0
$\begingroup$

diagram:

enter image description here

Assuming

1) the scanner can angle a beam anywhere on the circular off-axis mirror and

2) the off-axis mirror can be rotated as an the diagram to any angle needed,

can the beam shooting at any point on it from the scanner be made parallel to the middle beam bounced off of the off-axis mirror?

$\endgroup$
2
0
$\begingroup$

No. The final mirror on the scanner has a smaller surface area than the 'rotating off-axis mirror'. Consequently, for a beam of light to hit near the edge of the larger mirror, it must be diverging from the centreline connecting the centre of those two mirrors. As the light paths are diverging, they will not be parallel after the off-axis mirror, unless that mirror is not flat.

$\endgroup$
4
  • $\begingroup$ I need the beam orientation to be parallel, not the beams to be parallel in the sense of not diverging into each other. Sorry for being unclear. $\endgroup$ Apr 4 '18 at 14:49
  • $\begingroup$ But I don't understand your point "The final mirror on the scanner has a smaller surface area than the 'rotating off-axis mirror'" $\endgroup$ Apr 4 '18 at 14:50
  • $\begingroup$ I interpreted your question to mean "can the beam coming from the mirror (from any point - i.e. the point does not matter) be made parallel to the beam that is reflected from the centre of the mirror? Because the light is not collimated, the answer to that question is "No", If that's not the question you meant to ask, then please edit your question (and diagram) to be clearer. $\endgroup$ Apr 4 '18 at 22:19
  • $\begingroup$ collimation has nothing to do with whether a beam bouncing off of a flat mirror will be parallel or not to another beam bounced off at a different point. This is literally like adding a third very large scanner mirror to the setup. it's a tight beam but not a collimated laser-like beam with <1 mrad divergence. $\endgroup$ Apr 4 '18 at 22:21
0
$\begingroup$

Yes, but...

enter image description here

As you see, the outgoing part of the yellow beam is parallel to the reflected part of the red beam, just as you wished. But for that to happen, the galvo must be located in such a way that it will obscure a fragment of the reflected field - the green beam, reflected from the mirror, will strike the structure of the galvo, and the reflected image will have a big black blob of shadow of the galvo somewhere off to the side.

$\endgroup$
4
  • $\begingroup$ not sure I understand why $\endgroup$ Apr 4 '18 at 14:52
  • $\begingroup$ @MarkLegault: Angle of reflection is angle is equal to angle of incidence. If ray 1 (red) falls at angle x, reflected at same x, so the reflected part is at angle 2x from the incident part. Cast a ray 2 (yellow), at such an angle of incidence that it's equal to angle of reflection of the first. Its reflected part will be at -2x to the incident part. That means, in between them there exists a ray that falls - and is reflected - at angle zero, returning to the origin - hitting the galvo. $\endgroup$
    – SF.
    Apr 4 '18 at 15:04
  • $\begingroup$ But why is in your illustraton the galvo at the height or in front of the off axis mirror and not out of its way like in mine? Not that my illustration is in any way good (liek the angle off the off-axis mirror) $\endgroup$ Apr 4 '18 at 16:25
  • $\begingroup$ @MarkLegault: because the ray 2 would miss the mirror, simple. If you move it so that no normal to the mirror lies in line, it's not possible, period. $\endgroup$
    – SF.
    Apr 6 '18 at 21:03
0
$\begingroup$

This is a basic physics and math problem

and your very initial idea for solving it is wrong!

enter image description here

Light waves travel linearly. When they hit a reflective surface, they change direction according to entry-angle is equal to exit-angle, a shown above. The green-marked path hits the center. Unless the mirror is a curve or we actually use a laser, the outer bounds of the ray will be diverging - the Fisheye effect, shown with the red and yellow paths.

Now, let'S for simplicity assume we have a laser. by twisting the mirror by -2° on the red path and +2° on the yellow we get the same picture of divergent paths.

enter image description here

To keep the rays parallel we may not adjust the angle, but only the distance of our initial mirror: to hit more on the far side of our rotational mirror and get a stream that is parallel to the one when we hit the center, we need to move further away (red), to hit closer we need to move it shorter (yellow).

enter image description here

Instead of varying the distance, the mirror could also move on a curved path to create the shifted parallel rays needed:

enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.