Focal Length
Your telescope's focal length (the distance from the primary lens or
mirror at which it will form a prime focus image) is one of the major
factors in determining how much magnification you will get, both
visually and in a photograph. Most every amateur astronomer knows
that the visual magnification is figured by dividing the telescope
focal length (in millimeters) by the eyepiece focal length (also in
millimeters).
But how do you figure the photographic magnification? Well, for prime
focus photography, it's pretty simple. If you are using 35mm film,
just divide your telescope's focal length by 50. Why 50? Because your
telescope is serving as the camera's lens, and a 50mm lens is
considered "normal" (1x magnification) for a 35mm camera. If your
telescope's focal length is 1000mm, that's 20 times "normal" or 20x.
You might think it would be easy to figure the magnification for
eyepiece projection (by just using the same formula you would use for
visual -- telescope focal length divided by eyepiece focal length).
Unfortunately, it isn't that simple. The key word here is "projection"
and like any projection system, the image gets bigger the further the
"screen" is from the "projector." In this case, the projector is the
eyepiece and the screen is the film in your camera. There is a formula
that will let you figure the magnification based on the distance
between the eyepiece and the film (and other factors), but it's not
important enough to go into here. The important thing to know is that
your telescope's focal length is a major factor in determining the
magnification you get -- in fact, for prime focus photography, it's
just about the only factor.
The Perfect Focal Length for Astrophotography...
Sorry, people, there's no such thing. If you check out the section
about Bald Eagle Observatory you will see that we use a number of
telescope tubes at BEO, ranging in focal length from 400mm to 2032mm.
We also use camera lenses from 28mm to 205mm for piggyback
photography. Our "original" astrophotography scope was a Celestron
SP-C6, a 6-inch f/5 Newtonian with a focal length of 750mm (25.4*6*5).
This scope was fine for fairly large objects like the Andromeda Galaxy
and the Orion Nebula; but my first attempt to photograph the Ring
(M57) produced what looked like a fat, red star. The scope simply
didn't have enough magnification for such a small object. I have since
photographed it successfully, but I had to use an 8-inch f/10
Schmidt-Cassegrain (2032mm focal length) (25.4*8*10) telescope to do
it. That scope, on the other hand, is just "too much club" for the
North American Nebula (which spans almost four degrees of sky). The
best shot I've gotten yet of that object was taken with an 80mm f/5
refractor (400mm focal length) (80*5). (I guess CENTRON 1000mm 3kg is
4" F/10 i.e. 25.4*4*10 = 1016mm or 5"F/8 i.e. 25.4*5*Cool
But I Can't Afford More Than One Telescope...
Neither could I, when I first got started; but I've learned a few
tricks since then. For one thing, I've learned about "Focal Reducers".
A focal reducer is an auxiliary lens that reduces your telescope's
effective focal length. You start with a telescope that has a long
focal length, like an 8-inch f/10 Schmidt-Cassegrain (2032mm). Adding
an f/6.3 reducer (both Celestron and Meade sell these) brings the
telescope's effective focal length down to 1280mm. It's like having
two telescopes in one. Focal reducers are available for other types of
scope as well.
F/10, F/6.3...What Is This F/Number Thing, Anyway?
The "F/Number" is the telescope's Focal Ratio, which is calculated by
dividing the effective focal length by the diameter of the primary
objective (lens or mirror). Thus, an 8-inch (203mm) diameter telescope
with a focal length of 2032mm is an f/10, while my little 80mm
refractor with its 400mm focal length is an f/5. So why is the focal
ratio important? Well, for photographic purposes, the focal ratio
determines the overall brightness of the image. The lower the
f/number, the brighter, and the progression is logarithmic, not linear
(i.e. an f/5 telescope's image is nearly four times as bright as an
f/10 telescope of the same aperture. If you decrease the effective
focal length (by adding a reducer, for example), you also decrease the
f/number and get a brighter image, which means less exposure time is
needed to produce a picture.
One of the reasons you don't use eyepiece projection to photograph
dim, deep-sky objects is that by adding an eyepiece, you are
increasing the telescope's effective focal length, which in turn
increases the focal ratio (sometimes to as much as f/30 or f/40),
resulting in a dimmer image. Deep-sky photography really shouldn't be
attempted with focal ratios greater than about f/12.
By the way, this f/number concept is no different than the "f/stop"
used by photographers. If you hear a photographer talk about "stopping
down to f/11" it means he is going to increase the focal ratio of his
lens to the point where the focal length of the lens is 11 times the
diameter of the opening. The only difference is that astronomers
usually change the focal ratio by changing the effective focal length
of their telescopes; photographers do it by changing the diameter of
the lens opening (by means of a diaphragm built into the lens).
Mirror lens and DSLR
Recently there are some trials of using mirror lens with DSLR e.g.
Canon 10D, Canon 300D or Nikon D70. Because of the smaller CCD/CMOS
size, the focal length is multiplied by 1.5x/1.6x factor. So if you
are using a 500mm F8 mirror len with 10D/300D, you got a focal length
of 500*1.6 = 800mm. And with a 1000mm F11 plus 2x converter you got
1000*1.6*2 = 3200mm. It may be useful for taking photos for deep sky
object (DSO). However due to the design of DSLR, you may not able to
take a very long time exposure to capture bright images. Multiple
exposure with dark frame noise reduction may improve the situation.
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