Following on from a recent post in Beginners' Equipment...
I took some (very rough & kinda terrible) shots of the Moon, Mars and Omega Centauri tonight with both a little 80/600mm refractor and a 200/1000mm Newtonian reflector. Same exact camera & camera settings for each target. I switched OTAs between targets (meaning I took a shot, changed OTA, took the same shot, changed OTA, took second shot, changed OTA etc).
The ED80 had a 0.85 flattener on it so it was essentially f/6.4. The reflector is f/5. This SHOULD mean that the newt would be significantly brighter, like two thirds of a stop... but it clearly isn't.
Anyone have any suggestions why this should be? The secondary blocks a little of the light, but nowhere near 30% of it. Is a large part of the light "missed" by the secondary? Inherent vignetting by newts? Loss from the two mirrors? I just don't understand these results, it's weirding me out.
I think there are multiple factors involved. A major one is the effective difference in applied light per pixel on your sensor, (assuming it’s the same sensor) given the difference in focal length. Tracking errors will exacerbate this effect.
Cheers
I think there are multiple factors involved. A major one is the effective difference in applied light per pixel on your sensor, (assuming it’s the same sensor) given the difference in focal length. Tracking errors will exacerbate this effect.
Cheers
Not sure I follow.
If I take a photo of a featureless blue sky with any f/5 lens or OTA, their focal lengths will have no impact on the brightness of the image. That's quite literally what the f ratio means. If the aperture stays the same and the focal length increases, the f ratio changes and thus the brightness. If the ratio stays the same, that means both are receiving the same amount of light per pixel. Whether that light is from a larger or smaller patch of the sky doesn't change the net amount received.
Also not sure what difference the tracking can make. As I said, both OTAs were used with the same mount, camera and settings, within about 90 seconds of one another. Any impact from tracking would be identical between them, regardless of how accurate.
The only conclusion I can draw is that the newt is somehow less "efficient" and that its f/5 ratio is more in theory than practice. About 20% of the light is being lost somewhere, just curious as to where. Is some of the light from the primary going around the edges of the secondary? Is there something funky going on with the focuser assembly that 20% of the image isn't making it to the camera sensor? I'd really like to know!
My best guess is it's a combination of things:
- some light is blocked by the secondary (about 6% apparently);
- some light is escaping back out into space, ie going around the secondary
- some light is landing outside the camera sensor (so as to avoid vignetting around any images)
Bit of a shame that it all adds up to the thing being nearly a stop less efficient than advertised, though. It's a big strike against reflectors for me personally, especially when considering how inconveniently cumbersome and fiddly they are. Their main benefit is supposed to be their light-gathering capabilities, but they're considerably less efficient than they're supposed to be. (At least, the three I've owned are.)
The ED80 had a 0.85 flattener on it so it was essentially f/6.4. The reflector is f/5. This SHOULD mean that the newt would be significantly brighter, like two thirds of a stop... but it clearly isn't.
It's more complicated that just point and shoot. There's varied opnions on what is myth and what is fact...
Fantastic article, thank you!! Answered all my questions. I had the exact same impressions & experience; he actually went into the maths behind it and explained exactly why that's the case. It seems that reflectors are indeed significantly less efficient than refractors. I never realised by how much. 14% of the light is lost in each reflection! They sure don't advertise that fact on the boxes they come in...
There are two other aspects that you can take into consideration. Was the processing of the two images the same? I'm thinking more so with the cluster than the moon.
Another one is that of noise. Even if the images do look very similar, is there a difference in measured noise? Single exposures of the moon or of stars can be quite deceptive in evaluating brightness levels.
As has been mentioned already though, reflective surfaces do lose more light than refractive ones with modern day anti-reflective coatings.
There are two other aspects that you can take into consideration. Was the processing of the two images the same? I'm thinking more so with the cluster than the moon.
Another one is that of noise. Even if the images do look very similar, is there a difference in measured noise? Single exposures of the moon or of stars can be quite deceptive in evaluating brightness levels.
As has been mentioned already though, reflective surfaces do lose more light than refractive ones with modern day anti-reflective coatings.
Nope! Same exact processing (or lack thereof). Off the SD card into Lightroom with generic import settings, crop, export, upload. No visible difference in noise, apart from the whole crop factor thing. The reflector has twice the focal length so the noise takes on a different pattern, probably easier to smooth out, but there's no reduction applied here.
I'm sure in the $2-3k and upwards OTAs they're using better coatings that reduce the gap (or so I'd hope anyway), but here in the shallow end there is literally two thirds of an f stop lost.
It's all convincing me I need to become a refractor guy. Lighter, easier, less maintenance. Less focal length, but oh well. I'll happily leave the galaxies to the big boys with more time, patience and budget than me!
I've always found reflectors lose a bit of brightness compared to refractors. I love refractors but the cost I don't love so much. Bang for buck the reflector still wins, but it takes more aperture to do it.
Based on my own calculations, the loss of light gathering reduction due to the secondary mirror and spider veins is quite insignificant.
My example can be explained by these facts: My 8 inch f 5 Newtonian has a primary mirror diameter of 200mm and a focal length of 1000mm which equates to a f ratio of 5, as stated by the manufacturer. Now rounding it off to the nearest whole number after compensating the area taken up by the secondary mirror and spiders, is just that. Not really a misleading figure but rather a simpler number to remember.
By calculating the surface areas of each component in the optical path we can arrive with the true f ratio and in my case, this is f 5.33, hardly the one stop loss someone stated it could be.
According to a book I’m reading “ Telescopes, Eyepieces and Astrographs” the light blocked by the central obstruction in a Newtonian optical system is around 9% which is a minor issue.Diffraction spikes only become a major issue for apertures greater than 12 inches ( if that’s an issue at all) A bigger issue for Newtonian’s is off axis coma. I have a 10” f5 dob and 6” f6 OTA and have not had an issue with coma to this point at the edge of field both with visual observing and imaging
An 8" F5 reflector might have a T-stop of 6 or more making it a similar brightness to your F6.4 refractor (which will also have a T-stop but it won't be as lossy).
Based on my own calculations, the loss of light gathering reduction due to the secondary mirror and spider veins is quite insignificant.
My example can be explained by these facts: My 8 inch f 5 Newtonian has a primary mirror diameter of 200mm and a focal length of 1000mm which equates to a f ratio of 5, as stated by the manufacturer. Now rounding it off to the nearest whole number after compensating the area taken up by the secondary mirror and spiders, is just that. Not really a misleading figure but rather a simpler number to remember.
By calculating the surface areas of each component in the optical path we can arrive with the true f ratio and in my case, this is f 5.33, hardly the one stop loss someone stated it could be.
Again, I know what the f ratio is. I've been an amateur photographer for years - and am married to a professional photographer - so I really do understand the concept.
See the photos in my original post. There's very little difference in brightness between the two OTAs despite the reflector being ostensibly 2/3 of a stop faster.
There was an article posted here somewhere explaining how reflective coatings can be quite lossy. Add to that the 8% or so blocked by the secondary and you do indeed lose most of a stop.
Bottom line: Reflectors are advertised as being significantly faster, than refractors, but they (at least the SW/GSO/Saxons etc) aren't actually as fast as their f-ratio would suggest.
Yep, inch for inch the refractor wins every time. Last night I was visually comparing a 120mm F5 refractor with a 150mm F5 Newt. They were very close. Plus I was using a diagonal in the refractor and the refractor was an achro.
The ED80 had a 0.85 flattener on it so it was essentially f/6.4. The reflector is f/5. This SHOULD mean that the newt would be significantly brighter, like two thirds of a stop... but it clearly isn't.
Anyone have any suggestions ...
This is a common (astro)photography myth. The f/ ratio has NOTHING to do with absolute brightness.
Using the same ISO value and exposure time, a larger ABSOLUTE aperture in millimeters / inches will ALWAYS show fainter stars and therefore, the same stars appear brighter. Period.
But a smaller f/ratio ('brighter') with the same absolute aperture results in a larger FOV, so the entire image appears brighter because a larger portion of the sky (i.e. more stars) are on the image.
In your case, the newton should show much fainter stars if you use the same exposure time and ISO value, which is (20/8)2 = 6.25 times brighter which is about 2.6 stops or two stellar magnitudes.
As a terrestrial photographer for some 30 years prior to taking on astrophotography, I had great difficulty in reconciling the fact that there a different set of rules governing the imaging of stars. For extended objects such as nebulae or landscape, the traditional rules apply, but for stars it is different.
I haven’t attempted the math, but I believe that even the closest star would in reality be like a point source, occupying less than a pixel on any of your imaging systems. What we observe is the spread of the light caused by dispersion in the cosmic dust, the atmosphere, optical components, and our eyes. Stars with more brightness appear bigger through our imaging systems because of the wider dispersion of their light, not by their physical size.
This all becomes combined and confused with the other phenomena we observe such as nebulousity, and glow caused by scattered light reflected on cosmic dust.(extended objects) Gathering light caused by these effects obeys the same laws as for terrestrial photography. Ie, a lower F number for a given field of view, means more light gathered per unit of time.
I think this clicked for me when I was (as I usually am) viewing viewing in polluted skies. A combination of high thin cloud and usual smog. I compared views of the owl cluster in my 60mm Tak I was utilising as a finder scope and in the 18” Dob. Through the sky mush, with the Tak, and same FOV, I could barely make out the two brighter stars in the cluster, but with the extra aperture of the Dob, I could make out quite a few of stars. The conclusion being that the aperture of the Dob increased my star to mush brightness ratio considerably when compared to the smaller Tak. The background mush obeyed standard photographic laws with which most terrestrial photographers are familiar, but the star brightness changed as a result of larger aperture.
In the classic "Tools of the Astronomer", Harvard books...
p21 ""As sources of light, astronomical objects are of two kinds - point sources and extended sources. A star does not show any appreciable disk of its own but gives a more or less pointlike image in a photographic plate....Objects of the second type, such as nebulae and planets extend over appreciable areas. The telescopes which are required to image most efficiently these two types of objects are not the same.""
p109 "" The limiting magnitude depends on and increases with both the aperture and the f ratio. An f5 camera will photograph stars about three magnitudes fainter than an f1 camera.""
p110"" Thus either stars or nebulae may be selectively emphasized. One chooses a relatively long focal length and a large aperture to photograph the faintest stars, or a small f ratio to minimise the presence of stars in a photograph of nebulae"
