Answer
$L=1.55\mu m$
Work Step by Step
We know that;
$(N_2-N_1)\lambda=L(n_2-n1)$
We plug in the known values to obtain:
$0.5(620=L(1.65-1.45))$
$L=1550nm=1.55\times 10^3\times 10^{-9}=1.55\times 10^{-6}=1.55\mu m$
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