Why is the instensity reduced?

1. Jan 20, 2016

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 dont 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).

2. Jan 20, 2016

Staff: Mentor

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.

3. Jan 20, 2016

kent davidge

ohh can you explain me how to do that integration?

4. Jan 20, 2016

Staff: Mentor

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

5. Jan 20, 2016

kent davidge

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

dI = Io cos² dφ
I = Io0 cos² dφ = Io0 0.5 cos2φ = Io 0.5 (cos2 x 2π) - (cos2 x 0) = 0.5 Io

Is it correct?

6. Jan 20, 2016

Staff: Mentor

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.

7. Jan 20, 2016

kent davidge

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

8. Jan 20, 2016

Staff: Mentor

$$\cos^2 A\neq \frac{1}{2} \cos(2A)$$
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.
I have no idea what you are doing here, but it is wrong.

9. Jan 20, 2016

kent davidge

so where I start to find this relationship
?

10. Jan 21, 2016

Staff: Mentor

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.

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