effect of distance 2

Derek Robson

I slept on that  one and awoke with an optics headache, hmmmm.   why) the distance along the laser beam changes the measured dispersion”.  I’m not sure now.  I was thinking about what is it that we are seeing when looking at a LASER beam. I have a vague memory that you can’t or shouldnt be able to see them, except if the beam is blocked by something.  I think that the visible beams are due to reflections from illuminated particles in the air.  Am I right in assuming that the first order “beams” that we see, are not really there in the air (unlike the pointer beam)? That is, the first order beams are constructed from an infinitesimal number of diffractions from the same number of sources of illumination along the primary beam going outwards to the chimney pot? And the light is diffracted after the grating, so doesn’t happen in the air outside, but rather, in the camera?  On the parallelism, I started to think what the LASER beam would look like in the following two situations a) the beam is normal to and horizontal to the grating and sensor. I’d expect to see a dot on left and on right side, with dispersion distance “A”. But if we tilt the LASER to say 45 degrees upward (pointing to a roof top), how would the diffracted dots appear in the spectrum?  (the LASER would be hitting the grating at 45 degrees tilt.  Would the dots be dispersed by a distance B which was smaller or larger than A? or should A = B always (whatever the distance; but might differ with tilt)?  Is that what you meant?  It has me baffled.  Could geometry solve it?