Understanding the Beer-Lambert Law and Its Derivation

[Carlos de Meo]

Hi Guys
Studying some optical properties of materials and found this equation on Callister
T = I₀*(1-R)²*(exp(-βx)
Where T is transmission, R reflection, β the absorption coefficient and x the lenght.
In the same chapter, the author says that this equation´s derivation is homework
I tried to derive it but really can't understand why it is (1-R)²
My attempt:
I₀= I_R+I_A+I_T
= I₀*R +I₀^´*exp(-βx)+I₀^´*(1-exp(-βx))
I₀^´=I₀*(1-R)
Am i missing any variable here or forgetting some efect?

 

[Charles Link]

Hi Guys
Studying some optical properties of materials and found this equation on Callister
T = I₀*(1-R)²*(exp(-βx)
Where T is transmission, R reflection, β the absorption coefficient and x the lenght.
In the same chapter, the author says that this equation´s derivation is homework
I tried to derive it but really can't understand why it is (1-R)²
My attempt:
I₀= I_R+I_A+I_T
= I₀*R +I₀^´*exp(-βx)+I₀^´*(1-exp(-βx))
I₀^´=I₀*(1-R)
Am i missing any variable here or forgetting some efect?

I think the answer is, it got into the material (transmission factor of $(1-R)$ with fraction of energy $R$ being reflected), and then it traverses the material with exponential attenuation with distance, and then a factor (=another factor) of $(1-R)$ for the energy that gets out of the material. $\\$ It turns out, the energy reflection coefficient $R$ is the same regardless of whether the reflection begins from outside the material or inside of it. $R=(n_1-n_2)^2/(n_1+n_2)^2$. (The Fresnel reflection coefficient for the $E$ field is $\rho=E_r/E_i=(n_1-n_2)/(n_1+n_2)$ and intensity (energy) $I=nE^2$ (in units that the optics people use). The result is the energy reflection coefficient $R=I_r/I_i=E_r ^2/E_i ^2=\rho^2=(n_1-n_2)^2/(n_1+n_2)^2$ regardless of which order $n_1$ and $n_2$ become encountered.) $\\$ Additional item is that the problem assumes that no surface absorption occurs. Any absorption is a loss that occurs as the light beam traverses the material. It also makes the approximation that the contribution of transmitted beam resulting from multiple reflections is minimal. And it doesn't address at all the coherent case of multiple reflections that can occur where the surfaces are extremely parallel in which case a wavelength dependent interference can occur with the multiple reflections that is known as the Fabry-Perot effect.

 

[Carlos de Meo]

But doing the calculations and adding now the second reflection at the second interface, i still don´t get how Callister made the (1-R)². Now i get a value of (1-R)/(1+R) instead of the value above

 

[Charles Link]

But doing the calculations and adding now the second reflection at the second interface, i still don´t get how Callister made the (1-R)². Now i get a value of (1-R)/(1+R) instead of the value above

Energy transmission factor $T=1-R$. The material is considered to be a dielectric slab, perhaps 1" thick. You get one factor of $T$ as you enter the surface, (you lose a factor $R$ ), and one factor $T$ as it exits on the other side. Dielectric interfaces always cause partial reflections (and partial transmissions) as the light enters a new material=e.g. a glass air interface. When you pass the critical angle you can get total internal (100%) reflection, but that won't occur with parallel faces. Also, at the Brewster angle for parallel polarization there is 100% transmission, but otherwise, in general, dielectric interfaces cause partial transmissions and partial reflections. How did you get a $1/(1+R)$?. This one is really quite simple=it's simply a second (multiplying) factor of $T=1-R$. The attenuation factor, $e^{-\alpha x}$ is a transmission factor as well even though it is (loosely) called an attenuation factor. These factors are multiplied together. They do not add together. You multiply the 3 energy transmission factors to get the resulting energy transmission.

 

[Charles Link]

@Carlos de Meo I edited post #4. Please read the latest version.

 

[Carlos de Meo]

Oh, now that makes sense. Thank you very much @Charles Link