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Re: Solid Angles



Date:          Wed, 30 Aug 95 18:45:30 -0500
Reply-to:      radsafe@romulus.ehs.uiuc.edu
From:          "Dale E. Boyce" <dale@radpro.uchicago.edu>
To:            Multiple recipients of list <radsafe@romulus.ehs.uiuc.edu>
Subject:       Re: Solid Angles

Light etc. that is emitted such that it follows an invers square
law.  In this case the quantity of radiation or light per steradian
is a constant as a function of distance. (neglecting absorption).
Intensity per unit area is the quantity of interest in laser safety
i.e. watts per square meter.  If you know the output of the laser
in watts and its divergence you can calculate the watts per steradian
(which again is a constant as a function of distance).  If you want
something like the equivilant of the gamma constant you need to work
distance/area into the equation.  So you take the area of a sphere
at 1 meter (4 * pi * 1^2 m^@) and multiply by the watts per steradian.

Then to calculate the watts per meter squared for a given distance
you divide that calculated constant by the distance from the laser
squared.  Leaving the unit (which is really unitless) steradian in
the constant is confusing and I believe unecessary, but it gives the
reader a hint of what the constant means. 

Disclaimer, I am not a laser specialist, and my laser jock at work
is on vacation.  So, if I'm wrong...  Well consider the source [8-)

Dale E. Boyce
dale@radpro.uchicago.edu

I'm starting to become confused.  I'm not a laser type either, but I 
thought that laser light was distinct from isotropic light and 
therefore not subject to inverse square behavior. I can understand 
how inverse square might apply after reading the definition of a 
steradian( a conical angle defined by the area of the surface of a 
sphere of R^2 at a distance from the center of one radius( R)).  It 
seems that if the beam is coherent and divergence, i.e., scattering 
and attenuation are excluded, then the only thing that would affect 
the power density would be increasing the area over which the power 
is avaraged, i.e., increasing R. As R^2 increases  then power per 
subunit area of R^2 would decrease.  If I am understanding this 
correctly, then what's the purpose of the expression watts per square 
 meter per steradian?

             ***************************************
               Russ Meyer                            
               Internet: cmeyer@brc1.tdh.state.tx.us 
               tel: 512/834-6688                                                    
               fax: 512/834-6654
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