Answer
$1.71\leq n_{prism}< 2.02$
Work Step by Step
We can find the minimum and maximum index of refraction of the prism as follows:
$n_{min}=\frac{n}{sin\theta_c}$
We plug in the known values to obtain:
$n_{min}=\frac{1.21}{sin45^{\circ}}$
$n_{min}=1.71$
The maximum index of refraction of the prism is given as
$n_{min}=\frac{n}{sin\theta_c}$
We plug in the known values to obtain:
$n_{min}=\frac{1.43}{sin45^{\circ}}$
$n_{min}=2.02$
$1.71\leq n_{prism}< 2.02$
