Answer
$L=0.352\mu m$
Work Step by Step
We know that
$2L=(m+\frac{1}{2})({\frac{\lambda}{n}})$
This can be rearranged as:
$L=(m+\frac{1}{2})(\frac{\lambda}{2n})$
We plug in the known values to obtain:
$L=(m+\frac{1}{2})(\frac{624\times 10^{-9}}{2(1.33)})$
$L=(m+\frac{1}{2})(234.59nm)$
for $m=0$:
$L=(\frac{1}{2})(234.59nm)=117nm=0.117\mu m$
Therefore, for $m=1$:
$L=(1+\frac{1}{2})(234.59nm)=352nm=0.352\mu m$
