How does the Beer Lambert law work at non-normal incidences?

  • #1
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Main Question or Discussion Point

I want to calculate the intensity of a transmitted wave in a medium, but not at normal incidence. I want to consider the case of 45 degrees. What proportion of the intensity of the wave is lost by reflection?

Specifically I am looking at a the reflection off of a thin film of aluminum deposited on a glass substrate.
 

Answers and Replies

  • #2
DrDu
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That's not so much a question about Lambert Beer. Rather you have to use the Fresnel equation
http://en.wikipedia.org/wiki/Fresnel_equations
with some expression for the complex index of refraction of aluminium, to first determine the amount of reflected light from the aluminium surface. The latter one can be found e.g. in the Handbook of physics.
 
  • #3
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Thanks Dr Du. Since my light source is going to be unpolarized, the total reflectance is going to be (Rs+Rp)/2. Since both Rs and Rp are dependent on n2, would I just use the real index of refraction to calculate the amount reflected, and then when calculating what is attenuated, use the complex part as well?
 
  • #4
DrDu
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No, reflectivity also depends on the imaginary part,too.
 
  • #5
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Okay, so I calculated Rp+Rs / 2, and I got 93% reflectivity, but that doesn't take into account the thickness of the film. I think it's more appropriate as a way to model a thick film where practically nothing makes it through the aluminum film.

EDIT: Or is it just impossible, regardless of the film thickness?
 
Last edited:
  • #6
DrDu
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93% reflectivity sounds at least reasonable. The remaining 7% will get either absorbed or transmitted. Given the thickness of the film and the angle of refraction you calculated, you can calculate the effective path length in the aluminium film and use Lambert Beer to calculate what reaches the lower surface. There you should take reflectivity into account again, to calculate the final intensity in the glas.
 

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