Polarizing unpolarized light

[jdstokes]

Why does a polarizing filter transmit 50% the intensity of unpolarized light?

I would have thought that since only 0.5 of the electric field gets through, this would cut down the intensity by 0.5^2 = 0.25?

 

[Vanadium 50]

It may be easiest to think in terms of energy. If half the energy is in the horizontal polarization and half in the vertical, whhen I remove half, half is left.

 

[vanesch]

I would have thought that since only 0.5 of the electric field gets through, this would cut down the intensity by 0.5^2 = 0.25?

Well, not half of the E-field gets through. If you take a random orientation (uniformly distributed) between 0 and 90 degrees (noted by angle A) then the component that gets through is cos(A). We have to average cos(A) between 0 and 90 degrees then (or, in radians):

2/pi x integral(cos(A) dA between 0 and pi/2) = 2/pi.

 

[Creator]

Why does a polarizing filter transmit 50% the intensity of unpolarized light?

I would have thought that since only 0.5 of the electric field gets through, this would cut down the intensity by 0.5^2 = 0.25?

That's a great question;
the answer lies in Malus' law which states that the intensity of the transmitted light thru a polarizer varies as the square of the cosine of the angle...

I = I*[cos^2 (X)]

However, that is not the whole story. For an Unpolarized initial beam there are many linear polarization directions which are randomly oriented and so they have an average cos^2 value of 1/2.

Creator

 

[jdstokes]

Thanks all for your replies. I found a neat explanation that goes like this. Assume that the initial light is equally polarized in N random directions $\theta_i$. Then the intensity transmitted according to Malus' law is

$I = \frac{I_0}{N}\cos^2\theta_1 + \cdots = I_0 \frac{1}{N}\sum_{i=1}^N \cos^2 \theta_i$, which is just the average of cos^2 as N tends to infinity (i.e., 1/2).