Answer
$n_3 = 1.39$
Work Step by Step
By geometry, the incident angle of the ray on the interface between materials 2 and 3 is $\phi = 60.0^{\circ}$. It is given in the problem that this angle is the critical angle.
We can use Equation (33-45) to find $n_3$:
$\phi = sin^{-1}~\frac{n_3}{n_2}$
$sin~\phi = \frac{n_3}{n_2}$
$n_3 = n_2~sin~\phi$
$n_3 = 1.60~sin~60.0^{\circ}$
$n_3 = 1.39$
