Answer
We can rank the indexes of refraction:
$n_3 \gt n_2 \gt n_1$
Work Step by Step
We can write the expression for the critical angle:
$\theta_c = sin^{-1}~\frac{n_f}{n_i}$
In this expression, $n_f$ is the index of refraction of air, and $n_i$ is the index of refraction of the material.
We can rearrange this expression to find an expression for the index of refraction of the material:
$n_i = \frac{n_f}{sin~\theta_c}$
We can see that a smaller critical angle corresponds with a higher index of refraction.
In Figure 33-33, we can see that material 3 has the smallest critical angle. Therefore, $n_3$ is the highest index of refraction.
In Figure 33-33, we can see that material 1 has the largest critical angle. Therefore, $n_1$ is the smallest index of refraction.
We can rank the indexes of refraction:
$n_3 \gt n_2 \gt n_1$
