Answer
$\lambda=560nm$
Work Step by Step
We know that
$2L=(m)({\frac{\lambda}{n_2}})$
This can be rearranged as:
$L=(m)({\frac{\lambda}{2n_2}})$
For $m=1$ and plugging in the known values, we obtain:
$200=1\times (\frac{\lambda}{2\times 1.40})$
$\lambda=560nm$
