How is this variation of this ladder paradox resolved? I understand the ladder paradox but am confused by a variation of it where an action is taken to trap an object that is Lorentz-contracted.
The same question was asked here, but to the best of my understanding they don't address the actual question asked:
In the ladder and barn paradox what happens if we leave the doors closed?
The original scenario is confusing to me becuase it invokes limitations of object rigidity to explain why the door can't be closed to trap the lorentz-contracted object in the room. Also the answers in that post don't really seem to answer the question.
To avoid running into the same issue with this question I'll construct a new scenario that could be run in an actual laboratory (albeit an extremely sophisticated one), as opposed to requiring The Flash, barns and extremely rigid ladders.
Here is the scenario / experiment:
A rod travels in the direction of its lengthwise axis at relativistic speed. We set up a crystal and project a beam of light through it to create a spectrum of colours, such that the rod will pass through the red light first, and the violet last. Both of the rod and the wall eclipsed by the rod as it passes by measure the light landing on it. The beam is projected in such a way that all colors of the spectrum reach the path of the rod at the same time from the perspective of a stationary observer in the room.
The light is completely orthogonal to the path of the rod.
We project a pulse of light, such that, from the perspective of the room, the light hits the rod and some red and violet light lands on the sensors behind the rod, as it was shortened by Lorentz contraction due to its speed.
The orange, yellow, green and blue light landed on the rod.
From the rod's perspective, the experiment and spectrum were contracted. From the rod's perspective the light hit the rod at some point in time, but the red and violet light couldn't have passed through in front of and behind the rod, since they were too close. If the spectrum lands centered on the rod, then it will have eclipsed the full spectrum of colours.
What will we see when we retrieve the rod and inspect the sensors of both the rod and the sensors in the room? We can't both see the full spectrum landing on the rod and still have light landing on the sensors in the room...
 A: I think this answer will disappoint you since it doesn't provide much insight into the other question, but with respect to the rest frame of the rod, the light arrives at an oblique angle due to aberration. The cross-sectional width of the rod along the path of the light is narrower than the light beam, so the red and violet light will make it past. It will then proceed toward the wall, which is rapidly receding behind the rod. (The fact that the light must hit this moving target is one way of seeing that its path must be angled in this frame.)
A: I don't think the answer to your question is any different from the answer to the normal ladder paradox.
Your question says:

"The beam is projected in such a way that all colours of the spectrum
reach the path of the rod at the same time from the perspective of a
stationary observer in the room."

But that doesn't mean that all colours will reach at the same time from the perspective of the rod's frame. The Violet will reach the path of the rod before the rod even reaches that point and the red will reach the path of the rod when the rod has already passed that point.
I have here ignored the fact that the colours won't appear the same to the rod's observer as to the observer in the room.
