Answer
The maximum value that the index of refraction can have is $~~\sqrt{2}$
Work Step by Step
In part (a), we found that the index of refraction of the prism is $~~\sqrt{1.00+sin^2~\theta}$
The maximum possible value occurs when $sin~\theta = 1$
In this case, the index of refraction is $\sqrt{2}$
The maximum value that the index of refraction can have is $~~\sqrt{2}$
