Answer
$n_2=1.00030$
Work Step by Step
As we know that
$N_2-N_1=(\frac{2d}{\lambda})(n_2-n_1)$
This can be rearranged as:
$n_2=n_1+\frac{\lambda(N_2-N_1)}{2d}$
We plug in the known values to obtain:
$n_2=1.00+\frac{(500\times 10^{-9}\times 60)}{2\times 0.05}=1.00030$
