Which scatters light more: a sphere or a hemisphere?

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zheng89120
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In case the context, which this question is based needs to be provided, the context follows. A sub-project was being done on using sphere or hemisphere "scatterers" on TFSC (Thin-Film Solar Cells). These "scatterers" can be applied on the top of a TFSC in order so that the solar cells can direct light waves in a flatter "cone" after interaction, hence allowing the solar cells to be made thinner, and hence cheaper.

Now, back to the original question. Which shape might scatters light (regular solar light, centered around ~500 nm) better, according to classical Electrodynamics (and maybe some intuition), a sphere or a hemisphere? (Although, I suppose it might be harder for spherical "scatterers" to be "applied" to a TFSC than a hemispherical "scatterers".)
 
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In simple diagrams, which shape scatters light in a flatter "cone" (if that's the shape):

a) O<---
b) D<---

(O being a sphere, and D being a hemisphere)
 
zheng89120 said:
In case the context, which this question is based needs to be provided, the context follows. A sub-project was being done on using sphere or hemisphere "scatterers" on TFSC (Thin-Film Solar Cells). These "scatterers" can be applied on the top of a TFSC in order so that the solar cells can direct light waves in a flatter "cone" after interaction, hence allowing the solar cells to be made thinner, and hence cheaper.

Now, back to the original question. Which shape might scatters light (regular solar light, centered around ~500 nm) better, according to classical Electrodynamics (and maybe some intuition), a sphere or a hemisphere? (Although, I suppose it might be harder for spherical "scatterers" to be "applied" to a TFSC than a hemispherical "scatterers".)

My first reaction is to suggest that unless the radiation can somehow impinge on the "back half" of a sphere, there should be no difference--if both are positioned normal to the direction of radiation. A diagram would be helpful.
 
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As a loose guess, I would say the hemisphere since it has less focusing power.

Regardless, I don't think there is a straightforward answer (by straightforward I mean something that can be answered without some theoretical modelling).

Also, the material (dielectric, semiconductor etc.) will likely have a significant effect on the answer.

Claude.