Critical angle of ray ' 1 ': $\sin \; (c_1) \;=\; \frac{1}{\mu_1} = \; \frac{1}{1.35}$ $c_1 \;=\; \arcsin \left( \;\frac{1}{1.35} \right) = 47.79^{\circ}$ Similarly, critical angle of ray '2': $\sin \; (c_2) \;=\; \frac{1}{\mu_2} = \; \frac{1}{1.45}$ $c_2 \;=\; \sin^{-1} \left( \;\frac{1}{1.45} \right) = 43.6^{\circ}$ 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 $i$, so it will get refracted from the prism. Critical angle of ray ' 2 ' is less than that of $i$, so it will undergo total internal reflection.