The element sodium can emit light at two wavelengths, ${\mathbf{\lambda }}_{\mathbf{1}}\mathbf{=}\mathbf{588}\mathbf{.}\mathbf{9950}\mathbf{nm}$

and ${\mathbf{\lambda }}_{\mathbf{2}}\mathbf{=}\mathbf{589}\mathbf{.}\mathbf{5924}\mathbf{nm}$. Light from sodium is being used in a Michelson interferometer (Fig. 35-21). Through what distance must mirror ${\mathbf{M}}_{\mathbf{2}}$ be moved if the shift in the fringe pattern for one wavelength is to be 1.00 fringe more than the shift in the fringe pattern for the other wavelength?

$$\begin{gathered} {\mathrm{\lambda }}_1=588.9950\mathrm{nm} \\ {\mathrm{\lambda }}_2=589.5924\mathrm{nm} \end{gathered}$$

Fiber optic Michelson interferometer employs the same principle of splitting a laser beam and inserting the optical path difference between the arms.

Existence of different wavelength shows the change in position of the mirror $M_2$ because the path difference gives the distance at which the mirror is moved.

Express the relation for the distance at which the mirror $M_2$is to be moved.

$N_1-N_2=2L\left(\frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_2}\right)$

Here,$N_1$ is the number wavelength in 2L thickness of air by the wavelength is $N_2$, number of wavelengths in 2L thickness of air by the wavelength ${\lambda }_2$ and L is the distance moved by, mirror $M_2$.

Use the condition to express the distance moved by the mirror$M_2$.

$N_1-N_2=1$

Rearrange the expression for distance moved by $M_2$ in term of wave lengths.

$L=\frac{1}{2\left(\frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_2}\right)}$

Here L is the distance moved by mirror $M_2$

Substitute $\text{588.9950}\text{nm}$ for ${\lambda }_1$ and $\text{589.5924}\text{nm}$ for ${\lambda }_2$ to find L.

$\begin{array}{rcl}L& =& \frac{1}{2\left({\displaystyle \frac{1}{588.9950nm}}-{\displaystyle \frac{1}{589.5924nm}}\right)\left({\displaystyle \frac{1nm}{{10}^{-9}m}}\right)}\\ & =& 2.906\times {10}^{-4}\text{m}\\ & =& 290.7\mu m\end{array}$

Therefore, the distance moved by the mirror is $290.7\mu m$.