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Old 31-01-2012, 11:40 AM
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The_bluester (Paul)
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Explain something to me about focal ratios.

OK, this one even has my wife semi stumped (Trained as a professional photographer by the RAAF, who at that time even went into the chemical reactions going on in the film/paper to make it all work in film based photography) and she had to mumble about not quite recalling the techincalities of it.

F stop of a camera lens is easy to understand and easy to see why exposure times vary. Stop it down, reduce the apeture, reduce the amount of light falling on the media (Film, sensor, whatever) Open it up, more light, shorter exposure, simple.

How does it work with a scope? I can understand my CPC925 being an F10 system and that it will need exposures of X amount of time at prime focus etc. But how, if I get a fastar setup and convert it to F6 (I think) does it work to reduce exposures required to a fraction of the time when the objective that is gathering the light is precisely the same as before (Except it probably has a larger obstruction from replacing the secondary mirror with the fastar and the body of the camera)

Is it an inverse square thing where the light has had to travel further at F10 from the objective mirror to the camera media than with a fastar or other setup at a faster ratio? And if so, how does that work with folded optics like the 925? Apart from the fact that it just would not work, if I imaged at F2, straight off the objective, versus imaging at the prime focus at F10, the light clearly has not travelled five times as far!





I hate just taking things on faith! I am an ex tech so if anyone has a really detailed and techincal explanation hit me with it and I will go work it out if I do not follow it right away. This matters to me not at all, I am not really into imaging, but it is bugging me and I want to understand so I can forget it again!
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Old 31-01-2012, 01:01 PM
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rogerg (Roger)
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Quote:
Originally Posted by The_bluester View Post
OK, this one even has my wife semi stumped (Trained as a professional photographer by the RAAF, who at that time even went into the chemical reactions going on in the film/paper to make it all work in film based photography) and she had to mumble about not quite recalling the techincalities of it.

F stop of a camera lens is easy to understand and easy to see why exposure times vary. Stop it down, reduce the apeture, reduce the amount of light falling on the media (Film, sensor, whatever) Open it up, more light, shorter exposure, simple.

How does it work with a scope? I can understand my CPC925 being an F10 system and that it will need exposures of X amount of time at prime focus etc. But how, if I get a fastar setup and convert it to F6 (I think) does it work to reduce exposures required to a fraction of the time when the objective that is gathering the light is precisely the same as before (Except it probably has a larger obstruction from replacing the secondary mirror with the fastar and the body of the camera)
Yes, it works the same. To change the focal ratio of a telescope you put a focal reducer in, which takes the light from that same aperture put narrows it down to focus the same light on to a smaller area, reducing focal length (magnification) and making the focal ratio "faster".


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Originally Posted by The_bluester View Post
Is it an inverse square thing where the light has had to travel further at F10 from the objective mirror to the camera media than with a fastar or other setup at a faster ratio? And if so, how does that work with folded optics like the 925? Apart from the fact that it just would not work, if I imaged at F2, straight off the objective, versus imaging at the prime focus at F10, the light clearly has not travelled five times as far!
When you use something like Fastar, you are dramatically reducing the focal length, while keeping the aperture the same, so you are dramatically changing the focal ratio. A focal reducer might change the focal length by a multiplication of 0.8, but a Fastar option changes it by significantly more.

It's worth noting that, from what I understand, light sources which are a pinpoint at infinity do not become brighter when focal ratio is changed, so, the recorded brightness of a star will not change with focal ratio, it will only change with aperture. However a nebula will become brighter on the chip with a faster focal ratio. Now, this is what I've been told and I can see some logic there but I could also argue against it I was convinced (somewhat) of this when I started delving in to Occultations (point sources of light moving in front of each other) and being convinced a F/4 8" is not as capable as a F/10 16" telescope for such purposes because its the aperture which matters, faster focal ratio wouldn't have an impact. Like I say, I could still argue otherwise so obviously don't fully understand it

Roger.
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Old 01-02-2012, 06:28 AM
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Guys, its like this, without any mathematics (and assuming we are dealing with ideal optics - no absorption of light in glass or on reflective surfaces):

Firstly, the light from the objects is collected by the aperture of the objective.
So, in general, wider the aperture, brighter the image formed by objective.

Secondly, from this collected light the image is formed at the focal plane of the objective.
Further away the focal plane is (longer the focal length), the larger the image. Now, in case of nebulae (spread or extended objects) - all the available amount of light, collected by aperture is spread over the image area and if the image area is wider (longer focal length), the surface brightness of the nebula image (photons per square mm or whatever) will be lower (same amount of light on the wider image).
For very small objects (including stars, which are almost point sources) if the size of their image is smaller than the size of the sensor (pixel, halogen particle, rod in the eye), the surface brightness will not change because all the available light is concentrated on that one sensor or point.
If we go into mathematical analysis of the this situation, it can be shown that the surface brightness of the formed image of the spread objects depends only on the ratio between focal length and aperture diameter (assuming aperture is round, and yes, it is because of the square low). However, their total, integrated brightness depends on the aperture (area of the objective.. because it is the aperture that collects the light).

For visual telescope, the logic is the same, with the difference we have the magnification as a factor (higher magnification, wider spread - lower surface brightness of the extended objects, but point-like sources (stars) will not be affected, until the diffraction effects take place)
Ant that's all there is.. easy

Last edited by bojan; 01-02-2012 at 07:30 AM.
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Old 01-02-2012, 07:18 AM
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Thanks, that sorts me out nicely. And as a bonus, corrects my thinking in another area, that of the focal plane.

I have always tended to think of the optics as having a focal POINT, which I suppose is perfectly accurate, but it does not tell the whole story, that assumes objects are are a point source, plainly that is not the case for more or less anything but stars (Ignoring atmospheric effects) So to take M42 as an example, it is obviously an extended object and the image formed at the prime focus will not be a point but a similarly extended image. Increase the focal length (Leaving all else the same) increase the image size for the same amount of light captured so the light is distributed over a larger area and any given point will have less light falling on it.

Only leaves me one thing to ponder, if the image size formed at the focal plane by a given diameter objective is dependant on focal length (And therefore ratio) why does the image size not change when you stop a camera down?

I thought I would leave that question in although I reckon I have thought of the answer myself while typing it. Anyone who knows better please correct me.

In a camera lens we stop it down by reducing the apeture not increasing the focal length, so the image size would remain the same due to the focal length of the lens assembly, but the objective is effectively smaller so there is less light to be distributed in that formed image, we change a different parameter compared to doing something like sticking a fastar on my SCT, but the focal ratio is still the same between a 50mm lens at F2 and a 100mm at F2, the images formed will be different sizes but the INTENSITY of those images will be the same so the lesser amount of light captured by the 50mm is distributed in a smaller image and the exposure time will be about the same (Assuming like bojan that it is a perfect world, no light scattered sideways, none absorbed and turned into heat etc)

Nice to start my day by learning something totally irrelevant to me in my line of work!
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Old 01-02-2012, 09:21 AM
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Quote:
Originally Posted by The_bluester View Post
Thanks, that sorts me out nicely. And as a bonus, corrects my thinking in another area, that of the focal plane.
snip

Only leaves me one thing to ponder, if the image size formed at the focal plane by a given diameter objective is dependant on focal length (And therefore ratio) why does the image size not change when you stop a camera down?

I thought I would leave that question in although I reckon I have thought of the answer myself while typing it. Anyone who knows better please correct me.

snip

!
Your reasoning is correct. Recucing the f stop on a lens is effectively reducing the diameter of the lens. The focal lenght remains the same.
One important thing to remember is that the resolution of a lens is proportional to the diameter of the lens and the frequency of the radiation being imaged(radio waves need much bigger scopes than optical for the same resolution).
If the diameter of the scope gets too small then the resolution will become the limiting factor in image quality rather than other factors like seeing and pixel size. This is why bigger diameter scopes are needed.
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Old 01-02-2012, 09:17 PM
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I have always found it easier to understand these concepts with a picture, so I have put together a quick diagram which shows my understanding.

Hopefully this is accurate and useful.

Cheers,
Mark
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Old 02-02-2012, 05:56 PM
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Mark, your drawing is not correct.

See below, what it should look like (simplified drawing.. . the more accurate drawing from the geometrical optic point of view can be seen here (picture is from Wikipedia) : http://upload.wikimedia.org/wikipedi...-Lens3.svg.png))
webpage:
http://en.wikipedia.org/wiki/Lens_%28optics%29
The focal plane is presented with the vertical line on the right, focal length is the distance between the focal plane and the lens.

I found interesting applet that will give you even better insight as to what is going on here (it came from http://phet.colorado.edu/en/simulation/geometric-optics

http://phet.colorado.edu/sims/geomet...optics_en.html
Choose principal rays to see the same geometrical construction as on wiki page
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Last edited by bojan; 02-02-2012 at 06:21 PM.
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Old 10-02-2012, 12:41 PM
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Paul, one more thing to consider is the size of the pixels at the focal plane.

All other things being equal, a CCD with 10micron pixels operating at F5 will have the same sensitivity to extended objects as a chip with 20micron pixels operating at F10. It will also (in theory) have the same resolution.

The corollary to this is that if you keep the aperture and focal ratio the same, a chip with 20 micron pixels will have 4x the sensitivity of a CCD with 10micron pixels, but at half the resolution.
You can achieve this by 'binning'
Of course the quantum efficiency of the chip does not change when you operate it in this mode.
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