Question

If mirror ${\mathit{M}}_{\mathbf{2}}$in a Michelson interferometer (fig 35-21) is moved through $\mathbf{0}\mathbf{.}\mathbf{233}\mathbf{}\mathbf{mm}$, a shift of 792 bright fringes occurs. What is the wavelength of the light producing the fringe pattern?

Short answer

The wavelength of light producing is $\lambda =588\mathrm{nm}$.

Step-by-step solution

Given data

The mirror $M_2$ is moved by $\Delta x=0.233\text{mm} \text{or}0.233\times {10}^{-3} \text{m}$

The number of shifted fringes ${\text{n}}_s=792$

Concept of Michelson interferometer

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

Determine wavelength of the light producing the fringe pattern

From Michelson’s interferometer, the change in phase shift is

$\begin{array}{rcl}\Delta x& =& \frac{1}{2}{n}_s\lambda \\ \lambda & =& \frac{2\Delta x}{n_s}\end{array}$

So,

$\begin{array}{rcl}\lambda & =& \frac{2\times 0.233\times {10}^{-3} \text{m}}{792}\\ \lambda & =& 588.384\text{nm}\\ \lambda & =& 588\text{nm}\end{array}$

Hence, the wavelength of light producing is $\lambda =588\text{nm}$.