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I want to calculate if it is possible to photograph a subject lit only by starlight. I found one website claiming that a starlit scene is lv = -15 (daylight is lv 15, or 2^31 times brighter), but he has no references. However I'd prefer to confirm this before I take a one hour exposure in the middle of nowhere, far from light pollution.

So, what is the total flux from all star's light onto the earth?

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    $\begingroup$ Would photo.stackexchange.com be a better place for this question? $\endgroup$
    – Qmechanic
    Commented Jun 2, 2013 at 16:46
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    $\begingroup$ I think the question of "how much starlight?" is for us. Photo.SE can handle the followup of "is this enough for a picture?" That said, @user1512321, could you explain what "lv" is? It seems to be a logarithmic measure of brightness/intensity/luminosity/energy/flux, but I'm not sure how well-known the unit is. $\endgroup$
    – user10851
    Commented Jun 2, 2013 at 18:11
  • $\begingroup$ I was wondering where to ask, but I figured its a physics question because I want a flux value from the celestial bodies and I can figure out the rest. $\endgroup$
    – user16035
    Commented Jun 2, 2013 at 20:10
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    $\begingroup$ Lv is a log2 scale where lv 0 is defined as the light necessary to take a photo with f 1.0 t=1 ISO 100 $\endgroup$
    – user16035
    Commented Jun 2, 2013 at 20:13

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It is certainly possible. The human eye will be able to resolve objects in starlight sky and modern sensors are more sensitive than the human eye (permitting single photon detection even).

The illuminance will be dominated by a phenomenon known as airglow, an atmospheric illumination that exists even in the absence of light pollution

If you have a subscription to SPIE (I do not so I can't help here), there is a paper titled Night illumination in the visible, NIR, and SWIR spectral bands which will most likely give you the spectral radiance of the night sky under various conditions (full moon, no moon, with airglow, etc)

You will have to integrate the spectral radiance of the no-moon sky with airglow against the spectral response of your camera detector. Most camera detectors use silicon, however, your camera likely has an RGB filter, the specifics of which are camera dependent.

Your camera also most likely contains a broadband filter that cuts off IR light. This can be removed from most cameras with some technical know-how (and confidence). Removing the filters will increase your light collection ability. The response of a silicon detector goes out to 1 micron, whereas the human eye will cut off at about 700nm. Note that the IR light will register primarily as red on your camera (or as white if you remove the color filter).

With a sufficiently large lens (sufficiently low f/#) this will no doubt be possible.

This noteset indicates that the luminance under starlight will be on the order of $10^{-3} \ cd\ m^{-2}$, whereas sunlgiht is 10^5. 8 orders of magnitude difference.

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The apparent magnitude of the sun is -26.75, so 2.512^26.75 = 50,000,000,000 times brighter than the brightest star.

There are around 50 stars with a brightness within 1 magnitude (ie 50%) of the brightest star so roughly 1/billion-th the light from the sun for starlight.

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