I've found various sources on the web stating that the limiting apparent magnitude of the Hubble Space Telescope is about 31 or 31.5. Wikipedia lists that here and it has a cite to a pdf about it.

What I want to know is how is that calculated?

I'm aware of two methods found at the bottom of a table here, but they seems to be general calculations for ground-based telescopes, because if I plug in Hubble's numbers, I get something around 21. The difference between an apparent magnitude of 21 and 31 is actually something like ten thousands times (in terms of brightness).

Also, one key factor missing in their formulas is exposure time. I'm pretty sure the limiting magnitude has something to do with exposure time. Hubble Ultra Deep Field, for example, had an exposure time of about 1 million seconds.

So to sum up, how would we calculate the limiting magnitude of a space-based telescope, in visible light?

  • $\begingroup$ The 10k difference is due to the absence of the atmosphere in the case of Hubble. Absent that, the limiting factors are with the instrument itself. I would guess it's sensor noise that limits the magnitude. $\endgroup$ Jun 30, 2015 at 17:56
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    $\begingroup$ I would think the long exposures would have more effect on the limiting magnitude than the absence of atmosphere. $\endgroup$ Jul 1, 2015 at 16:28

1 Answer 1


The table you linked to gives limiting magnitudes for direct observations through a telescope with the human eye, so it's definitely not what you want to use.

The quoted number for HST is an empirical one, determined from the actual "Extreme Deep Field" data (total exposure time ~ 2 million seconds) after the fact; the Illingworth et al. PDF you linked to explains how it was done. Briefly, they decided the limit would be defined by a signal-to-noise ratio of 5, and then figured out what the noise level in the final combined images was by integrating the total flux in many circular apertures of diameter 0.35 arc seconds placed in regions of blank sky background, and then computing the standard deviation ($\sigma$) of these measurements. Five times $\sigma$ was then the limiting flux. Knowing the photometric calibration (the conversion between observed counts and apparent magnitude; see the paper for specific details), they converted this $S/N = 5$ limit into an apparent magnitude.

If, instead, you want try to estimate in advance what the limiting magnitude should be, then you have to consider several factors. Simplifying a bit, the equation for long-exposure, background-limited $S/N$ is $$ S/N \approx \frac{s t}{ \sqrt{ s t + n_{\rm pix}(b t)} } $$ where $s$ = detected flux from the source (star, galaxy, etc.) in electrons (i.e., detected photons) per second, $t$ is the integration time in seconds , $b$ is the flux from the background in electrons per pixel, and $n_{\rm pix}$ is the number of pixels you integrate over for the measurement. $s$ can be computed from the combination of the object's apparent magnitude, the diameter of the telescope's mirror and the overall efficiency of the telescope + detector system. $b$ can be determined in a similar fashion, using the estimated background brightness (which, for a telescope like HST, is mostly sunlight reflected from dust in the inner solar system). $n_{\rm pix}$ is a number chosen as a compromise between maximizing the amount of light from the source (higher $n_{\rm pix}$) and minimizing the noise from the background (lower $n_{\rm pix}$). This depends on the telescope's resolution (better resolution = smaller images of stars and other point-like sources = fewer pixels needed); the empirical calculations by Illingworth et al. used an aperture size amounting to about 100 pixels.

  • $\begingroup$ Thanks but what is the formula for s? I couldn't find it in the pdf. $\endgroup$
    – DrZ214
    Jul 3, 2015 at 4:14
  • $\begingroup$ What I described in the second part of my answer is the general approach to estimating a S/N ratio before you make an observation. That's not what that paper (Illingworth et al.) did. Instead of trying to estimate sigma as sqrt(st + n_pix bt), they measured it directly by making many small measurements (each with an area of ~ 100 pixels) on the images they already had. Then they multiplied that sigma by 5. (If you define your limit as a S/N = (s t)/sigma = 5, then S = 5 sigma.) $\endgroup$ Jul 3, 2015 at 20:37
  • $\begingroup$ Okay but can you give me a way of estimating s? Lowercase s, not capital S. As it stands now, I can't use your S/N formula because i have no way of plugging in values for s and b. $\endgroup$
    – DrZ214
    Jul 3, 2015 at 21:54
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    $\begingroup$ s = number of photons detected per second = F * A * T. F = flux in photons at the telescope (number of photons per second per square meter; e.g. astro.umd.edu/~ssm/ASTR620/mags.html); A = effective collecting area of the telescope; T = "throughput" of telescope (a number between 0 and 1) -- what fraction of light entering telescope is actually detected. $\endgroup$ Jul 4, 2015 at 16:04

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