# Light polarization

1. Apr 20, 2015

### skrat

1. The problem statement, all variables and given/known data
Elliptically polarized light, where the rotation of the ellipse is $\pi/6$ and its major axis is $2E_0$ and minor axis $E_0$, is left through a polarization foil. The foil transmits light in $x$ axis (a) and in $y$ axis (b). Calculate the ratio of transmitted light (intensity) in both cases (a) and (b).

2. Relevant equations

3. The attempt at a solution

Ok, If my ellipse is rotated by an angle $\pi/6$ than in my original coordinate system:
$E_x={E_x}'\cos \varphi +{E_y}'\sin\varphi$ and
$E_Y=-{E_x}'\sin\varphi +{E_y}'\cos \varphi$.

Knowing that ${E_x}'=2E_0$ and ${E_y}'=E_0$ the amplitudes above can be written as: $$E_X=E_0(\sqrt 3+1/2)$$ and $$E_y=E_0(\sqrt{3}/2-1)$$.

The intensity before the polarization foil is $(2E_0)^2+E_0^2=5E_0^2$. Than the ratio is

(a) $\frac{|E_x|^2}{5E_0^2}=0.9964$ and
(b) $\frac{|E_y|^2}{5E_0^2}=0.00358$.

Or is this completely wrong?

2. Apr 20, 2015

### ehild

It is not completely wrong.

I think "rotation of the ellipse" means the angle its mayor axis makes with the positive x axis. In that case, your Ex and Ey are not correct, but otherwise, your method is right.

3. Apr 21, 2015

### skrat

Hmmm, should it be:

$E_x={E_x}'cos\varphi +{E_y}'sin\varphi$ and $E_y={E_y}'cos\varphi + {E_y}'sin\varphi$ ?

4. Apr 21, 2015

### ehild

No. The x components of the electric vectors parallel and perpendicular of the main axis of the ellipse add, and so is with the y components. Make a drawing, and you will see.

5. Apr 21, 2015

### skrat

I am staring at this picture but I can't figure out what's the problem.

6. Apr 21, 2015

### ehild

The elliptically polarized light has components along the mayor and minor axis of the ellipse, they are Er and Eq, respectively. Eq and Er are 90 degrees out of phase in case of elliptically polarized light, so Er=2Eo cos(ωt), Eq=Eo sin(ωt).
The x component of Er and the x component of Eq add up to give EX, the component of the field along the X direction of the polariser. That component will be fully transmitted in case a), and the intensity is the time average of EX. In the same way, the Y components of Er and Eq add up to give EY, the Y component of the electric field, which will be fully transmitted in case b).

Last edited: Apr 21, 2015
7. Apr 22, 2015

### skrat

Huh,

I think I understand now.

So $E_x=E_r\cos\theta -E_q\sin\theta$ and $E_y=E_r\sin\theta +E_q\cos\theta$. Is this ok?

8. Apr 22, 2015

### ehild

It is, but remember that both Rr and Eq depend of time, and their phase difference is 90 degrees.