Natural Light and Linear Polarizer

[Frank Einstein]

Hi, In the first place I want to apologize for my bad English. In the second, I have a concept doubt, If I make natural light to go on a linear polarizer of angle 0º, the intensity, or irradiance here, It would be I₀/√2 and it's Jones vector would be (1, 0) (as column). What would happen If I put infinite polarizators which angles were 0º for the first and Pi/2 for the last, changing infinitesimal angles in between?

Would I have light which irradiance would be 0 at the end?

That's what I think that might happen, because the first and last polarizators have an angle of Pi/2 between them, so the irradiance would be 0.

 

[Drakkith]

Nope, you'd actually get light through. You can see this by putting a polarizer at a 45 degree angle in between to polarizers set to 0 and 90 degrees. How much gets through I'm not certain.

 

[A.T.]

Would I have light which irradiance would be 0 at the end?

I think that, if the polarizators are ideal, you would still have I₀/√2 at the end.

 

[Frank Einstein]

Yes, that's right; if you use the fact that I_i+1=I_i*cos^2(b) where b is the differencie between the angles of polarization, adn is constant, you get an infinite sum which result is one.

 

[M Quack]

The electric field gets reduced by cos(b). For one filter at 45 deg with respect to the previous one that is 1/sqrt(2), and for two steps 1/2.

The intensity is proportional to E^2, so if E gets reduced by 1/2, the intensity drops to 1/4. If you then assume that the original light was unpolarized the first filter already cuts out 1/2, and the final intensity is 1/8 of the unpolarized intensity.

Yes, that's right; if you use the fact that I_i+1=I_i*cos^2(b) where b is the differencie between the angles of polarization, adn is constant, you get an infinite sum which result is one.

It is not a sum, it is a product. Since all factors are the same the result is trivial.

If you have N steps, than at each step the E field gets reduced by cos(90 deg/N), i.e. (cos(90deg/N))^N in total for the E-field, and (cos(90deg/N)^(2N) for the intensity, with the additional factor 1/2 for the unpolarized incident light.

 

[Andy Resnick]

Nope, you'd actually get light through. You can see this by putting a polarizer at a 45 degree angle in between to polarizers set to 0 and 90 degrees. How much gets through I'm not certain.

The transmitted intensity is 25% of the incident intensity.