The light rays get "straighter", closer to the normal direction of the boundary plane, when they travel from lower $n$ (water, $1.3$) to higher $n$ (glass, $1.5$).
However, the photons sent by the objects you see in the reflection are also reaching the glass-air boundary, as John Rennie pointed, out, and if they continued, they would travel from a higher $n$ (glass, $1.5$) to a lower $n$ (air, $1$). For angles (between the normal direction and the light ray's direction) greater than the critical angle
$$\theta_c=\arcsin(1.00/1.50) = 41.8^\circ$$
the photons can't get through and they will reflect from the glass-air boundary back to the glass and some of them will make it to the water (there may also be a total internal reflection when going from glass to water! the critical angle is $\arcsin(1.3/1.5)=60.1^\circ$). These photons that get through the water back to your eye (the water-air critical angle is $\arcsin(1/1.3)=50.3^\circ$) will communicate the reflected image of the object that emitted the light we were tracing.
The last interface, the water-to-air transition, is also the reason why you don't see through the sides of the glass: the photons going from those directions (that would allow you to see through the side walls) reflect (down) back to the water from the horizontal surface of the water by TIR.
So what’s to prevent you from describing a laser that's arbitrarily tight, at some arbitrary distance, and learning that the required angle is very close to 0?
The half angle of divergence is given by
$$\theta = \frac{\lambda}{\pi w_0}$$
where $w_0$ is the beam diameter at its narrowest point (the waist, or focal point), and $\lambda$ is the optical wavelength.
Typically with a laser the waist point is at the output aperture of the laser cavity, and the beam diverges from there. If you built your laser with a converging output, you'd push the waist point out along the z direction (the direction of propagation) but you'd also reduce the waist diameter, so ultimately increase the divergence angle.
So you can't choose to produce an arbitrarily small divergence angle unless you're prepared to build a laser with an arbitrarily large output aperture.
To use some concrete numbers, suppose the target is a lightsail 1 light year away, and it is 1 Mm in diameter.
1 light year is about $10^{16}$ meters. So you need a divergence angle on the order of $10^6 / 10^{16}$, or $10^{-10}$ radians. You need a beam waist of
$$ w_0 > \frac{\lambda}{\pi \theta} $$
If your wavelength is 500 nm, this means a waist of at least 1600 m. In practice I expect there would be "unique engineering challenges" in designing optics close enough to ideal to achieve this kind of divergence. I've never heard of beam divergence being measured in units smaller than milliradians, but I don't know what's been achieved in hero experiments.
Best Answer
Yes. Focus the radiant heat from a thermal reservoir onto a spot that is hypothesized to be raised to a higher temperature through its concentration into a smaller area. Now connect heat engine - a Carnot engine - between the hot spot as the engine's heat intake and the original reservoir as the heat exhaust. Now the engine will run, outputting work. Your hypothesis means that you have an heat engine system spontaneously converting the heat in the thermal reservoir to work and there's your perpetual motion machine (of the so-called second kind).
Obligatory in any conversation of this kind is Randal Munroe's Fire From Moonlight article.
One way to understand all this is to note that optical systems are reversible, so that if light can pass from point A at the input to point B at the output, light can equally well go the other way. So if a hot body directs its radiant heat at another object through a lens system, the temperature of the latter will naturally begin to rise. That means that the second body will radiate back to towards the first body. If the second body became hotter than the first, it would be returning a higher heat power to the first along the reverse paths whence the incident heat came. Therefore, heat transfer will stop before the second body reaches the temperature of the first.
The second law of thermodynamics in optics is equivalent to the non-decreasing of étendue, which is the volume of a system of rays representing a light field in optical phase space and thus a measure of entropy. If étendue cannot be decreased, this means that density of rays in phase space cannot be increased; in turn this means that the divergence angles of a set of rays must increase if the area they pass through is shrunken down. This means that the light from any point on a hot body cannot be made brighter at the point where it reaches the target body.
This also is why a laser works differently if we try to reason as above. If energy reaches a body through a laser, the incident light paths taken have near to zero étendue - there's hardly any beam spreading at all. The second body will get hotter and hotter, but the radiant heat from the hot second body is all spread out in all directions (this is fundamental to blackbody radiation - there's no such thing as collimated blackbody radiation). So hardly any of the radiated light is accepted back along the extremely narrow range of paths back to the laser. Laser light is highly nonequilibrium light - it is the optical equivalent of thermodynamic work, rather than heat.
As well as by thermodynamic arguments, one can show that étendue is conserved very generally in passive optical systems using the Hamiltonian / symplectic geometry formulation of Fermat's principle. I discuss this in more detail in this answer here. Fermat's principle means that propagation through inhomogeneous mediums wherein the refractive index (whether the material be isotropic or otherwise) varies smoothly with position corresponds to Hamiltonian flows in optical phase space; mirrors, lenses and other "abrupt" transformations as well as smooth Hamiltonian flows can all be shown to impart symplectomorphisms on the state of the light in phase space, which means that they conserve certain differential forms, including the volume form. All these things mean that the volume of any system of rays in optical phase space is always conserved when the rays are transformed by these systems. This is the celebrated Liouville Theorem.
There is a clunkier but more perhaps accessible way to understand all this in optics. We linearize a system's behavior about any reference ray through the system, and write matrices that describe the linear transformation of all building block optical systems. It may seem that linearization involves approximation and thus something not generally true, but hold off with this thought - this is not the case. This is the Ray Transfer Matrix method and these linear transformations describe the action of the system on rays that are near to the reference (the "chief ray") ray of the light field in optical phase space. These matrices act on the state $X$ of a ray at the input plane of an optical subsystem:
$$X = \left(\begin{array}{c}x\\y\\n\,\gamma_x\\n\,\gamma_y\end{array}\right)\tag{1}$$
where $(x,\,y)$ is the position in the input plane of the ray, $(\gamma_x,\,\gamma_y)$ are the $x$ and $y$ components of the direction cosines of the ray's direction and $n$ is the refractive index at the input plane at the reference ray's position. The quantities $n\,\gamma_x$ and $n\,\gamma_y$ are the optical momentums conjugate (in the sense of Hamiltonian mechanics) to the positions $x$ and $y$; interestingly, they are indeed equivalent (modulo scaling by the constant $\hbar\,\omega/c$) to the $x$ and $y$ components of the photonic momentum $\hbar\,\vec{k}$, where $\vec{k}$ is the wavevector, but this fact is an aside. (1) describes our points in optical phase space.
Now we write down the matrices that represent the linearized action of every optical component we can think of; for example, a thin lens (representing the paraxial behavior of an optical surface) will impart the matrix:
$$\left(\begin{array}{cccc}1&0&0&0\\0&1&0&0\\-\frac{1}{f}&0&1&0\\0&-\frac{1}{f}&0&1\end{array}\right)$$
If you study this matrix's action, you'll see that it transforms a collimated beam into one that converges to a point a distance $f$ from the input plane.
A key point to take heed of is that this matrix has a determinant of 1. If you go through the list of every possible passive optical component, you'll find that the matrices that describe their paraxial behavior all have unity determinant (they are unimodular). So they all multiply together to give a unimodular ray transfer matrix of the overall system built from these subsystems chained together.
This determinant is the Jacobian of the general, non-linearized, non approximate transformation that the system imparts on any system of rays. We can imagine recalculating a matrix from every neighborhood of every chief ray in an arbitrary, noninfinitessimal volume of rays in phase space. These matrices will all be unimodular, so what we've shown is the key idea:
This means that if we work out the volume $\int\mathrm{d}V$ of a system of rays in phase space, then the volume of their images $\int\,J(X)\,\mathrm{d}V$ will be exactly the same for any passive optical component. So we've shown the exact version of the law of conservation of étendue for optics without needing the full machinery of symplectic geometry and Hamiltonian mechanics.