[Laser] Spherical geometery

[Stelios Passas]

Why you want this? It is simpler.
Since the bulb is a nearly isotropic source, it radiates all its power
"uniformly" in a sphere. The area of the pupil of your eye catches a part of
this energy.
Calculate what percentage is the area of your eye-pupil to the area of a
sphere with radius the distance between you and the bulb. That will give you
how much of the radiated 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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