Applying Snell's law at M,

$$n = {{\sin \alpha } \over {\sin {r_1}}} \Rightarrow \sqrt 2 = {{\sin 45^\circ } \over {\sin {r_1}}}$$

$$ \Rightarrow \sin {r_1} = {{\sin 45^\circ } \over {\sqrt 2 }} = {{1/\sqrt 2 } \over {\sqrt 2 }} = {1 \over 2}$$

$${r_1} = 30^\circ $$

$$\sin {\theta _c} = {1 \over n} = {1 \over {\sqrt 2 }} \Rightarrow {\theta _c} = 45^\circ $$

Let us take $${r_2} = {\theta _c} = 45^\circ $$ for just satisfying the condition of TIR.

In $$\Delta PNM$$,

$$\theta + 90 + {r_1} + 90 - {r_2} = 180^\circ $$

or $$\theta = {r_2} - {r_1} = 45^\circ - 30^\circ = 15^\circ $$

Note If $$\alpha$$ = 45$$^\circ $$ (the given value). Then, r₁ > 30$$^\circ $$ (the obtained value)

$${r_2} > {\theta _c}$$ (as r₂ $$-$$ r₁ = $$\theta$$ or r₂ = $$\theta$$ + r₁)

or TIR will take place. So, for taking TIR under all conditions $$\alpha$$ should be greater than 45$$^\circ$$ or this is the minimum value of $$\alpha$$.