Simple Problem Involving Light Absorption
aruzinsky
problem before? I independently thought of it because it is a
simplification of a practical problem I have.
Consider two flat, infinite, facing, parallel surfaces, labeled A and
B, with a point source of light between them. The reflectivity of A
and B are respectively a and b . What fractions, alpha, beta, of the
light are respectively absorbed by A and B?
My solutions are:
alpha = 0.5*(1 - a)*(1 + b)/(1 - a*b)
beta = 0.5*(1 - b)*(1 + a)/(1 - a*b)
Helmut Wabnig
eventually, after numerous reflections between the surfaces,
virtually all of the light will be absorbed.
Methinks your problem description is missing something.
w.
jbriggs444
> On Thu, 3 Sep 2009 07:36:52 -0700 (PDT), aruzinsky
>
Nope. He's presented problem and solution correctly. My solution
matches his. I'll let you work on proving that the two expressions
add up to 1. [Not too hard -- lots of nice cancellation]
He's assuming that the reflectivity is independent of the angle of
incidence and that the point source emits symmetrically.
So half the light goes toward a and then bounces from a to b to a to b
until eventually being absorbed.
And half goes toward b and bounces in a similar fashion.
The amount of light absorbed by a is
50% * (1-a) of the emitted light (absorbed on first strike)
+ 50% * (1-a) * a*b (absorbed on third hit)
+ 50% * (1-a) * (a*b) ^2 (absorbed on fifth hit)
+ ...
+ 50% * b * (1-a) (second hit)
+ 50% * b * (1-a) * a*b (fourth hit)
+ 50% * b * (1-a) * (a*b)^2 (sixth hit)
+ ...
An infinite series with first term x and ratio r has sum x * 1/(1-r)
So that gives us ( .5 * (1-a) + .5 * b * (1-a) ) / (1 - ab )
= .5 * (1+b)(1-a) / (1 - ab)
By symmetry, side b absorbs .5 * (1+a)(1-b) / ( 1 - ab )
aruzinsky
> On Thu, 3 Sep 2009 07:36:52 -0700 (PDT), aruzinsky
>
> >Please, check my problem statement and solution. Have you seen this
> >problem before? I independently thought of it because it is a
> >simplification of a practical problem I have.
>
> >Consider two flat, infinite, facing, parallel surfaces, labeled A and
> >B, with a point source of light between them. The reflectivity of A
> >and B are respectively a and b . What fractions, alpha, beta, of the
> >light are respectively absorbed by A and B?
>
> >My solutions are:
>
> >alpha = 0.5*(1 - a)*(1 + b)/(1 - a*b)
>
> >beta = 0.5*(1 - b)*(1 + a)/(1 - a*b)
>
> eventually, after numerous reflections between the surfaces,
> virtually all of the light will be absorbed.
alpha + beta =
0.5*[ (1 - a)*(1 + b) + (1 - b)*(1 + a) ] / (1 - a*b) =
0.5*[ 1 + b - a - a*b + 1 + a - b - a*b ] / (1 - a*b) =
0.5*[ 2 - 2*a*b ] / (1 - a*b) = 1
So, what was your point?
>
> Methinks your problem description is missing something.
>
> w.
Should I specify that there is no fluorescence, no phosphorescence and/
or no luminescence at the surfaces?
Helmut Wabnig
The ratio of absorption of A and B will be inverse
to their reflection factors a and b.
The side which reflects better will absorb less.
No need to calculate complicated series.
The problem is either mis-stated or it is a pseudo-problem.
Note that he did not define mirrors, which would
pass a percentage of light through,
he just defined 2 surfaces which absorb more or less.
In sum A and B absorb 100 %, and the ratio is b/a.
w.
aruzinsky
>
> - Show quoted text -
Thanks, that is identical to my derivation. I should have specified
no transmittance.
The practical application is growing plants under artificial light in
a reflective environment. Photosynthetically Active Radiation (PAR)
(see http://en.wikipedia.org/wiki/Photosynthetically_active_radiation)
is measured in non-reflective environments and lamps for plant growth
are selected by PAR watts. I had the epiphany that, in a perfectly
reflective environment, all wavelengths of light are completely
absorbed by plants and therefore PAR is relatively meaningless in such
an environment. For example, green light, which is ~20% absorbed by
plants in a nonreflective environment, is 100% absorbed in a perfectly
reflective environment. Since a perfectly reflective environment
doesn't exist, I formulated this problem to model the situation in
which plants are grown under broad canopy of fluorescent lamps with
reflectors behind them.
Helmut Wabnig
<jbrig...@gmail.com> wrote:
The ratio of absorption of A and B will be inverse
to their reflection factors a and b.
The side which reflects better will absorb less.
No need to calculate complicated series.
The problem is either mis-stated or it is a pseudo-problem.
Note that he did not define mirrors, which would
pass a percentage of light through,
he just defined 2 surfaces which absorb more or less.
In sum A and B absorb 100 %, and the ratio is b/a.
This is, because he puts a point source lamp symmetrically
between the plates. Quite different is the situation when we
have 2 parallel mirrors and we shine a laser beam sideways
between the plates under a certain angle and the beam
undergoes a reflection first on A, then on B then on A again
and so forth.
w.
Helmut Wabnig
Understood.
Differs of course from shining a single laser beam
at a certain angle between two parallel mirrors,
which was my first idea.
w.