Optics problem --- absorbtion index

[marcis]

Hello
I have an optic problem

I want to calculate absorption index

I know transmittance by 90% ;refraction index to material 1.586 ; thickness 2 mm ; power of source 21 lm

 

[ehild]

The transmittance of a weakly absorbing slab can be approximated by

T=\frac{(1-R^2) e^{-\alpha d}}{1-R^2 e^{-2\alpha d}}
where
$R=\left(\frac{n_0-n}{n_0+n}\right)^2$
and α is the absorption coefficient. It is related to k, the imaginary part of the complex refractive index : $\alpha = \frac{4 \pi k }{\lambda }$

 

[marcis]

The transmittance of a weakly absorbing slab can be approximated by

T=\frac{(1-R^2) e^{-\alpha d}}{1-R^2 e^{-2\alpha d}}
where
$R=\left(\frac{n_0-n}{n_0+n}\right)^2$
and α is the absorption coefficient. It is related to k, the imaginary part of the complex refractive index : $\alpha = \frac{4 \pi k }{\lambda }$

thank you for your answer but you are not specific

 

[ehild]

thank you for your answer but you are not specific

What is not clear? no is the refractive index of the ambient (air). n is the refractive index of the slab, d is its thickness. λ is the wavelength. k is the extinction coefficient. If you know the transmittance, the refractive index and the thickness, and taking no=1, you can calculate α. If you know α and the wavelength, you can get k. I think, it is you want to calculate.
You need to measure the transmittance with as high accuracy as possible. At least 0.1% accuracy is needed.

 

[marcis]

What is not clear? no is the refractive index of the ambient (air). n is the refractive index of the slab, d is its thickness. λ is the wavelength. k is the extinction coefficient. If you know the transmittance, the refractive index and the thickness, and taking no=1, you can calculate α. If you know α and the wavelength, you can get k. I think, it is you want to calculate.
You need to measure the transmittance with as high accuracy as possible. At least 0.1% accuracy is needed.

now is clear thanks