Veil + Bright Doughnut

[Ian]

thing is, a dust bunny is a different thing to an out of focus star image. You didn't answer the question though. Is there a piece of planar glass 2.625mm from the CCD (give or take)?

[Mac]

how thick was the glass that you replaced the IR filter with, if thats 2.5ish mm then that would give you your 5mm ish distance,

the relection being from the top of the ccd, relfecting back off of the internal top side of of the replacement glasscover.


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total distance ~5mm

Mac

[MarkS]

Ian, Mac,

Sorry - I didn't answer the question.  The glass is approx 3mm thick and it may be about 2.625mm from the CCD (give or take).

But how does that create an out of focus star image?

Mark

[Ian]

Actually I got my sum wrong. The image would be 9mm away from the focus to be that size, or 4.5mm assuming it's a single reflection. If I hadn't got my sums wrong I'd have said Mac hit the nail on the head. Thinking about it, there the possibility that we're looking for two planar surfaces 4.5mm apart, one of may or may not be the sensor. But since the sensor cover glass is unlikely to be coated that'd be a prime candidate, together with the fact that the image is very nearly at focus and very bright (comparatively) as it reaches the sensor cover glass making the reflection more visible.

It's an out of focus star image because it's formed by light reflecting off on a surface and then reflecting back onto the ccd. because of the additional distance the light has travelled, by the time it finally gets to the ccd it's no longer in focus.

A dust bunny is a diffraction pattern caused by a mote of dust and it's size is related to the distance from the imaging plane and the size and shape of the mote itself, the fact it looks a bit like a doughnut or out of focus star image is coincidence. Refractors also show doughnut dust bunnies, but an internal reflection wouldn't look like a doughnut.

[Mike]

Some lenses are made up of sandwiches of different pices of glass. Could it be that?

[Ian]

it could be, but I wouldn't look there at the moment for two reasons.

To keep things sensible in the thinking department, I would keep to planar surfaces as reflections from non-planar ones are much harder to calculate (and impossible for us for all intents and purposes as we would need to know the radius of curvature).

In a lens there are rarely planar surfaces, let alone two parallel ones, and secondly in good lens construction the designer would have gone to some lengths to avoid internal reflections using cements, placement and coatings.

[MarkS]

Ian,

I still disagree with your 9mm I think it should be 15mm.  F6.3 means that the cone of light will be a particular shape: in this case the height is 6.3 times the width.  The doughnut is 2.4 mm wide so the "height" is 6.3 times this i.e. 15mm.

So 15mm is the extra path length if the cause is reflections off planar surfaces. But  I can't see anything that could account for 15mm.

My best guess is the following:  light is being reflected off my new glass "filter" back up to the inside glass face of the focal reducer (2-3 inches away) and then back to the CCD.   The glass face of the focal reducer is non-planar so it could act to refocus the light for its retun journey to the CCD.   Now the focal reducer is anti-reflective coated but maybe, for instance, it is not optimised for infrared.  Under this hypothesis, infrared is coming through the scope and focal reducer; reflected off the planar glass back to the reducer; reflected off the non-planar reducer and back to the CCD where it is not completely in focus. As discussed elsewhere, the RGB pixels are all sensitive to infrared so the doughnut appears whitish.

This hypothesis is easily tested by putting an IR filter between the scope and the reducer.

If the test is positive, the solution is either
1) Put an IR filter in the optical train
or
2) Re-mod my camera with something specifically designed for the purpose i.e. not cheap, uncoated glass!!

Mark

[Ian]

Mark, the optical system is not f6.3. It's f10 for about 2.5m and then gets significantly shortened after the FR to give an apparent f6.3. The geometry of the light cone is not that straightforward.

Did you take a look at that link I posted? The subtended angle of the light cone after the focal reducer is tan-1(1/(2NR)) where N is the natural focal ratio of the scope (f10) and R is the reducers focal multiplier (0.63).

Substituting: angle=tan-1(1.2/x) where 1.2 is the radius of the out of focus star image and x is the distance from the focal point gives an increased light path of 9mm and as it'd folded back on itself (coz you can see it) the light is traversing 4.5mm twice before finally reaching the sensor.

I got the sum wrong the first time because I used diameter, not radius in the second expression.

[Fay]

My point exactly!     

[RobertM]

Mark, I agree with you.  To me it looks like a magnification of an out of focus star image which could only occur from a curved optical surface.  The doghnut moving non linearly with the star image backs up that hypothesis.  The only curved optical surface is the Reducer Corrector and if that was the case then the size of the reflection would depend on the accuracy of focus rather than being something that could easily be calculated i.e. dust bunnie.  This could easily be proven by changing the focus slightly between images.

[Ian]

Actually Robert that's a good point. We can work out if the reflection is off of a planar surface by looking at the way it moves across the image plane in relation to the star image. It would be superimposed if the star is on the optical axis and then as the star moves away the reflection will be 2x as far from the axis but in the same direction. If the reflection moves in a non-linear fashion, it'll be off of a curved surface. Then all bets are off regarding distance from the imaging plane.

Obviously bear in mind the optical centre of the telescope may not correspond with the centre of the sensor, but you could work that out too

[MarkS]

Ian,

I still don't understand where 9mm comes from.   

Let me work from your own equations:
(1)  angle = arctan(1/(2NR)) = arctan(1/(2*6.3))
(2)  angle = arctan(1.2/x)

setting (1) equal to (2):
arctan(1.2/x) = arctan(1/(2*6.3))

Hence,
1.2/x = 1/(2*6.3)

So,
x = 1.2*2*6.3 = 2.4*6.3 = approx 15mm !!!

So your complicated method gives the same answer as my simple method.

So, yes, the geometry of light cones really are that straightforward.

[Ian]

yep, you're right, I did the substitution wrong. 

We've successfully proved that the addition of a focal reducer changes the geometry of the light cone to be effectively the same as an equivalent scope of the modified focal length.

And that's why I don't earn my living doing sums. I'd be skint. 

However, we've still got the question of what is adding 15mm to the light path and that would be two surfaces 7.5mm (check my maths on that too) apart. What do you think about my postulation regarding the curvature or not of the surfaces?

[Mike]

OK this thread went way over my head a long time ago 

[Mac]

Sorry - I didn't answer the question.  The glass is approx 3mm thick and it may be about 2.625mm from the CCD (give or take).

amended ray diagram


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depending on the light path the reflection image will be either 6mm or 12mm
red incoming
green first reflection
blue second reflection.

if your glass is .5mm thicker and the distance between ccd and glass is .5mm difference that would allow
you the calculated 14mm
as each light path reflection is extended by 7mm.

The front of the CCD / glass is probably in the region of about 25mm from the back of your focal reducer, which means the reflected light path
would be in the order of 50mm, so you could probably rule this out as the source of your reflection.

Mac.