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I'd like to know how much magnification I would get by this telescope. The main specs are as follows:

  • Focal length: 2563 mm
  • Focal ratio: F/7.2
  • Number of lenses: 2
  • Optical diameter of Primary mirror: 355.6 mm
  • Diameter of Secondary mirror: 165 mm

Based on this simple class material by NASA (for kids?), the magnification may be computed as a ratio of the focal length of object lens and that of eyepiece. However, the datasheet of this telescope does not provide the focal length of eyepiece. In the first place, I'm doubtful that this simple computation scheme applies to this telescope with mirrors and lenses... Does anyone know more about how to compute the magnification of this kind of telescope?

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    $\begingroup$ the eyepiece you use with it will define the magnification - check out astronomy.tools/calculators/magnification $\endgroup$
    – Aaron F
    Commented Jan 10, 2022 at 23:27
  • $\begingroup$ Thanks for your comment. This may be a very basic question but how do I know the spec for the eyepiece? Is it possible to estimate from the spec information in the link? $\endgroup$
    – Ogiad
    Commented Jan 10, 2022 at 23:31
  • $\begingroup$ high-end telescopes like that one don't come bundled with any eyepieces - the manufacturer would assume that if you're spending that much on a telescope then you probably already own some high-end eyepieces $\endgroup$
    – Aaron F
    Commented Jan 10, 2022 at 23:37
  • $\begingroup$ (here's an example of a high-end eyepiece: shop.nikon-image.com/front/ProductBFB20040 ) $\endgroup$
    – Aaron F
    Commented Jan 10, 2022 at 23:39
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    $\begingroup$ Thanks, Aaron! I'll dig into that! $\endgroup$
    – Ogiad
    Commented Jan 10, 2022 at 23:44

2 Answers 2

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Telescopes such as this are often described as "astrographs" since they are designed specifically for astrophotography, and specifically astrophotography at the prime focus rather than visual use. The advanced optical engineering in such a telescope is so that a relatively large CCD sensor can be used at the prime focus and have a "flat field" (uniformly in focus) and uniformly illuminated across the sensor.

Prime focus astrophotography doesn't use an eyepiece like visual observing; what determines the magnification is the visual scale at the focus, expressed in something like degrees per millimeter. The visual scale is determined entirely by the focal length, since it is the arctangent of 1/focal length.

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  • $\begingroup$ I noticed that they list focusing mechanisms under "Often Purchased Together" which suggests that the back of the telescope is just a flat metal plate with a big hole in it (and some threaded screw holes) and purchased alone one couldn't really do anything with it except point it at the Sun and start fires :-) $\endgroup$
    – uhoh
    Commented Jan 12, 2022 at 0:17
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The calculation of the magnification is as simple as $$magnification = \frac{focal\;length\;of\;telescope}{focal\;length\;of\;eyepiece}$$

The NASA paper is showing a simple objective (lens) and simple eyepiece, each made from one element. Each element has a known focal length which leads to the formula given for the magnification.

In many telescope designs, the light gathering portion is a combination of mirrors and lenses, each of which have an inherent focal length. When used together, the telescope has an "equivalent" focal length dependent on the design. The focal length of the telescope in question is 2563 mm without any details of how that length is obtained through a combination of curved mirrors and/or lenses.

Likewise, a quality eyepiece has several elements which together give an effective focal length as indicated on the eyepiece. The formula for the magnification does not mention to use the "effective focal length" because it is implied.

If the telescope in the question is used with an eyepiece, then you know the two focal length values that are necessary to calculate the magnification.

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