Answer
The index of refraction of the prism is $~~\sqrt{1.00+sin^2~\theta}$
Work Step by Step
Let $\theta_2$ be the angle of refraction when the ray enters the prism.
Let $n_p$ be the index of refraction for the prism.
We can use Snell's law to find an equation:
$n_p~sin~\theta_2 = n_1~sin~\theta$
$n_p~sin~\theta_2 = 1.00~sin~\theta$
$n_p^2~sin^2~\theta_2 = sin^2~\theta$
Then the incident angle as the ray leaves the prism is $~~90^{\circ}-\theta_2$
We can use Snell's law to find an equation:
$n_p~sin~(90^{\circ}-\theta_2) = n_1~sin~90^{\circ}$
$n_p~cos~\theta_2 = 1.00$
$n_p^2~cos^2~\theta_2 = 1.00$
We can add both sides of both equations:
$n_p^2~sin^2~\theta_2+n_p^2~cos^2~\theta_2 = 1.00+sin^2~\theta$
$n_p^2~(sin^2~\theta_2+cos^2~\theta_2) = 1.00+sin^2~\theta$
$n_p^2 = 1.00+sin^2~\theta$
$n_p = \sqrt{1.00+sin^2~\theta}$
The index of refraction of the prism is $~~\sqrt{1.00+sin^2~\theta}$
