Answer
$d = 36.2~nm$
Work Step by Step
In part (a), we found that the path length difference is $5d$
For the rays to be exactly out of phase, the path length difference must be $(m+0.5)~\frac{\lambda}{n}$, where $m$ is an integer
To find the least value of distance $d$, we can let $m = 0$:
$5d = (m+0.5)~\frac{\lambda}{n}$
$5d = \frac{(0.5)(500~nm)}{1.38}$
$d = \frac{(0.5)(500~nm)}{(5)(1.38)}$
$d = 36.2~nm$
