Actually, I think the point is that total internal reflection can't occur in this situation.
Picture the object as a block of acrylic (for example) submerged in a fluid. If the object is visible from above the fluid, then there must be a path for a ray of light to travel via refraction through the fluid, strike the block at some angle ##\theta## (measured relative to the surface normal), and reflect upward back towards the surface. If ##\theta## is too small, then the light will (mostly) refract through the block rather than reflect, rendering the block (mostly) "invisible" at this angle.
So the first thing we want to do is calculate this "critical angle" below which light will tend to refract through the block. Let ##n_1## and ##n_2## denote the indices of refraction of the fluid and block, respectively. If ##\alpha## is the angle at which light is refracted through the block, Snell's Law gives ##n_1 \sin(\theta) = n_2 \sin(\alpha)##, or ##\sin(\alpha) = (n_1 / n_2) \sin(\theta)##--in particular, if ##\theta > \theta_c = \sin^{-1}(n_2 / n_1)##, then no refraction is possible because Snell's Law has no solution for ##\alpha##. (This, by the way, is the condition for total internal reflection that Charles Link alluded to.)
Next, we want to show that any ray of light refracted from above the fluid's surface will strike the block at an angle less than the critical angle ##\theta_c## (again, measured from the surface normal). Suppose we have a ray that strikes the block at angle ##\phi > \theta_c##. Assuming the block's surface is parallel to the fluid's, simple geometry implies that the ray must have been refracted at the same angle ##\phi## through the surface of the fluid. This means that the angle of incidence ##\phi_0## of our hypothetical ray on the fluid must obey the following relation: $$\sin(\phi_0) = n_1 \sin(\phi) > n_1 \sin(\theta_c) = n_2$$ (I'm assuming the ambient environment to be air or vacuum, so that the index of refraction outside the fluid is roughly 1.) If ##n_2 > 1## (and it would be strange indeed for the block to have an index of refraction less than that of air), then this relation is impossible to satisfy.
To summarize: Any ray of light passing from the air above the fluid to the block submerged in the fluid will strike the block at an angle less than ##\theta_c##, precluding the possibility of total reflection from the block. However, this is not quite the end of the story.
Even without total reflection, some light will still be reflected by the surface of a dielectric material--at least up to a second "critical angle" called Brewster's angle. Without going into details, Brewster's angle is given by ##\theta_B = \tan^{-1}(n_{ref}/n_{inc})## for light reflected from a medium with index ##n_{ref}## and traveling through a medium with index ##n_{inc}##. Light incident on the reflecting medium will be totally refracted (i.e., with no reflection) if the angle of incidence is less than ##\theta_B##. So once again, we could ask whether any ray of light can strike the block at an angle greater than Brewster's without violating Snell's Law.
Unfortunately, the answer seems to be yes--it is possible to have ##\theta_B < \theta_c##, and moreover the equation ##n_1 \sin(\theta_B) = \sin(\phi_0)## does seem to have solutions. (I'll let you think about when and why this might be true.) In this case, the block should be visible for a small band of incident angles .
