Why Does a Single Polarizer Reduce Light Intensity by Half?

[kent davidge]

When unpolarized light is polarized with two polarizers, the intensity is reduced in according to Malu's law. But... when unpolarized light is polarized with only one polarizer, the intensity is reduced to half the intensity of the unpolarized light. Why? I've read that unpolarized light had electric-field vectors in all possible directions, but I don't understand the idea of resolve them in only two components, one parallel to polarizer's axis and the other perpendicular to it. (sorry my bad english).

 

[mfb]

Which fraction other than 1/2 would you expect?

You don't have to choose a basis for resolving them into components: If you assume all polarization directions have equal intensity, and integrate, you get 1/2 as well.

 

[kent davidge]

ohh can you explain me how to do that integration?

 

[mfb]

Use Malus' law.
$$\frac{1}{\pi} \int_0^\pi sin^2(x) dx = \frac{1}{2}$$
The 1/pi is the normalization.

 

[kent davidge]

Ok, however, I find the same result by a different way:

dI = I_o cos² dφ
I = I_o ∫₀^2π cos² dφ = I_o ∫₀^2π 0.5 cos2φ = I_o 0.5 (cos2 x 2π) - (cos2 x 0) = 0.5 I_o

Is it correct?

 

[mfb]

Should be $\cos^2 \phi ~d\phi$, and "cos2 x 2π" does not work (cosine of 2? or of 4 pi?), and there are missing brackets. The prefactor in the first equation is wrong and happens to cancel the integration problems in the second line.

 

[kent davidge]

Sorry, why should I use φ. dφ ?
I rewrite my expressions in the correct form:

 

[mfb]

$$\cos^2 A\neq \frac{1}{2} \cos(2A)$$
You are missing a constant there. Only this constant leads to a non-zero integral later, the cosine term doesn't contribute to the integral.

cos 2(2π-0)

I have no idea what you are doing here, but it is wrong.

 

[kent davidge]

so where I start to find this relationship

cos2ϕ dϕ

?

 

[mfb]

That is not a relationship, and the equation you posted was wrong in the first place, so I don't get the point of your question.

$\cos^2 (d\phi)$ (what you posted earlier) does not make sense, that is not a proper integrand.