Re: Light gathering cone?


In article <1169828484.613273.227470@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Boxman <boxman@xxxxxxxxxxx> wrote:


On Jan 26, 9:07 am, "Boxman" <box...@xxxxxxxxxxx> wrote:
1) Does this device have a name?

2) Has anyone analyzed the problem? While there is no "magic angle",
it would seem that for a given ratio of exit size to entrance
size (i.e., gathering all of the light falling on a cirlce 2cm
across and concentrating it onto a detector 0.5cm across) there
is probably an "adequate angle".

Just a note on the angle you mention here. In the 2D section, the
angle over wich 100% of the light will be delivered to the target plane
can be found by taking the arcsin(target diameter/input diamter) which
in your example would be around 14.47 degrees. Rays greater than that
angle will be reflected back out.

Understood. But suppose the CPC is made of solid glass (yes, reflective
losses on the entrance surface, but let's say that some sort of glass
surface was inevitable anyway, so that doesn't represent an additional
loss). Now, all rays that exist inside it are at an angle less than the
critical angle. So I can make this the design acceptance angle without
losing anything. According to this book...

http://www.powerfromthesun.net/Chapter9/Chapter9new.htm#9.2%20%20%20%20Compound%20Parabolic%20Concentrators%20(CPC)

....the concentration ratio is one over the sine of half the acceptance
angle, which is exactly the reverse of how I just calculated the acceptance
angle from the index of refraction, so the concentration ratio is the
same as the index of refraction. I won't be able to concentrate from a
2cm circle onto a 0.5cm detector, but I can get the light to be more
intense without sacrificing any of it. I believe that the book figures
the concentration ratio assuming a linear trough; if I have a round CPC I
think the cocnentration ratio should be the square of that.

In other words, if my detector was forced to be behind glass anyway,
and is not free to track the light source, then I think I could get a
geometric concentration ratio equal to the square of the index of the
glass without sacrificing any light that I would not have lost anyway.
Does my reasoning make sense?

I figure that in practice, the concentration ratio could be tightened
up further by assuming that any light hitting the glass at a shallow
angle is reflected so badly (even with an AR coating) that it can be
written off with no great loss; then the acceptance angle can be made
a little narrower.

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