Heading over to [PhET's Bending Light Simulator][1], you can try out this experiment yourself. As it turns out, you cannot get the light to be infinitely contained in the glass if you shine light from the outside. But it is possible if the light source is within the glass, as is shown below.

[![light not trapped if source outside circular glass][2]][2]
[![light trapped if source within circular glass][3]][3]

But why is this true?

To answer this question, let us first try to find out the necessary conditions for the light to have an angle of incidence equal to the critical angle at its second incidence (when it goes from glass to air).

[I’ll be using “the first incidence of light” to refer to when the light goes from the air to the glass,  and “the second incidence of light” to refer to when it goes from the glass to the air.]

Have a look at the following image.

[![enter image description here][4]][4]

First you must understand that the normals drawn at points B and C have the radii of the circle as their part, i.e., if extended, they intersect at the centre always. Why is this true? Because the normals are by definition perpendicular to the tangents to the circular glass drawn at those points — B and C. But it is known that the radius of a circle is the line segment that is perpendicular to a tangent drawn at any point on the circumference. Therefore, the normals must be extensions of the radii of the circle.

If you didn’t get the above part, you may ignore it; just know that OB and OC are radii of the circle, hence they are equal. This would imply that triangle BOC is isosceles, and so ∠OBC = ∠OCB.

Now we know by Snell’s law that $\sin{i} \propto \sin{r}$. In other words, more the angle of incidence, more the angle of refraction, and vice versa, because the sine function increases from $0°$ to $90°$, and those are the only angles we are considering here. Thus, if we look at the first incidence, to maximise $r_{1}$ (∠OBC), we need to maximise $i_{1}$ (∠EBA).

But $i_{1}$ cannot be any bigger than $90°$, because if it were greater, the light would basically have its source within the glass, and we know that the light does get infinitely contained if its source is in the glass.

Now what is the angle of refraction $r_{1}$ corresponding to a $90°$ $i_{1}$? Let’s find that out using Snell’s law! Assuming $1.5$ to be the refractive index ($\mu$) of glass,

\begin{align}
& \frac{sin(i)}{sin(r)} = \mu \\
& \implies \frac{sin(90°)}{sin(r_{1})} = 1.5 \\
& \implies \frac{1}{sin(r_{1})} = 1.5 \\
& \implies r_{1} = sin^{-1}(1/1.5) \\
& \implies r_{1} \approx 41.8103149°
\end{align}

[Which is exactly equal to the critical angle for glass and air, in accordance with the principle of reversibility of light.]

$\therefore ∠OBC = ∠OCB \approx 41.81°$

But this wasn’t our true objective was it? We wanted the light to get totally internally reflected, but for that, $r_{2}$ must be greater than $90°$. Hence $i_{2}$ must be greater than $41.81°$ (because $i \propto r$, by Snell's law), which implies that $r_{1}$ must be greater than $41.81°$ (because $r_{1}$ = $r_{2}$), which means that $i_{1}$ must be greater than $90°$ (again, because $i \propto r$). That is impossible, unless the source be in the glass itself, thus proving our observations.


  [1]: https://phet.colorado.edu/sims/html/bending-light/latest/bending-light_all.html
  [2]: https://i.sstatic.net/ef0cy.jpg
  [3]: https://i.sstatic.net/b5xER.jpg
  [4]: https://i.sstatic.net/s9F2P.jpg