[Laser] Spherical geometery

[Stelios Passas]

Doing the same for the laser is easy too.
Since the laser beam is not isotropic, it is actually a cone. again, all the
energy falls at the base of the cone that exists where you stand. Calculate
the area of the pupil of your eye to the area of the base of this cone at
your distance from the laser source, and again you have how much power
enters your eye.
Stelios Passas

----- Original Message -----
From: <G0MRF@aol.com>
To: <laser@mailman.qth.net>
Cc: <G0MRF@aol.com>
Sent: Monday, September 16, 2002 3:12 PM
Subject: [Laser] Spherical geometery


Hi.

A week ago someone asked me what would the equivalent output power of a
light
bulb be when compared to my 5mW Laser when the bulb and laser were viewed at
a distance.

Having rehashed the question a little, the member of my amateur radio club
was trying to find out the power of a normal household filament bulb ( near
isotropic source) which would appear as bright as my 670nm laser at a
distance.

Ignoring, for the moment, the different eye sensitivities to wavelength, I
have tried without success to find a formula relating the divergence of a
laser to the area of a sphere. I hope to use it to calculate the percentage
area illuminated and hence make a comparision between filament bulb and
laser.

So, my question: Does anyone on the list have a formula which relates
divergence in milli-radians to the area of a sphere?

(In future I'll just say "no idea" )



73

David  G0MRF
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