Hi
Quote:
Originally Posted by bgilbert
G'day all.
Why do we observe 6 diffraction spikes in a Newtonian with 3 support vanes to support the secondary mirror, and only 4 spikes if we add another vane, making 4 vanes ? Assume all vanes are equally spaced.
Barry
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Hi Barry,
I believe a clue to the answer is to be found in looking at / induction from the diffraction spike pattern of 1 and 2 vane spiders and comparing with what is observed in 3 and 4 vane spiders..... And really questioning what one is seeing with the 4 vane spider....is it 4 spikes or 2 sets of four spikes superimposed on top of one another?
What you are really observing is diffraction of light around the edges of the vane, this creates an interference pattern / diffraction pattern which is perpendicular to the edge/s so that ...
No vanes
will simply have a round (airy disk) diffraction pattern around a star
1 vane
will have round (airy disk) pattern around the star & a spike either side of the star like a line passing through it, which may appear as a dotted/dashed line due to interference or possibly not so dashed depending on how well the scope is collimated and focused
2 vanes - in line
will have round (airy disk) pattern around the star & a spike either side of the star like a line passing through it, which may appear as a dotted/dashed line due to interference and slightly more pronounced than the 1 vane pattern
2 vanes - at 90degrees to each other
This is the kicker in terms of understanding....
will have round (airy disk) pattern around the star & a 4 sided spike evenly distributed around the star. All else being equal it will appear to be a slightly more pronounced pattern than the 2 vane spider with vanes at 90 degrees, because there are more edges (of vanes) in the way of the light and hence more diffraction patterns superimposed.
3 vanes - at 120degrees to each other
will have round (airy disk) pattern around the star & a 6 sided spike evenly distributed around the star.
Each of the vanes will create a diffraction pattern (line) perpendicular to it. Looking at the OTA end-on we have 3 vanes: one at say 0 degrees (TOP), another at 120 degrees and another at 240 degrees around its circumference. For simplicity say we have a star at the dead centre of the OTA, then
taking Vane 1 (at 0degrees on the OTA circumference) it creates a diffraction pattern (dashed line) perpendicular to it which passes through the star and therefore goes towards 90 degrees on one side and 270 degrees on the other side of THE IMAGE.)
taking Vane 2 (at 120degrees on the OTA circumference) it creates a diffraction pattern (dashed line) perpendicular to it which passes through the star and therefore goes towards 30 degrees on one side and 210 degrees on the other side of THE IMAGE.)
taking Vane 3 (at 240degrees on the OTA circumference) it creates a diffraction pattern (dashed line) perpendicular to it which passes through the star and therefore goes towards 150 degrees on one side and 330 degrees on the other side of THE IMAGE.)
This results in a 6 pointed star shaped diffraction spike pattern. With spikes at 30, 90, 150, 210, 270 and 330 degrees being seen IN THE IMAGE relative to the position of Vane 1 ON THE OTA, measuring the angle clockwise.
4 vanes - at 90degrees to each other
will have round (airy disk) pattern around the star & a 4 sided spike evenly distributed around the star. It will appear to be a more pronounced pattern than the 2 vane spider, because there are more edges (of vanes) in the way of the light and hence more diffraction patterns superimposed. The reasoning behind there being 4 spikes has to do with the diffraction pattern of any one vane being at perpendicular to it and culminates in 4 diffraction spikes only at 0, 90, 180 and 270 degrees in a similar reasoning to that used in considering the 3 vane spider, detailed above, except that the patterns can be thought of as being superimposed on one another and hence only showing up at 4 locations.
5 vanes (evenly distributed @72 degrees)
10 pointed spike pattern
and so on..... OR NOT
.....(kidding)
A drawing may have been easier, but it needed some explanation (at least to me)
Hopefully no typos or worse.
Best
JA