Understanding Polarizing Angle: Explained by an Engineering Student

[kdm06d]

Hey--

Im an engineering student having trouble with the polarization concepts.

So let's say you have a ray of light traveling and it travels through a polarizing film with its respect to the horizontal, now only the x-component of the light vector passes through...

I believe my understanding of this is wrong..But understand I do know the idea behind the ray being an oscillating wave etc...

So first off, how can the polarizing angle only allow an x-component of something which is oscillating up and down, without preventing further oscillation?

Secondly, in the classic case of three polarizers with the middle being rotated to allow light to pass through all three, my confusion deepens.

If my previous theory were correct (which I do not believe it to be)... Then the first film would allow let's say only the x component through. Then does the middle 45 degree tilted film somehow rotate the x-direction moving vector of light to more of an x=y type slope, then the third allows only the y-direction vector through?

I do not believe I am going about thinking of this in the proper manner, I was hoping someone on here could understand my view point and perhaps see where I am misunderstanding...

Thanks for any help

 

[kuruman]

Think of a linear polarizer as allowing only the component of the electric field parallel to its axis to go through. Also remember that the intensity of a light wave is proportional to the square of the magnitude of the electric field. It's best to see what's going on with some examples. In all cases we have a polarizer lined up in the $x$-direction. This means that only the $x$ component of the electric field will pass through, that is all the light that passes through is polarized in the $x$-direction. The question is how much light makes it through?

Case I. Incident light is already polarized and the electric field vector forms angle $\alpha$ relative to the $x$-axis.
Let $E_0$ be the electric field vector magnitude. Then the initial intensity is proportional to $E_0^2$. The component of light that makes it through is $E=E_0\cos\alpha$ and the intensity is $I=E_0^2~\cos^2\alpha$ which can be written as $I=I_0 ~\cos^2\alpha.$ This equation is known s Malus' law. Clearly when $\alpha =90^o$, $I=0$.

Case II. Unpolarized light.
Unpolarized means you have a uniform distribution of electric field vectors over all angles. To find the intensity after the polarizer, we have to take an average over all angles. The transmitted intensity of electric field vectors between $\alpha$ and $\alpha+d\alpha$ is $dI=I_0~\cos^2\alpha~d\alpha.$ Averaging over all angles,$$I=\frac{\int_0^{2\pi} I_0~\cos^2\alpha~d\alpha}{\int_0^{2\pi}d\alpha}=\frac{1}{2}=I_0$$ Clearly, if you start with unpolarized light and you pass it through a polarizer, you can always define the polarizer axis as "the $x$-axis".

Case III Unpolarized sent through 3 polarizers. Light passes first through P1, then P2 and then P3.
P1 defines the $x$-axis; P2 is oriented at $45^o$ relative to P1; P3 is oriented at $45^o$ relative to P2 and $90^o$ relative to P1. Note that if P2 is removed, no light will make it through because P1 and P3 are crossed. What does the presence of P2 do?
If the initial intensity is $I_0$, after P1 light is polarized along x and has intensity $I_1\frac{1}{2}I_0.$
After P2 light is polarized at $45^o$ relative to the $x$-x is and has intensity $$ I_2=I_1 \cos^2(45^o)= \frac{1}{2}I_1=\frac{1}{4} I_0.$$Note that the light between P2 and P3 is not polarized perpendicular to the axis of P3. The light the emerges through P3 is polarized along the $y$-axis and has intensity$$ I_3=I_2 \cos^2(45^o)= \frac{1}{2}I_2=\frac{1}{8} I_0.$$

See how it works? Every time you put a polarizer at $45^o$ relative to the previous one you allow half of the intensity through and toss out the other half.