1. Snell’s Law at Outer Surface

To ensure “all” light enters the liquid, we must consider the extreme case: a ray incident at grazing angle ($90^\circ$) on the outer surface. If this ray enters the liquid, all other rays (with smaller angles of incidence) will also enter.

Let the refractive index of air be 1, glass be $\mu$, and liquid be $\mu_l$. For the grazing ray, the angle of refraction $r_g$ at the outer surface (radius $r_o$) is given by: $$ 1 \cdot \sin(90^\circ) = \mu \sin(r_g) $$ $$ \sin(r_g) = \frac{1}{\mu} $$

2. Geometry of the Concentric Shells

Consider the triangle formed by the centre $O$, the point of entry on the outer surface $A$, and the point of impact on the inner surface $B$. By the Sine Rule in $\Delta OAB$: $$ \frac{r_o}{\sin(180^\circ – i)} = \frac{r}{\sin(r_g)} $$ Where $i$ is the angle of incidence at the inner surface (radius $r$). $$ r_o \sin(r_g) = r \sin(i) $$ Substituting $\sin(r_g) = 1/\mu$: $$ r_o \left(\frac{1}{\mu}\right) = r \sin(i) \implies \sin(i) = \frac{r_o}{\mu r} $$

3. Condition for Entering the Liquid

For light to enter the liquid rather than reflect back (Total Internal Reflection), the angle of incidence $i$ must be less than the critical angle $C$ for the glass-liquid interface. $$ \sin(i) \le \sin(C) $$ The critical angle is defined by: $$ \sin(C) = \frac{\mu_l}{\mu} $$ Substituting our expression for $\sin(i)$: $$ \frac{r_o}{\mu r} \le \frac{\mu_l}{\mu} $$ $$ \frac{r_o}{r} \le \mu_l $$ $$ r \ge \frac{r_o}{\mu_l} $$

Conclusion

The minimum internal radius of the capillary is: $$ r_{min} = \frac{r_o}{\mu_l} $$