# How many lumens is required for this?

1. Dec 9, 2003

### zoobyshoe

A light is visible to the unaided human eye at a distance of 70 miles. How many lumens is required for this?

Stipulating the light is equally bright in all directions, can we calculate from the above lumens a reliable estimate of the joules needed to produce a light of that strength?

-zoob

2. Dec 10, 2003

### suyver

Lumen is a fysiological quantity. That means it's defined by the eye-sensitivity. Therefore, you'd need to include the wavelength of the light.

For the energy required to produce the light: you'd have to include how the light was made and what the efficiency of the process was.

3. Dec 10, 2003

### zoobyshoe

What's the non-physiological scale of measurement I need here, then, candle-power?
Don't have that info.

Thank you.

4. Dec 10, 2003

### suyver

That would be the candela.

Then you really cannot say anything about the required power. For example: consider the electroluminescence of silicon and the thermal (black-body) radiation of a metal at about 1000 K: there is a factor of ~1,000,000 in power required to obtain equal brightness...

5. Dec 10, 2003

### zoobyshoe

Thanks, suyver.

This, then, is the wrong tree, and I must go bark elsewhere.

6. Dec 11, 2003

### Nereid

Staff Emeritus
Re: Lumens

If you turned the question around, you can get an estimate of the minimum amount of energy required (actually power, since you didn't say how long the light was visible for).

A sample calculation:
- assume the light is 'red', say 650 nm
- assume the human eye has a quantum efficiency of 5% (only 1 in 20 photons incident on the eyeball result in the firing of a neuron)
- assume the extinction of a column of 70 miles of air is x% (of the red photons which are emitted, only 1-x reach the eyeball)
- assume the human threshhold for detection is [see below]
- assume the source can turn input energy into red photons with 100% efficiency
- assume the light is tightly collimated, and spreads to a circle of radius 7cm at distance 70 miles (in other words, hits both eyeballs, and surrounding part of the face, but not much else)

... just plug in the numbers, with the appropriate formulae, and the answer will fall out. You can then modify the assumptions to suit whatever case you're interested in investigating (isotropic emission, solar-spectrum light source, eyes which aren't dark adapted, ...).

I've only a dimly remembered 'rule of thumb' that can get you to the human eye threshhold: at the top of the atmosphere, the number of photons from a zero magnitude star with an A spectrum is 10,000 per square cm per angstrom per second (probably V or B band). More reliably, the visual threshhold of the fully dark-adapted human eye (under truly dark skies) is 6.5 mag. If my 'A star' number is more or less correct, it's quite straightforward to calculate the number of photons, across the visible spectrum, that an eyeball needs to receive before the brain perceives a light.

Oh, if your source isn't a 'point', there are several effects which will need to be added to the above calculations.

Does this help?

7. Dec 11, 2003

### zoobyshoe

That's exactly the sort of thing I was looking for. I was certain there had to be ways to derive meaningful information from the fact of a light visible to the unaided eye at this distance. Your method of attacking the minimm required is really what I was hoping for. Thanks much.

8. Dec 12, 2003

### suyver

However, this lower limit will be very low indeed. Example: a 2 W output red laser (input: ~250 W) will be easily visible at 70 miles if there is no smog in the air.

Even in the non-columated case, a ~few kW halogen / xenon lamp (such as those used in IMAX theaters) will be visible over this distance under clear conditions. Just think about lighthouses...

By the way: if you want to assume equal distribution of the light in all directions, then you can simply use the following:

$$I(r)=\frac{I_0}{4\pi r^2}$$

where $$I(r)$$ denotes the intensity at distance $$r$$ from the source, if the source has an intensity $$I_0$$.

Cheers,
Freek Suyver.

9. Dec 12, 2003

### zoobyshoe

Point taken.

Thanks for the formula. That will be needed.

-Zoob