Question

100 lumen of light are incident on an interface which has 25% reflectance. At a point 3cm into the new medium, the amount of transmitted light is measured as 50 lumen. What is the absorptance of the material per cm?
0.75 * exp(-k*L) = 50/100 = 0.5

L = 3 cm
The 0.75 is the fraction that penetrates the surface.
Solve for k
exp(-kL) = 0.667
kL = 0.4055
k = 0.135 cm^-1

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