Answer
The index of refraction of the liquid is $~~1.40$
Work Step by Step
We can find $n_p$, the index of refraction of the plastic:
$phase~difference = n_p~L-1.00~L = 60~\lambda$
$n_p-1.00 = \frac{60~\lambda}{L}$
$n_p = 1.00+\frac{60~\lambda}{L}$
$n_p = 1.00+\frac{(60)(400~nm)}{40.0~\mu m}$
$n_p = 1.60$
The index of refraction of the plastic is $~~1.60$
We can find $n_l$, the index of refraction of the liquid:
$phase~difference = 1.60~L-n_l~L = 20~\lambda$
$1.60-n_l = \frac{20~\lambda}{L}$
$n_l = 1.60-\frac{20~\lambda}{L}$
$n_l = 1.60-\frac{(20)(400~nm)}{40.0~\mu m}$
$n_l = 1.40$
The index of refraction of the liquid is $~~1.40$
