Laser Spectroscopy: Calculating Transmitted Intensity

[senobim]

Homework Statement

Unpolarized light of intensity I0 is transmitted through a dichroic polarizer
with thickness 1mm. Calculate the transmitted intensity when the absorption
coefficients for the two polarizations are α|| = 100cm −1 and α ⊥ = 5cm −1

Homework Equations

Is it possible to say what would be transmitted intesity for unpolarized light from this info?

The Attempt at a Solution

Is it valid to say that α(unpolarized) = α || + α ⊥?

 

[gneill]

You've presented three questions but no Relevant equations or solution attempt.

What equation(s) are relevant to intensity for polarizers, especially dichroic polarizers? Check your course text or notes, or perhaps do some research on the web to find out.

 

[senobim]

I know how to calculate transmited intensities in horizontal and vertical polarization (Bouguer Law), just not sure about unpolarized light case

I = I_{0}\varepsilon^{-\alpha l}

 

[gneill]

The unpolarized (or natural) light source is treated as a mix of two mutually incoherent beams of equal amplitude that are polarized in mutually orthogonal directions. If you choose these directions to coincide with the || and ⊥ axes associated with the coefficients of your polarizer then you can apply your Beer-Lambert-Bouguer law.

Take a look at page 850 of this reference (Thanks to Google Books)
Principles of Optics: Electromagnetic Theory of Propagation, Interference ...
By Max Born, Emil Wolf

In particular note the exponents of e that contain the coefficients and how the resulting intensities add.

 

[senobim]

Thank you very much!

What does these coefficiants means exactly?

 

[gneill]

Thank you very much!

What does these coefficiants means exactly?

They set the rate of extinction of the light intensity with distance through the medium (the polarizer in this case). They're your absorption coefficents mentioned in the problem statement.

 

[senobim]

I = I_{0}\varepsilon^{-2\alpha' l}

why there is 2 before these? It is mentioned in your ref book, that's there I've lost ;]

 

[gneill]

I = I_{0}\varepsilon^{-2 \alpha' l}

why there is 2 before these?

I am not certain as I haven't attempted a derivation. But I suspect that it has something to do with splitting the total intensity into two.