# How do polarizing filters work

1. Apr 25, 2010

### HorseBox

If I'm not mistaken polarized light consists of light waves vibrating only in a specific direction because the waves vibrating in other directions have been filtered out. I watched a video on youtube and the guy showed how putting 2 polarizing filters together in a certain alignment so as they block the light altogether. Then he demonstrated something weird he put a third polarizing filter in between the other 2 filters and for some reason this counteracted the opacity of the series of filters so that the final filter actually let light through. The inner and outer filters where at 90 degrees to each other while the middle filter was at 45 degrees.

My first guess would be that the filter in the middle changes the vibration plane of the waves that pass through the first filter so that they are no longer blocked by the final filter. (EDIT: Watched that video again and he claims thats how it works)

Heres the video

What I don't understand is how it works. If I'm not mistaken light consists of waves oscillating in every single one of the 360 degrees but how many of those degrees does the polarizing filter block? I assume if it blocked everything except waves oscillating on 1 specific plane then the 2nd filter would only allow light to pass if it was aligned to the exact same angle of the 1st filter.

Last edited by a moderator: Sep 25, 2014
2. Apr 25, 2010

### Born2bwire

You can always decompose polarized light into the superposition of two polarizations that are normal to each other. For example, if I have light that is polarized along the 45 degree angle, it is the same as the superposition of light along the 0 and 90 degree angles (since when you add the vectors they will always point along the 45 degree angle).

When light passes through a polarizer, we filter out all but components of polarization along the polarizer. So if it only lets through 90 degree light, then only components of the 90 degree will pass through. Thus, if we pass a 45 degree polarized light through our 90 degree filter, since we can show that we can decompose the 45 degree light into a 90 and 0 degree superposition, we will see that the 90 degree component of the superposition will pass through while the fully normal 0 degree component gets suppressed.

Simply put, the light is polarized along the filter and the intensity of the light that passes through is proportional to the projection of the original polarization onto the filter's polarization (dot product). So if the original light is:
$$\hat{p} = \frac{1}{\sqrt{2}} \left( \hat{x}+\hat{y} \right)$$
The polarizer is along
$$\hat{f} = \hat{y}$$
Then the resulting light's intensity that passes through will be proportional to: (the intensity will be the square here, so the light is reduced by half)
$$\hat{f}\cdot\hat{p} = \frac{1}{\sqrt{2}}\hat{y}$$

So yes, as long as the light is not polarized perfectly normal to the filter, then some of the light will pass through polarized along the filter. However, the more the light is normal to the filter, the less the intensity of the resulting light that passes through.

$$\hat{p}_{final} = \hat{p}_{initial}\cdot\hat{p}_{90^\circ}\hat{p}_{90^\circ}\cdot\hat{p}_{45^\circ}\hat{p}_{45^\circ}\cdot\hat{p}_{0^\circ}\hat{p}_{0^\circ}$$
$$\hat{p}_{final} = \hat{p}_{initial}\cdot\hat{y}\hat{y}\cdot\frac{1}{\sqrt{2}} \left( \hat{x}+\hat{y} \right)\frac{1}{\sqrt{2}} \left( \hat{x}+\hat{y} \right)\cdot\hat{x}\hat{x}$$
$$\hat{p}_{final} = \frac{1}{2}\left(\hat{p}_{initial}\cdot\hat{y}\right)\hat{x}$$

EDIT: EDIT:

After watching the video I have no idea why the poster thinks that this is unexplained in classical physics. This is perfectly within the explaination of classical electromagnetics.

Last edited: Apr 25, 2010