I=lo Nair=1 R NAbs=3 Oc Lambertian scatterer
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![Figure 1: Schematical view of an absorber layer and back reflector
NAir=1
R
L
Nabs=3
Oc
Oc
Lambertian scatterer
The absorber layer shown in Figure 1 has a Lambertian scatterer at the back side. The absorber layer
thickness is equal to D = 1 [µm]. The critical angle for light propagating from the absorber layer to air is
indicated in the figure by Oc.
A) What fraction of Ig is scattered into the critical angle Oc,Abs-Air ? Give a percentage %.
For the angles smaller than the critical angle, light can escape the absorber layer. The area under which light
can escape the absorber layer is called the escape cone, indicated by the red area in Figure 1.
B) Calculate the fraction of angles under which light can escape the absorber layer. Give a percentage. (Hint:
the fraction of angles is equal to the ratio of the area of the escape cone, to the surface area of the full
hemisphere with radius L into which light can be emitted.)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5713e60e-4214-429c-84eb-536c7ecdc24f%2F9a294b37-87ff-4963-bb46-1d9772780bb9%2Fmyes1zvb_processed.png&w=3840&q=75)
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