91Ó°ÊÓ

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

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

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

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

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

Here, ${\mathit{λ}}_{\mathbf{0}}$ is the base wavelength of light, ${\mathit{λ}}_{\mathbf{1}}$ 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

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

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

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

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 $\mathrm{λ}$ and $0.6\text{m}$ for $\mathrm{L}$ in equation (2).

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

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

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

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

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,

$\begin{array}{c}\mathrm{Δ²Ô}=1,895,737.6−1,895,731.6\\ =6\end{array}$

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

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

That can then be inserted in for the ${â‹„}_1$ in equation (3).

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

The ${\mathrm{λ}}_0$ 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 ${\mathrm{λ}}_1$ to be larger than ${\mathrm{λ}}_0$, the denominator will have to be smaller. Thus 1 will be subtracted from the $n$ for the $n_1$ :

$\frac{2\mathrm{L}}{\mathrm{n}−1}−{\mathrm{λ}}_0>\mathrm{d}$

That can be rearranged to get rid of the fraction:

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

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

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

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

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

Or just:

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

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

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

That can be converted to :

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