Focal Ratio vs aperture size.

[narrowbandpaul]

Hi gang.

It was mentioned a while ago that it was not clear whether focal ratio or aperture was more important for viewing/imaging deep sky objects.

I managed to derive an eqaution for the flux per square arcsecond, that clearly shows only the focal ratio is important.

The flux per sq arcsec is the surface brightness. Using the equation derived in my pdf, one can look at your image and work backwards to obtain a decent estimate for the brightness of the nebula.

results rom IC1795 show a Ha surface brightness of about 3x10^-17 W/m2/arcsec^2

further proved was the strength of the S[iI] line increases in area's of compression. This result was previously thought.

Enjoy the derivation, its actually very straightforward.....no integrals, I promise.

paul

Nebula Surface Brightness.pdf

[ngc2403]

nice long proof of something you were taught in first year astronomy

perhaps prof paul could share some more first year astronomy with us sometime soon?

[narrowbandpaul]

what like ray diagrams for a completely out of focus sun image?

[ngc2403]

yes please,

you know perhaps i should add the pdf of shot noise and SNR? maybe it 's too soon

[roundycat]

Paul, have a look at this, seems Stan Moore has the opposite view, focal ratio is irrelevant. f-ratio myth

Dennis

[narrowbandpaul]

i had a look, but im not convinced. the above derivation is quite simple and elegant, and physics likes that sort of thing.

thanks for that though. keeps me honest :-)

paul

[Kaptain Klevtsov]

It seems Stan also fails to overlook the fact that the two sample images can't possibly be at the same scale, or have the same processing and so aren't comparable in any useful way.

As NBPaul has shown above, the 'scope collects light shining at us at a (relatively) constant rate in photons per arc second square of sky. This light lands in the 'scope or misses, so a bigger 'scope catches more. The bigger the area at the front of the 'scope, the more photons go in it. That bit seems obvious to me at least.

Then what the 'scope does, is focus these photons onto the CCD sensor at an image scale related to the focal length of the 'scope. A longer focal length 'scope will spread the photons more thinly so that the CCD might get the light from one, several, or several hundred of those square arcseconds of sky. If the focal length is long, only a small amount of sky corresponds to the area of the CCD sensor, what we call the field of view, so only those photons coming from that bit of sky hit the CCD, the bits at the sides miss the sensor altogether.

Now if we use instead a short focal length 'scope, of the same aperture as before, the CCD represents a bigger area of sky. More sky = more sources of photons, so the CCD gets more photons altogether. The photons that hit all the CCD in the long FL 'scope example are crowded into the centre of the CCD and new extra ones hit the edges.

The two systems collect the same amount of light, by having the same aperture, but the long FL system has more "spillage" where some of the light misses the CCD.

Hope that makes sense?

Kaptain klevtsov

[narrowbandpaul]

well said KK

[The Warthog]

Maybe someone could further clarify here. I have been told reepeatedly that for visual astronomy, an 8" scope is an 8" scope, regardless of focal length, but in photographic use, the f/ratio makes a difference as anyone who has taken pictures manually would expect.

[ngc2403]

the difference in star brightness is nil for similar aperture telescopes, because they are point like.

however nebula are extended objects and so unlike the unresolved star the nebula's light is spread over an area in the focal plane. the amount of light inceases with the aperture however the focal length then spreads this light. more focal length thinner spreading of light, shorter focal length thicker spreading.

[ngc2403]

just a final point on that PDF, look at the final image comparision, anyone see a problem with the two images that he has posted?

the f/12 looks to be taken with less focal length than the f/4???????

[The Warthog]

So if I look at a star through an 8" scope, it won't matter what the f/l is, the star will look the same, but looking at a nebula, the nebula should be dimmer to my eyes through a longer f/l scope? I am assuming the same magnification for this question.

[ngc2403]

ah, magnification will do the same effect as focal ratio for imaging. the smaller the magnification the brighter the nebula.

This is complicated by the sky glow and best seen if you look at a Bright PN and then increase the magnification, you will still be collection the same number of photons from the nebula but you will be spreading them over a larger and larger area in the EP.

Hope this covers it

[The Warthog]

OK. I think I answered my own question when I specified the same magnification, therefore the same exit pupil.

[Kaptain Klevtsov]

The magnification is the key for visual stuff. If the magnification is the same you'll probably not be able to tell the two apart, but using the same eyepiece, you get different magnifications. The higher magnification makes things look dimmer (the light is more spread out as in the CCD example).

I wonder what a dark sky would look like at a magnification of 1X through an 8" 'scope?

Kaptain Klevtsov

[mcrossley]

The same as it does to the naked eye I would say! Think about the size of the exit pupil

[mcrossley]

Aren't Paul and Stan arguing about different things? Stan does not say the flux remains the same, but the SNR - so long as you remain in within the linear response of the CCD and sufficiently above the read noise etc. In reality those sources of noise will degrade the long FL image slightly, but maybe not to the extent that many people would have expected.

[dph1nm]

Which means that N(DN), the signal you record, is only dependant on the focal ratio, not the aperture. A 1m F5 telescope will record the same N(DN) as a 0.1m F5 telescope!

This is only for a fixed p in metres of course, in terms of arcseconds, one pixel will be 1/10 the size of the other - looking at it from the opposite perspective, the same equation shows that, for the same area of sky in arcseconds, the 1m will collect 100x as much flux.

NigelM

[acey]

Paul, you derive an equation which you say is independent of aperture, but it includes the term N(DN), which is dependent on aperture.

When I look at the Andromeda galaxy in a dark sky I see it as a faint patch. When I look at it through 10x50 binoculars I see it as a very bright patch. This is because an aperture's worth of photons are being directed into my 6mm pupils. The only way that can happen is if the image is magnified. If the binoculars were of different focal length I'd catch the same number of photons, but see an image of different magnification.

I suspect that visibility of deep sky objects is very largely to do with the physiology and psychology of sight, rather than just photon number. With imaging it's a different matter of course.

Andrew

[ngc2403]

True the N(DN) is dependent on the aperture but the area over which the light is spread is dependent on the focal length squared... so just back to 1 over f ratio squared?

[narrowbandpaul]

N(DN) is a meaningless number when it comes to the human eye response. The human eye was never covered in this, only image sensors. But there are parallels with regard to focal ratio and surface brightness

[acey]

Actually you started by speaking of "viewing/imaging deep sky objects", and I confess it was the "viewing" side that made me read the thread. But I think we can agree these findings have no relevance to visual observation, where aperture is the all-important factor.

It's still the case, though, that N(DN) is dependent on aperture. So I don't see how, for imaging, you've demonstrated a dependency solely on focal ratio.

[narrowbandpaul]

True the N(DN) is dependent on the aperture but the area over which the light is spread is dependent on the focal length squared... so just back to 1 over f ratio squared?

this turns it back to f ratio

[MartinB]

I have just taken an image of M63 with an identical scope ( 10" LX200ACF) and CCD (KAF3200) as Roundycat. He was imaging at F10 whereas my image was at around F7. My image had 2 hours of data whereas Dennis had 3 hours. If you adjust the 2 images so that the galaxies are the same size it is clear that Dennis has a better S/N ratio because of the extra 50% extra exposure time. The resolution of my image may actually be marginally better but that would related to seeing conditions and possibly the benefit of 8 guide cycles per sec from AO.

My simple take on this is that the benefits of the faster ratio are undone by the enlargemet of the galaxy relative to the F10 image. I think this is the jist of Stan's argument.

[roundycat]

reading the PDF again there is a term in there pi D squared. Is this a new way of calculating area or have I missed something?

Dennis