When you collimate a point source using an off-axis parabolic mirror (OAP) with a circular shape, the beam area of the collimated light becomes more and more elliptical (x-dir. is smaller than y-dir.) as the off-axis angle is increased. Is there a reason why this happens? And is there a way to quantify this? My best guess is that it should be proportional to the cosine of the angle, but I can't see why it's physically happening. Shouldn't every point on the mirror collimate a cone of small radius from the point source? Here is an image that gives a better idea of what's happening - note that the light source P is at the focal point of the mirror.

enter image description here

  • $\begingroup$ While your diagram suggest that there is a parallel wavefront emerging from the elliptical mirror, that's not the case for off-axis systems. There is a detailed discussion of optical effects in physics.stackexchange.com/questions/74894/…, which may or may not help. My guess is, that the effects are second order, so for a few degrees it wouldn't matter, but the errors will be huge for large angles. If your mirror shape is not elliptical, then the beam will be elliptical. If you don't like that, you can always add an aperture. $\endgroup$ – CuriousOne Oct 11 '14 at 1:14
  • $\begingroup$ @CuriousOne It looks like he briefly mentions the off-axis orientation in his answer to the question that you linked. He mentions that the 1/sqrt(2) factor comes out because the beam spreads out farther in the x-direction than the y-direction. I'm still not clear as to how this affects the output collimated beam. Shouldn't each point on the circular mirror collimate light in exactly the same fashion? I'm trying to size out an off-axis parabolic mirror, so I need to determine the effective size of the beam that is being reflected from the mirror. $\endgroup$ – Kimusubi Oct 11 '14 at 22:06
  • $\begingroup$ It seems to me, that the shape and size of the the mirror should be given by the projection of the parallel beam (which you would like to be round, I suppose?) on the mirror plane (is that the plane perpendicular to the reflected center beam?). That may be an approximation, though. $\endgroup$ – CuriousOne Oct 12 '14 at 1:01

Your Answer

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

Browse other questions tagged or ask your own question.