Polarizing Elliptically Rotated Light: Calculating Intensity Ratios

[skrat]

Homework Statement

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).

Homework Equations

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?

 

[ehild]

Homework Statement

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).

Homework Equations

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?

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.

 

[skrat]

Hmmm, should it be:

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

 

[ehild]

Hmmm, should it be:

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

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.

 

[skrat]

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.

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

 

[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 E_X, 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 E_X. In the same way, the Y components of Er and Eq add up to give E_Y, the Y component of the electric field, which will be fully transmitted in case b).

 

[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?

 

[ehild]

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?

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