Answer
$\lambda=509nm$
Work Step by Step
We know that
$2L=(m)(\frac{\lambda}{n_2})$
This can be rearranged as:
$\lambda=\frac{2n_2L}{m}$
We plug in the known values to obtain:
$\lambda=\frac{2(1.34)(380)}{m}$
$\lambda=\frac{1018}{m}$
For $m=1$
$\lambda=\frac{1018}{1}=1018nm$
For $m=2$
$\lambda=\frac{1018}{2}=509nm$
Therefore,$509nm$ is in the visible range.
