Re: Excuse me Sir, can you see the laser beam?

chibitul <[EMAIL PROTECTED]> wrote in message news:<[EMAIL PROTECTED]>...
> Hi,
> 
> I have a laser diode, looks red, probably about 660 nm, but I don't know 
> the power. The laser spot is well colimated and easily visible in 
> average lit room. Haven't had the chance to try outside in sunlight.
> 
> Now this colleague of mine is worried that after reflexions on several 
> mirros (which are optimized for a different wavelength) the intensity 
> becomes too weeak and the laser beam is not visible anymore when it hits 
> the target. I estimate the reflectivity of these mirrors at the laser 
> wavelength to be 40%. So after 2 mirrors, I have 16% left. I would argue 
> the beam is easily visible, since the eye has such an incredible dynamic 
> range (11 or so???). But he really wants to do the experiment to make 
> sure the beam is still visible. Fine. He has a point. But my question is:
> 
> How do you defind the visibility treshold of a laser beam onto a target? 
> When is the beam too weak to be considered visible by most observers?
> 
> rant on!

Others will no doubt have better sources, but just off my shelf:
"Optics" 1988, by K.D. Moller states (when speaking of the human eye)
on page496 "...The ultimate sensitivity ... is given as 6 x 10^(-13)
W/m^2, corresponding to 100 light quanta in a second."  Peak spectral
sensitivity is ~555nm.

Extremely rough calculation:

Assume a 1mW laser diode = [1 x 10^(-3)]W
2 reflections reduces this to 1.6 x 10^(-4) W

What is the material onto which the collimated laser beam is
impinging?  You didn't say.  Let's suppose it absorbs 50% of the
light, diffusely scatters the rest.

Power becomes 8.0 x 10^(-5) W due to absorbtion.

Now, suppose this power is uniformly scattered over 2pi steradians.
[This does not represent a real case.  A perfect Lambertian diffuser
will scatter as a function of cosine of the scattering angle, as
measured from the normal. Real scatterers do other things.]

But, just *suppose* the scattering is perfectly uniform over half a
sphere.

The question is, how close is the observer, with say a 5mm diameter
eye pupil? (As I recall, the normal fully dilated human eye pupil has
a diameter of ~7mm)

Assume the observer is 1 meter away.  A 5mm diameter circle is ~3.1 x
10^(-6) times half the area of a 1 meter radius sphere. So, the power
the eye receives is 3.1 x 10^(-6) * 8.0 x 10^(-5)W ~2.5 x 10^(-10)W,
or about 400 times the earlier cited threshold.

But, the human eye response at 670nm is less than at 555nm, so the
factor is smaller.

But, for a more realistic diffuse reflection, it depends on what angle
relative to normal the observer is viewing the spot.  If the observer
is viewing at anywhere within ~~~45 degrees from normal incidence, the
intensity is higher, so the factor will be higher. If she's near
grazing incidence, the factor will be much lower.

But, this does not take into account ambient conditions which will no
doubt reduce the signal to noise ratio a lot unless you are in total
darkness except for the laser.

But, supose the oberver is much closer or much farther than 1 meter.

But, suppose your laser is 10mW.

But ...

Okay, enough already! You can see we get to "if to the fifth" really
fast here. Because the very rough calculation only gives less than 2
orders of magnitude safety factor above the bare minimum sensitivity,
I sure wouldn't count on an estimate if more than a few nickels were
at stake. Do the experiment! :)

Spencer
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Spencer D. Luster, Managing Member
LIGHT WORKS, LLC
333 N. 14th Street
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