Most of the answers tackle the classical problem described in the first part of the question involving a laser and mirrors. I'm going to focus on the part of the question about the uncertainty principle.
infinite square well
There's a "toy" system that is studied by many introductory quantum mechanics students. The system is a single quantum particle trapped in a rigid 1D box. The potential energy of the particle is described as an infinite square well.
The position wavefunction of the particle is exactly zero outside of the box, so the particle remains trapped. Photons are massless, but for the time being, lets imagine a massive particle in the box. The energy eigenstates of the system are
$$ E_n = n^2 \frac{\pi^2\hbar^2}{2m L^2},$$
where $m$ is the mass of the particle and $L$ is the length of the box. In the minimum energy (or ground) state $n=1$, and $E_1 \ne 0$. The particle cannot stop moving. It must have some energy.
In an energy eigenstate the particle's energy is exactly known, but its position is in a mixture state. The position wavefunction is non-zero in many places inside of the box. We could approximate the position uncertainty, the width of the wavefunction, as the size of the box:
$$\Delta x \approx L$$
(It's possible to exactly calculate $\Delta x$, and the Wikipedia article linked above works it out)
If the particle is in a definite energy state, then its energy is known exactly. Since $E = \frac{p^2}{2m}$ it might seem that its momentum must be known exactly too. But because momentum is squared in the equation, its momentum can be either positive or negative. The particle could be moving to the left or right. In the ground state
$$ p = \pm \sqrt{2m E_1} = \pm \frac{\pi\hbar}{L}. $$
We can approximate the momentum uncertainty as
$$\Delta p \approx \frac{\pi\hbar}{L}$$
(again, the Wikipedia article works it out exactly)
Using our approximations:
$$\Delta x\, \Delta p \approx \pi \hbar \ge \frac{\hbar}{2}$$
The size of the box $L$ cancelled out. It doesn't matter how small the box is the uncertainty relation holds.
In the energy equation we can see that if we decrease $L$, the minimum energy increases. This in turn increases the momentum uncertainty. When we decrease the position uncertainty the momentum uncertainty grows to compensate.
As others point out, there is an energy cost to shrinking the box. To decrease $L$, we must increase the energy of the particle. As $L$ gets smaller and smaller, the energy cost blows up. It would take an infinite amount of energy to decrease $L$ to zero. We just can't do it.
a photon in a box
The same idea holds for a photon.
The infinite square well is a perfectly reflective box.
The photon wavefunction is a standing wave for the energy eigenstates, so $L = \frac{n}{2}\lambda_n$.
The energy eigenstates are
$$E_n = \hbar \omega_n = \frac{2\pi\hbar c}{\lambda_n} = n \frac{\pi \hbar c}{L}$$
We use the relativistic relation between the energy and momentum of a particle. For a massless photon, we see
$$E^2 = p^2 c^2 + m^2 c^4 \quad\implies\quad E = \pm pc.$$
We have the same direction ambiguity in the momentum associated with an energy eigenstate of the photon.
In fact using our simple approximation, we get the same
$$\Delta x \approx L \quad\text{and}\quad \Delta p \approx \frac{\pi\hbar}{L}$$
in the ground state!