in equation (3).

$\begin{array}{l}{\mathrm{\lambda }}_1-{\mathrm{\lambda }}_0>\mathrm{d}\\ \frac{2\mathrm{L}}{{\mathrm{n}}_1}-{\mathrm{\lambda }}_0>\mathrm{d}\end{array}$

The ${\lambda }_0$ can stay as is because the value of that is known. The ${\mathrm{n}}_1$ will differ from the base value of $n$ by just $1$, to show that an additional sized wavelength fitting inside the resonant cavity. In order for the ${\lambda }_1$ to be larger than ${\lambda }_0$, the denominator will have to be smaller. Thus 1 will be subtracted from the $n$ for the $n_1$ :

$\frac{2\mathrm{L}}{\mathrm{n}-1}-{\mathrm{\lambda }}_0>\mathrm{d}$

That can be rearranged to get rid of the fraction:

$\mathrm{L}>\left(\mathrm{n}-1\right)\left({\mathrm{\lambda }}_0+\mathrm{d}\right)$

However, the $n$ can be filled in with equation (2) for the number of the original wavelengths that will satisfy the standing wave condition:

$2\mathrm{L}>\left[\left(\frac{2\mathrm{L}}{{\mathrm{\lambda }}_0}-1\right)\right]\left({\mathrm{\lambda }}_0+\mathrm{d}\right)$

Then multiply that out, and collect the $\mathrm{L}\text{'}\mathrm{s}$:

$\begin{array}{l}2\mathrm{L}>2\mathrm{L}+\frac{2\mathrm{Ld}}{{\mathrm{\lambda }}_0}-{\mathrm{\lambda }}_0-\mathrm{d}\\ {\mathrm{\lambda }}_0+\mathrm{d}>\frac{2\mathrm{Ld}}{{\mathrm{\lambda }}_0}\\ \left(\frac{{\mathrm{\lambda }}_0}{2\mathrm{d}}\right)\left({\mathrm{\lambda }}_0+\mathrm{d}\right)>\mathrm{L}\end{array}$

Or just:

$L<\left(\frac{{\lambda }_0}{2d}\right)\left({\lambda }_0+d\right)$

And that gives what the length of the resonant cavity should be so that no longitudinal modes appear. To evaluate that, just fill $6.33\times {10}^{-7}\text{m}$ for the ${\mathrm{l}}_0$ and $1\times {10}^{12}\text{m}$ for $\text{d}$:

$\begin{array}{l}\mathrm{L}<\left[\frac{\left(6.33\times {10}^{-7}\text{m}\right)}{2\left(1\times {10}^{-12}\text{m}\right)}\right]\left[\left(6.33\times {10}^{-7}\text{m}\right)+\left(1\times {10}^{-12}\text{m}\right)\right]\\ \mathrm{L}<0.2003\text{m}\end{array}$

That can be converted to :

Hence, the length of the resonant cavity would have to be .