Answer
The mirror must be moved through a distance of $291\mu\,\mathrm{m}$.
Work Step by Step
Let L be the distance through which the mirror $M_2$ is moved.
If $N_1$ is the number of wavelengths in this length of wavelength $\lambda_1$ and $N_2$ that of wavelength $\lambda_2$, then we can write,
$2L=N_1\lambda_1$
$2L=N_2\lambda_2$
If the shift in the fringe pattern for $\lambda_1$ must be 1 fringe more than that for $\lambda_2$, we must have $N_1-N_2=1$.
I.e.,
${2L\over \lambda_1}-{2L\over \lambda_2}=1$
Or,
$L={\lambda_2\lambda_1\over 2(\lambda_2-\lambda_1)}$
Substituting the given values of the wavelengths, $\lambda_1=588.9950\times 10^{-9}\mathrm{m}$ and $\lambda_2=589.5924\times 10^{-9}\mathrm{m}$, we get,
\begin{align*}
L&={589.5924\times 10^{-9}\times 588.9950\times 10^{-9}\over 2(589.5924\times 10^{-9}-588.9950\times 10^{-9})}\\
&={347266.9\times 10^{-9}\over 1.1948}\\
&=290.6\times 10^{-6}\\
&=291\mu\,\mathrm{m}
\end{align*}
Hence the mirror $M_2$ must be moved through a distance of $291\mu\,\mathrm{m}$.
