Step 1: Picture the prism. The prism is right-angled with two equal sides ($PQ = QR$). The two equal base angles are each $45^\circ$ and the angle at the top is $90^\circ$. The light hits one face, then strikes the slanted face.
Step 2: Identify the key event. At the second face the ray turns back inside the prism instead of leaving. This means total internal reflection is happening there.
Step 3: Recall the rule for total internal reflection. Total internal reflection happens only when the ray hits the face at an angle equal to or greater than the critical angle $C$. Here the ray meets the face at $45^\circ$, so we need $45^\circ \geq C$.
Step 4: Connect critical angle to refractive index. The critical angle is linked to the refractive index by $\sin C = \dfrac{1}{\mu}$.
Step 5: Apply the limit. Since $C$ must be at most $45^\circ$, we have $\sin C \leq \sin 45^\circ = \dfrac{1}{\sqrt{2}}$. So $\dfrac{1}{\mu} \leq \dfrac{1}{\sqrt{2}}$.
Step 6: Solve for the smallest index. Flipping the inequality gives $\mu \geq \sqrt{2}$. The smallest refractive index that still gives total internal reflection is therefore $\sqrt{2}$. \[ \boxed{\sqrt{2}} \]
