I am planning to buy celestron nexstar 127SLT and I want to know what can I see using my telescope.

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    $\begingroup$ Do you have a specific one (or more) in mind? Then please add the link to its specifications. $\endgroup$
    – user1569
    Oct 25, 2018 at 13:35
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    $\begingroup$ This question does not give enough detail to answer. It is like saying "I'm buying a car with a top speed of 200 km/h. What are some places I can visit?" The magnification only one factor, and not the most important one. $\endgroup$
    – James K
    Oct 25, 2018 at 17:16
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    $\begingroup$ @JamesK That is the perfect simile. Excellent comment. $\endgroup$ Oct 25, 2018 at 19:17
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    $\begingroup$ @PalliVishal welcome to Astronomy! 1. don't buy that telescope yet! 2. Have a look around this site, especially other questions about telescope purchases and magnification (there's a search feature near the top of the page) 3. Have a look at the Help Center here and how to ask good questions. Your question is likely to be closed because there's not enough information in it. You can still edit your question and improve, and it may be reopened. But check other questions here first. $\endgroup$
    – uhoh
    Oct 25, 2018 at 22:56
  • $\begingroup$ I've made some small grammar edits to the title and added some better tags. You can just click the amateur-observing tag and start reading those, or narrow it down by adding additional terms like "magnification" astronomy.stackexchange.com/… $\endgroup$
    – uhoh
    Oct 25, 2018 at 23:00

1 Answer 1


First of all, the high end of magnification is generally not as important as several other factors. However, considering that no additional information was given in the question beyond "reflecting", I will make some assumptions and only give you a very general estimate.

The book "Fundamental Astronomy: H. Karttunen et al.", page 51, formula (3.5) gives a relationship between the diameter of a telescope's aperture and it's useful magnification.
This relation is as follows ($e$ is the resolving capacity of the human eye, $D$ is the aperture diameter, $ω$ is the maximum magnification and $λ$ is the wavelength of visible light, used in the formula for the telescope's resolving capacity:

$$θ ≈ 1.22×λ/D ≈ λ/D,$$

$$ω_{max} = \frac{e}{θ} ≈ \frac{eD}{λ} = \frac{(5.8 × 10^{-4} × D)}{(5.5 × 10^{-7} m)} ≈ \frac{D}{1 mm}.$$

That's a key concept for you, because any additional magnification above that can be thought of as "empty magnification". It exists mathematically, but the image is so blurry that you will not see any additional detail, and probably be quite disappointed in your purchase if you paid a lot of money for that magnification!

Let's try backtracking from 500x.

If this is the true maximum useful magnification this telescope can achieve, then its aperture diameter would be about 50cm. This aperture allows for $D^2/d^2$ brighter stars to be observed ($d$ is the diameter of the human pupil). When this is calculated and converted to magnitudes, it turns out you should be able to see stars up to magnitude $9+6=15$.

However this is a very rough estimation and you certainly will not be able to see magnitude 15 stars due to atmospheric extinction, seeing and other factors ... my estimate is you'll see stars about the 12th magnitude on a good night.

If the 500x telescope you are looking at has a much smaller aperture than 50 cm (and probably that's true) then that 500x is not going to be so useful in reality.

Take some time and read further about telescopes. This site has many existing questions and answers on the subject.

  • $\begingroup$ Entrance aperture has nothing to do with achievable magnification. $\endgroup$ Oct 25, 2018 at 17:53
  • $\begingroup$ @CarlWitthoft "Fundamental Astronomy: H. Karttunen", page 51, formula (3.5) connects objective aperture with maximum useful magnification. I changed the word "possible" to "useful" in my comment. I am certain maximum useful magnification does indeed have much to do with objective aperture. $\endgroup$
    – Tosic
    Oct 25, 2018 at 18:05
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    $\begingroup$ thank you for doing that. You are correct that the useful/useable magnification does depend on aperture sizes (same as when dealing with microscope objectives). $\endgroup$ Oct 25, 2018 at 18:07
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    $\begingroup$ I have found that telescope makers who talk about magnification are usually at the budget (ie low cost) end of the market. If I’m correct then a diameter of 500mm is out of the question. We are probably look at a reflector with a mirror of 115 to 125 mm, for which a max useful magnification (under very good conditions) would be about 200. $\endgroup$
    – Dr Chuck
    Oct 25, 2018 at 19:02
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    $\begingroup$ @Tosic your answer is basically correct, I've just expanded it a bit and adjusted the formatting. If you can find that equation and put it into the answer itself, that would be great! If you need help with MathJax formatting for equations you can ping me, or someone else may help as well. $\endgroup$
    – uhoh
    Oct 25, 2018 at 23:13

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