Why is unpolarized light halved in intensity when polarized?

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1. Oct 27, 2015

Elmer Correa

I would appreciate if the explanation didn't derive this phenomenon using Malsus' Law exclusively and would go into the horizontal and vertical components of polarized light, and how the blocking of the horizontal components results in a halved intensity...so more of a geometric explanation would be preferable. Also, if this has anything to do with it, as I understand it the amplitude of an polarized wave after passing through a polarizer is equal to cosθ where θ is the angle at which the wave is being polarized, and I don't quite understand why this is. On top of this amplitude squared is proportional to intensity, making the intensity then cosθ squared, the average of which is somehow 1/2. If this all ties into the geometric explanation please let me know. Also, how would I justify that the intesity is halved using a diagram?

2. Oct 27, 2015

andrewkirk

It can't be properly answered without considering components. Let $\vec{p}$ be the polarization direction of the polarizer, and assume that is perpendicular to the direction of propagation. For a given ray R let $\theta_R$ be the angle the ray's polarisation vector makes with $\vec{p}$. Assume $\theta_R$ has a uniform distribution in $[0,2\pi]$. Then integrate the component $A_R\cos\theta_R$ over the range of $\theta_R$.

3. Oct 27, 2015

robphy

http://www.wolframalpha.com/input/?i=integral(cos^2 x, x=0, 2*pi) equals pi
So, http://www.wolframalpha.com/input/?i=(1/(2*pi))*integral(cos^2 x, x=0, 2*pi) equals 1/2

To see this, note that
http://www.wolframalpha.com/input/?i=plot(sin^2+x,+x=0,+2*pi)
http://www.wolframalpha.com/input/?i=plot(cos^2+x,+x=0,+2*pi)
have the same area under the curve (and so these integrals are equal to each other)

Since http://www.wolframalpha.com/input/?i=integral(1, x=0, 2*pi) equals 2*pi (didn't need alpha to do that)
and $\cos^2 x+ \sin^2x=1$,
we must have
http://www.wolframalpha.com/input/?i=integral(cos^2 x, x=0, 2*pi) equals pi

4. Oct 28, 2015

blue_leaf77

Another alternative is to use the identifying relations for an unpolarized light. The perpendicular components of an unpolarized light satisfies
$$\langle E_x(t) E_x^*(t) \rangle = \langle E_y(t) E_y^*(t) \rangle = \langle A(t) \rangle$$
$$\langle E_x(t) E_y^*(t) \rangle = \langle E_y(t) E_x^*(t) \rangle = 0$$
Therefore the cycle averaged intensity of an unpolarized light is
$$I = \langle |E_x(t) + E_y(t)|^2 \rangle$$
$$= \langle E_x(t) E_x^*(t) \rangle + \langle E_y(t) E_y^*(t) \rangle + \langle E_x(t) E_y^*(t) \rangle + \langle E_y(t) E_x^*(t) \rangle = 2\langle A(t) \rangle$$
That's why if you blocked one component, say $y$, it will be $I' = \langle |E_x(t)|^2 \rangle = \langle A(t) \rangle = \frac{1}{2} I$.

Wouldn't that give zero?

5. Oct 28, 2015

andrewkirk

Yes, I forgot the bit about squaring it. I was in a bit of a rush.