Answer
$\Delta \phi=411.4^{\circ}$
Work Step by Step
We know that:
$N_2-N_1=(\frac{L}{\lambda})(n-1.00)$
We plug in the known values to obtain:
$N_2-N_1=(\frac{2000}{700})(1.400-1.00)=1.142857$
Now,
$\Delta \phi=(N_2-N_1)\times 360$
$\Delta \phi=1.142857\times360=411.4^{\circ}$
