Why is the intensity of polarised light equal to half of the incident light?

[nolanp2]

the intensity of polarised light is equal to half of the intensity of the light before it has been polarised, and this has been explained to me as being a result of half of the incident light being parallel and half perpendicular to the plane of polarisation, and so half passes through and half is blocked.

This does not make sense to me, as i would have thought that most of the light would be incident at all angles between 0 and pi, and only a very small amount would happen to be parallel to the P of P.

Can somebody help me out with a better explanation please?

 

[Meir Achuz]

The passed intensity varies like cos^2. The simple half and half argument works because the average value of cos^2 is 1/2.

 

[nolanp2]

i see. so can you explain exactly where the cos squared functions come from, possibly in terms of the electric field direction of the light?

 

[jtbell]

Suppose the light travels in the z direction. Then the electric field \vec E is perpendicular to the z-direction, at some angle \theta with respect to the x-direction.

Now suppose the polarizer axis is along the x-direction. Then the component of the electric field along the x-direction (E \cos {\theta}) makes it through, but the component along the y-direction (E \sin {\theta}) is blocked.

Finally, in general, the intensity of the light is proportional to the square of the electric field in the wave.

 

[daniel_i_l]

You can split the EM radiation of light into x and z components. by symetry these components should be equal. as jtbell said, the amount blocked by the polarizer on each axes is relative to the angle of the polarizer and the intensity is the square of the radiation.