1. Set up the prism geometry

   Let the right-angled prism have angles:

   • $\angle R = 90^\circ$
   • $\angle Q = q$
   • $\angle P = 90^\circ - q$

   The light is incident on face $PQ$ from air at angle $a$ (with the normal).

2. Refraction at face $PQ$

   Let the angle of refraction inside the prism at face $PQ$ be $r$.

   By Snell's law: $\sin a = n \sin r$ since outside medium is air and prism refractive index is $n=\sqrt{2}$.

   For the minimum value of $a=45^\circ$ at which total internal reflection just begins at face $PR$, the ray inside must strike face $PR$ at the critical angle.

3. Critical angle for prism-air interface

   $\sin C = \frac{1}{n} = \frac{1}{\sqrt{2}}$ $C = 45^\circ$

   So at the limiting condition, angle of incidence on face $PR$ inside prism is $i_{PR} = 45^\circ$

4. Find $r$ using Snell's law at the given minimum incidence

   Given $a=45^\circ$, $\sin 45^\circ = \sqrt{2}\sin r$ $\frac{1}{\sqrt{2}} = \sqrt{2}\sin r$ $\sin r = \frac{1}{2}$ $r = 30^\circ$

5. Relate prism angle and incidence at face $PR$

   The normals to faces $PQ$ and $PR$ make an angle equal to the angle between the faces themselves, i.e. angle at $P$: $\angle P = 90^\circ - q$

   Therefore, for the ray inside prism, $r + i_{PR} = \angle P = 90^\circ - q$

   Substituting $r=30^\circ$ and $i_{PR}=45^\circ$, $30^\circ + 45^\circ = 90^\circ - q$ $75^\circ = 90^\circ - q$ $q = 15^\circ$

6. Check options

   • A: $15^\circ$ ✅
   • B: $22.5^\circ$ ❌
   • C: $30^\circ$ ❌
   • D: $45^\circ$ ❌

Therefore, the correct answer is A.