Probably that means the prism is an isosceles triangle but that doesn't influence the propagation of the ray. Anyway you can actually prove the statement I made in post #10. Denote the exit angle out of the second side as ##\eta##. In the second surface, Snell's law gives ##n \sin \delta = \sin \eta##. Then ##\beta = 90^o-\delta## and Snell's law in the first side gives
$$
\begin{aligned}
\sin \alpha &= n \sin\beta = n \sin(90^o-\delta) \\
&= n\cos \delta = n \sqrt{1-\sin^2 \delta}
\end{aligned}
$$
Then by substituting Snell's law from the second side for ##\sin \delta##,
$$
\sin \alpha = n \sqrt{1-\frac{\sin^2 \eta}{n^2}} \\
\sin \alpha = \sqrt{n^2 - \sin^2 \eta}
$$
The maximum value for the left side is unity, while the minimum value for the right side with ##n=2## is ##\sqrt{2^2-1} = \sqrt{3}##. This means, for any incident angle ##0\geq \alpha \geq 90^o##, there is no real solution for the exit angle ##\eta##. Physically, this means that the rays, irrespective of the incident angle, will undergo total internal reflection at the second surface.