mg0shtisha, no I do not think that is correct. Feynman is attempting to explain his path integral theory for QED. In path integrals, each path contributes the same amount to the probability density for the wave function. So each path that he calculates always has the same magnitude. What changes is the phase of the path. This phase is represented by his stopwatch, progressing through the 360 degrees of phase. In a path integral, the phase component is directly proportional to the action of the desired path, that is it is the time integral over time of the path's Lagrangian.
In classical physics, the path of motion for a particle in a given system is given by the stationary path of the action. About the stationary point, there will be very little change in the action. This means that the phase around the classical path of Feynman's path integral will vary slowly. This is shown by his stopwatch moving slowly near the classical path. It also means that the paths around the classical path contribute the most to the path integral since the phases do not cancel out.
Away from the classical path, the action is highly oscillatory. This means that the phases of paths that are very close to each other will cancel out. Feynman talks about his stopwatch being at 12 o'clock and then suddenly going to 6 o'clock resulting in a net contribution of zero. So the adding of the vectors from the stopwatch, going from tip to tail, would be another way of adding up the phase contributions of various paths. If the phase contributions cancel out, then the addition of all the stopwatch vectors would end up back at the tail of the first phase vector. This is just a vectoral representation of the addition of phase contributions.
This is very much like using the techniques of steepest descent or stationary phase when integrating an integral. If we have an integral that is highly oscillatory in phase, but constant in magnitude, then the points where the oscillation in the phase is the slowest will have the greatest contributions to the final result. We can get a good estimate of the integral by simply integrating around the stationary points as opposed to the entire length of the integral. With a path integral, we need to consider every path that can be made from point A to point B. Obviously, this is a monumental task to integrate over for even the simplest of situations. However, having knowledge about the stationary points, the classical path that is, allows us to truncate the paths that we need to consider since we know paths that deviate far from the classical path will be highly oscillatory and thus will not contribute greatly to the final path integral.
In regards to the OP:
The path for the reflection is the classical path of that problem. If we said that light goes through the minimal path length, then we would not get any reflected path, light would only go from point A to point B directly. A reflected path comes about due to Fermat's Principle:
http://en.wikipedia.org/wiki/Fermat's_principle . This principle allows for more paths than just the direct path to be valid and, as expected, is the same as Feynman's path integral in the classical limit. Fermat's principle will give you the classical paths that I have mentioned, so Feynman's path integral will naturally pick out the same classical paths as the main paths that contribute to the wave function. The proof of this would be to work out the Lagrangian and find the stationary paths of the action from the Lagrangian. This is purely a mathematical exercise.
It has been a while since I read QED but I do not recall any real variation in the length of the hands. I would expect them to always be the same length since the contribution from each path is the same, only the phase (represented by the direction of the hands) is different between the paths.
EDIT: I think this would obviously be more appropriately situated in the Quantum forum.
