In a Michelson interferometer, one mirror is moved by $\Delta x=0.1\text{ mm}$. If the light source has a wavelength of $\lambda =500\text{ nm}$, how many fringes $N$ will shift across the field of view?

Solution

1. Given the mirror displacement $$\begin{array}{rl}\Delta x& =0.1\text{ mm}\\ & =10^{-4}\text{ m}\end{array}$$ and the wavelength $$\begin{array}{rl}\lambda & =500\text{ nm}\\ & =5\times 10^{-7}\text{ m}\end{array}$$.
2. In a Michelson interferometer, moving a mirror by $\Delta x$ changes the total path difference by $2\Delta x$.
3. The number of fringes $N$ that shift is given by the formula $N=\frac{2\Delta x}{\lambda }$.
4. Substituting the values, $$\begin{array}{rl}N& =\frac{2\times 10^{-4}\text{ m}}{5\times 10^{-7}\text{ m}}\\ & =400\end{array}$$.
5. Hence the answer is (C).