How far is a candle visible to the human eye?

Posted by barb on Feb 8, 2005 in Science Musings |

One of the questions I got last week through the Ask a High Energy Astronomer website was asking if a candle, burning in a vacuum (we’ll let that part slide) would be visible from a great distance. The idea is that if there is nothing in the way to absorb the light, then we should be able to see the candle.

However, this does not take into account the decreasing flux from that candle. There are only a finite number of photons emitted by that candle, and those are emitted isotropically (equally in all directions). The number of photons, then, impinging on a surface, say, your eye, will decrease as the distance of the candle increases.

Out of curiosity I decided to do a quick calculation to see how far away the candle could be before it was just detectable by the human eye. I didn’t send this off with my answer, because I wasn’t sure of the calculation, and the person indicated their level as “novice”, so they probably wouldn’t have been enlightened by the calculation anyway. Just for posterity, here is my “scratch pad” calculation.

A few quantities to start with:

  1. From Lumens, Illuminance, Foot-candles and bright shiny beads.,
    the output of one candle is: 1/685 W per steradian
  2. From Light Measurement Handbook: The Power of Light, the threshold of the human eye is: 3.58e-18 W

From (1), we can first get the total power output by that candle, assuming isotropic emission, and using the fact that there are 4*π steradian in a full sphere.

Lcandle = 4π/685 W

The flux of that candle at a distance, R, is then:

Fcandle = Lcandle/(4πi*R2) = 1/(685*R2) W

Now from (2), we can find the threshold flux that the eye can detect. Assume that the eye has a light-collecting area equivalent to a circle with diameter 1 cm, then:

Feye, threshold = 3.58e-18 W/π*(0.005 m)2 ~ 4.56e-14 W/m2

Finally, by equating the two, we can find the distance at which the candle will be just detectable:

Feye, threshold = Fcandle

So:

4.56e-14 W/m2 = 1/(685*R2) W
R2 = 3.2e10 m2
R ~ 1.8e5 m

That’s just 180 km, which is less than the distance between New York City and Washington DC.

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5 Comments

Mush
Feb 8, 2005 at 5:34 pm

Joo are zo sexy, darlink!


 
Kayhan Gultekin
Feb 9, 2005 at 2:00 pm

You must be Barb Mattson. Am I right? I came across your blog while googling for Hubble poetry. Your postings about how much you hated being a grad student were what gave it away. That and the windowed cow reference.
Just to nerd it up a little, the diameter of the human pupil after it is dark adapted is 8mm, which translates almost to a factor of 2 in the final distance down to 115km.
PS I have to enter a valid email address?


 
Kayhan Gultekin
Feb 9, 2005 at 2:02 pm

Whoops! I entered an invalid URL.


 
Barb
Feb 9, 2005 at 5:27 pm

Yup, it’s me, Barb Mattson…I forgot that I don’t have any links to my other webpages from the blog. I meant to add some when I redesigned a few weeks ago.
Not sure that you have to post a valid e-mail address…probably not. Though I’ll certainly check out your blog now that I know it exists.


 
Kayhan
Feb 10, 2005 at 10:47 pm

I felt like a moron when I went to airynothing.com. I should have checked there first. It doesn’t have to be a valid email address but I can’t put in “kayhan -aht- gmail -daht- com” like I often do in online forums. My blog is nothing special. It falls under the category of “what I had for breakfast blogs” and under the subcategory of “what color my son’s poop was.”
Congratulations on getting married. The pictures were great.


 

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