Glass and water

So, for a ray of light traveling in water to enter the liquid in the capillary, it must first refract into the glass and then into the liquid. At the water/glass interface, the largest angle of incidence occurs when the ray is tangent to the glass, creating an incident angle of $90^\circ$ .

Using the Law of Refraction: $n_i\sin \theta_i = n_r\sin \theta_r$ we get: $1.33\sin90^\circ = 1.55\sin\theta_2$ which gives $\theta_2 = 59.1^\circ$

Now, at the glass/liquid interface, the liquid has a lower index of refraction, requiring that a ray must have an incident angle less than the critical angle in order to enter the liquid. To calculate this critical angle, we use the Law of Refraction again, while setting the refracted angle to $90^\circ$ :

$1.55\sin \theta_c = 1.45\sin90^\circ$ which gives $\theta_c = 69.3^\circ$

To ensure that all rays impingent upon the glass enter the liquid, we want the refracted ray found above to enter at the critical angle.

From the diagram above, and the triangle $\Delta$ ABC , we can use the Law of Sines $\frac{a}{sinA} = \frac{b}{sinB}$ to find r:

$\frac{r}{sin59.1^\circ} = \frac{2.5 mm}{sin(180^\circ - 69.3\circ)}$ which gives r = 2.29 mm