Critical angle of ray ' 1 ':

$\mathit{s}\mathit{i}\mathit{n}\left({\mathit{c}}_1\right)=\frac{1}{{\mathit{\mu }}_1}=\frac{1}{1.35}$
${\mathit{c}}_1=\mathrm{arc}\mathrm{}\sin \left(\frac{1}{1.35}\right)={47.79}^{\circ }$
Similarly, critical angle of ray '2':

$\mathit{s}\mathit{i}\mathit{n}\left({\mathit{c}}_2\right)=\frac{1}{{\mathit{\mu }}_2}=\frac{1}{1.45}$
${\mathit{c}}_2=\mathit{s}\mathit{i}\mathit{n}{}^{-1}\left(\frac{1}{1.45}\right)={43.6}^{\circ }$
Ray ' 1 ' and ' 2 ' will fall on the side AC at an angle of incidence (i) of ${45}^{\circ }$. Critical angle of ray ' 1 ' is greater than $\mathit{i}$, so it will get refracted from the prism. Critical angle of ray ' 2 ' is less than that of $\mathit{i}$, so it will undergo total internal reflection.