Answer
The index of refraction of the plastic is $~~1.40$
Work Step by Step
According to the graph, the phase difference would be zero if the total length of the hollow is $L = 60.0~\mu m$
We can find the index of refraction of the plastic:
$phase~difference = n~L-1.00~L = 60~\lambda$
$n-1.00 = \frac{60~\lambda}{L}$
$n = 1.00+\frac{60~\lambda}{L}$
$n = 1.00+\frac{(60)(400~nm)}{60.0~\mu m}$
$n = 1.40$
The index of refraction of the plastic is $~~1.40$
