I'm trying to estimate the distance and power I'd need for a green LED to appear visually roughy as bright as a relatively bright star - say a visual magnitude of zero. Here is what I have so far.

Be warned I am just ballparking it here.

The sun is visual magnitude -27, and five visual astronomical magnitudes are a factor of 100, so a zero magnitude star should appear to be a factor of $100^{-27/5} \approx 1.6 \times10^{-11}$ as bright as the sun.

The FWHM of the sensitivity of human vision is about 100nm and peaks roughly in the green part of the spectrum, however the center changes between about 550nm and 500nm depending on photopic or scotopic conditions.

At sea level, direct sunlight is about $ 1.3 \ W/m^2/nm$, so for a 100nm wide bandpass that's $130 \ W/m^2 $. A zero visual magnitude object should then produce $ 2.1 \times10^{-9} \ W/m^2$.

If I have a say 555nm green LED with 30% external quantum efficiency, then $0.1 \ A$ of current should produce $0.22 \ W \times 0.3 \approx 0.067 \ W$ of light. If it is roughly uniform over a cone with a half-width of 10°, then the LED produces $ 0.7 \ W/Sr$, or $ 0.7/r^2 \ W/m^2$ at a distance of $r$ meters.

That means I would have to move my 100 mA, 30% eQE 555nm LED with a 10° half-angle 18 kilometers away for it to look roughly as bright a 0 visual magnitude star!

Have I made some fundamental mistake here? Or - baring atmospheric absorption - could I actually see a green LED ~20km away (or on a balloon 20km straight up) on a dark night?

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  • 1
    $\begingroup$ I really like this question. I did a back of the envelope calculation via different methods and come to the same conclusion. $\endgroup$
    – Myridium
    Commented Sep 24, 2016 at 16:33
  • $\begingroup$ @Myridium That's good to know, thanks for the independent checking! $\endgroup$
    – uhoh
    Commented Sep 24, 2016 at 16:49

1 Answer 1


This is not a full answer, as I don't know how to compare an led with a candle, and I am not sure than the "answer" makes sense, even in rough estimate terms. Please just treat it as background and I will then delete it.

Candle Study Abstract

Using CCD observations of a candle flame situated at a distance of 338 m and calibrated with observations of Vega, we show that a candle flame situated at ~2.6 km (1.6 miles) is comparable in brightness to a 6th magnitude star with the spectral energy distribution of Vega. The human eye cannot detect a candle flame at 10 miles or further, as some statements on the web suggest.

Wikipedia says, as you already know:

A magnitude 1 star is exactly a hundred times brighter than a magnitude 6 star, as the difference of five magnitude steps corresponds to $(2.512)^5$ or 100.

But this does not make much sense either, especially when you go up another magnitude scale to mag 0.

Comparing units of light power brings us back to the dark ages (sorry).

One candlepower is the radiating power of a light with the intensity of one candle. This unit is considered obsolete as it was replaced by the candela in 1948, though it is still in common use. 1 candlepower is equal to about 0.981 candela.

candela The standard unit for measuring the intensity of light. The candela is defined to be the luminous intensity of a light source producing single-frequency light at a frequency of 540 terahertz (THz) with a power of 1/683 watt per steradian, or 18.3988 milliwatts over a complete sphere centered at the light source.

  • $\begingroup$ This is helpful! The candle is radiating in a wide range of directions - say $2 \pi \ SR$, while the LED is focused to only $0.1 \ SR$. The candle is almost all heat and IR, the LED is wavelength optimized for maximum sensitivity. So I may in fact not be wrong! The paper is interesting reading as well. I like it when people can still ask, and then answer really basic questions. $\endgroup$
    – uhoh
    Commented Sep 24, 2016 at 15:14
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    $\begingroup$ I remember 100 W bulbs! Actually this is what I was really looking for - a sanity check, or a "could this actually be right" check rather than the math. Thanks for your help! $\endgroup$
    – uhoh
    Commented Sep 24, 2016 at 15:32

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