physics

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?

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  1. 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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    posted by drwls

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