ABC is an isosceles right angled triangle with the angle BAC 90^o. A ray of light parallel to the base BC strikes the face AB.  The refracted ray in the prism strikes the base BC. If the refractive index of the material of the prism is $\sqrt{2}$, find the total angle of deviation of the final emergent ray

$\mathrm{sin\alpha }=\frac{\mathrm{sini}}{\mathrm{\mu }}=\frac{\sin \left({90}^{\mathrm{o}}-{45}^{\mathrm{o}}\right)}{\sqrt{2}}=\frac{1}{2}\Rightarrow \mathrm{r}={30}^{\mathrm{o}}$

Now, ${45}^{\mathrm{o}}+\left({90}^{\mathrm{o}}+\mathrm{r}\right)+\mathrm{\theta }={180}^{\mathrm{o}}\Rightarrow \mathrm{\theta }={15}^{\mathrm{o}}\therefore {\mathrm{i}}^1={90}^{\mathrm{o}}-\mathrm{\theta }={75}^{\mathrm{o}}$

Now critical angle at glass-air interface, $\mathrm{c}={\sin }^{-1}\frac{1}{\mathrm{\mu }}={\sin }^{-1}\frac{1}{\sqrt{2}}={45}^{\mathrm{o}}$

Since ${\mathrm{i}}^1>\mathrm{c}$, the ray incident on the base BC will suffer total internal reflection and the final emergent ray through the face AC will be parallel to the base BC as shown in the figure. Hence total angle of deviation is zero.