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by Bart Arondson

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In the past, I've asked about taking photos of luciferase. Now I'm curious how weak of a light source I can detect. From How many photons per second is one Lumen? on Physics Stack Exchange, I can determine how many photons/sec makes one lumen. What then is the minimum amount of lumens needed for a camera to detect?

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I think this is already answered at photo.stackexchange.com/questions/16427/… –  James Youngman Mar 27 '12 at 21:00
    
@JamesYoungman. Good point. Although I'm now curious about an imperfect sensor. –  bobthejoe Mar 27 '12 at 23:03

2 Answers 2

up vote 12 down vote accepted

Note: The dark adjusted human eye can detect a single photon!

Short: About 5 picolumen per pixel with the best commercial DSLRs such as a Nikon D3s.

Long :-) :

Minimum detectable light source will depend on camera and how much of the image area the source occupies. For best detectability, a source will be "brightest" if all it's energy arrives in a one pixel area. The image will not be very interesting in most cases :-).

But, to attempt to put a very approximate empirical answer to the question:

I'll make various assumptions along the way and summarize them at the end so they can be adjusted as desired.

1 EV is a bit above bright Moonlight and is correctly exposed at ISO 100 at f1 for 1 second.

1 EV = 1 lux = 1 lumen per square meter.

I'll avoid the temptation here to leap into steradians and candela and stick with more intuitive empirical terms :-).

Let's assume you are using a Nikon D3s which has a 12 megapixel sensor that can just about see in the dark with no photons at all.
At about 100,000 ISO and an exposure of one second at f1 at 1 EV and dark field subtraction you may perhaps have difficulty detecting whether a given pixel was illuminated or not as even a D3s is getting somewhat noisy. At around 12800 ISO there would be little doubt.

If you set your camera to image 1 square meter then then the 1 EV lighting will be providing 1 lumen total so the 12 million pixel sensor will be accepting ~1/12,000,000 th of a lumen per pixel.

That's at f1 and ISO 100 and 1 second exposure.
Increase ISO to 12800 as above and you can detect 1/12800th less light again.
1/12 million x 1/12800 ~= 6.5 x 10^-12 lumen = 6.5 picolumen.
I don't think I've seen picolumen used before :-)

So, if, all of:

  • You use an f1 lens

  • Your camera can image at ISO 12800 for one second at 1 lux or 1 EV and produce a discernible change in a given pixel

  • You have a 12 megapixel sensor

Then you can DETECT about 5 picolumen **in a single pixel area.
A Nikon D3S should do thus with relative ease.
Longer exposure times will produce increased sensitivities but in time noise will catch up with even a D3s.

Over the whole 12 megapixel sensor his corresponds to 78 microlumen which is 1/12800 th of a lumen total which is no surprise as it is just the inverse of the ISO setting when imaging a square meter at 1 lumen per square meter.

If you vary imaged area, aperture, ISO, sensor pixels, exposure time or camera capability then the answer will vary accordingly.

The biggest gain you can make with a given sensor is to cryo cool it.
And then there are advanced photo multiplying sensors that take the question away from the realm of "normal photography". eg Electron Multiplying CCD, Frame Transfer CCD, Intensified CCD, ...

See also:

Wikipedia astrophotography


Added: Sensor dynamic range -

It's been suggested that some cameras are superior to the D3s in light detection ability per pixel. For cameras which are actually available to take real photos wih, The D3s is still the king of low ISO. See DxOMark's evaluations here.
They explain their reasoning and method on the above site.

In per pixel terms raher than whole image the closest contenders are actually worse than shown by a factor of square_root(Mp/12) where Mp is the megapixel rating of the comparative camera and 12 = D3s 12 Mp. eg 36 Mp D800 is worse than shown by a facor of sqrt(36/12) = 1.7 on a per pixel basis.

enter image description here


Added:

For those who have the time and patience to wade through a long thread, this DPReview user discussion discusses sensor dynamic range and MUCH more related material. There are several quite capable and high powered people there banging heads together rather solidly but they largely seem to come to a good degree of agreement overall.

Max dynamic range of a sensor is one of their easy topics. It is generally agreed that gains due partially due to dithering allow up to +1.8dB more dynamic range than ADC bits.

Note that you can get more again IF the sensor is better than the resolution of the ADC used BUT if the ADC is either more precise than its number of bits ie LSb = 1.0 or 1.00 and not just 1. in actual accuracy OR if the ADC is stable in its results, regardless of actual accuracy. In such cases the addition of controlled noise of designed characteristics can allow more bits to be extracted from the ADC than it notionally possesses.

Here is a NatSemi (now TI) application note which covers the subject well.
TI literature Number: SNOA232. Here -> Improving A/D converter performance using dither

National Semiconductor Application Note 804, Leon Melkonian, February 1992

See figure 12, page 5, where the addition of optimum dither noise allows recovery of a 1/4 LSB amplitude signal - effectively adding 2 bits to the ADC!!!.

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We have some details on how steradians bring geometry into the equation here: photo.stackexchange.com/a/14721/124. –  jrista Mar 28 '12 at 14:59
    
We have much more in-depth information about steradians and luminance here: photo.stackexchange.com/a/10183/124 –  jrista Mar 28 '12 at 15:00
    
+1 Excellent answer. :) Might want to add a couple notes about sensor INefficiencies. Technically speaking, you would only be able to gather 1/12800th of a lumen assuming perfect behavior, perfect, even distribution of light, etc. There are quantum efficiencies to consider, the random nature of photon shot noise, etc. that would eat into sensitivity well before electronic noise affected things, and there is still electronic noise to consider. –  jrista Mar 28 '12 at 15:30
    
This is pedantic, but both common usage (including Wikipedia and commercial dictionaries) and the NIST standard on SI disagree with you over the notion that 'lumen' is plural. –  mattdm Mar 28 '12 at 17:44
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@jrista - Thanks - extra material useful. I tried to keep the answer in the empirical and experience based area rather than very much of the underlying physics. The D3s seemed like a good reference as it has the best low noise sensor at the pixel level so far available for DSLRs and I (even though I've never even hel one :-) ) can make some reasonable assessments of how it would perform in the current context. –  Russell McMahon Mar 28 '12 at 23:07

It all depends on your exposure length! I've successfully photographed nebula with a telescope attached to my camera. Using imaging sensors I've measure the black of an LCD display. (Not with any huge accuracy but hey, the sensor was picking up photons...) Exposures were on the order of minutes to 30 seconds.

I read your earlier question on luciferase. I take it you've not been able to shoot it yet? Can you ask those who did and see what they did or is this question just for general knowledge?

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Thanks for asking. I have been able to shoot it. Now I just want to be able to shoot less of it. –  bobthejoe Mar 27 '12 at 22:53
    
Longer exposures. Modern sensors. My GF's Canon 7D can easily do ISO 2000 without too much noise, my older, Canon 40D balks around 400-800. And the new cameras are even better. Do you have any sample shots, I think I'm not alone in wanting to see some! –  Paul Cezanne Mar 27 '12 at 23:37

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