How Does Rotating Polaroid Filters Affect Light Intensity?

[bowma166]

Homework Statement

Two sheets of polaroid are aranged as polarized and analyzer. Suppose that the preferential direction of the second sheet is rotated by an angle $$\phi$$ about the direction of incidence and then rotated by an angle $\alpha$ about the vertical direction. If unpolarized light of intensity $I_0$ is incident from the left, what is the intensity of the light emerging on the right?
http://img22.imageshack.us/img22/8189/polaroiddiagram.jpg
Sorry about the crappy diagram quality.

Homework Equations

$$I=I_0\cos^2\phi$$
or
$$E=E_0\cos\phi$$

The Attempt at a Solution

Well the intensity drops down to $\frac{1}{2}I_0$ after the first polaroid. The second one is hard. The way I'm thinking about it, it basically turns into a geometry problem... Say you're looking at the second polaroid head-on from the first direction. It's turned an angle $\phi$, making the component of the light that can pass through smaller. It's then turned about the vertical axis by an angle $\alpha$, which, when you look at the 2D projection of the filter you see when you look at it head-on, makes the lines seem more vertical and should allow more light to pass through, right?
Here's my even crappier hand drawn diagram of the situation sort of:
http://img254.imageshack.us/img254/9007/polaroid2.jpg

So, again, I think the problem is: look perpendicularly at the plane made by the top and bottom two vectors in the above diagram when $\alpha=0$, then rotate it away by $\alpha$ and determine what angle you see the diagonal line makes with the bottom line. Maybe? I just don't know how to do the geometry of that. Sorry for the wordiness and not being very clear, it's hard to express my thoughts on this problem very well.

These points seem clear to me:

at $$\alpha=\frac{\pi}{2}$$, $$I=\frac{1}{2}I_0$$

at $$\alpha=0$$, $$I=\frac{1}{2}I_0\cos^2\phi$$

Thanks.

 

[bowma166]

Anyone?

 

[nlightner]

I can help you with the solution to this problem. The intensity of the light emerging on the right can be calculated using the Malus' Law, which states that the intensity of polarized light transmitted through a polarizer is proportional to the square of the cosine of the angle between the polarization direction of the incident light and the polarizer.

In this problem, we have two polarizers, one acting as the polarized and the other acting as the analyzer. The first polarizer will reduce the intensity of the incident light by half, as you correctly pointed out. The second polarizer, which is rotated by an angle φ, will further reduce the intensity of the light by a factor of cos^2φ.

Now, for the second rotation by an angle α, we can use the Malus' Law again. The angle between the polarization direction of the incident light and the polarizer is now α+φ, as shown in your diagram. Therefore, the intensity of the light emerging on the right will be:

I = I_0 * (1/2) * cos^2φ * cos^2(α+φ)

I = I_0 * (1/2) * cos^2φ * (cos^2α * cos^2φ - sin^2α * sin^2φ)

I = I_0 * (1/2) * cos^2φ * (cos^2α * cos^2φ - (1 - cos^2α) * sin^2φ)

I = I_0 * (1/2) * cos^2φ * (cos^2α * cos^2φ - sin^2φ + cos^2α * sin^2φ)

I = I_0 * (1/2) * cos^2φ * (cos^2α * cos^2φ + cos^2α * sin^2φ)

I = I_0 * (1/2) * cos^2φ * cos^2α * (cos^2φ + sin^2φ)

I = I_0 * (1/2) * cos^2φ * cos^2α

I = I_0 * (1/2) * cos^2α * cos^2φ

Therefore, the final intensity of the light emerging on the right is given by the above equation. I hope this helps you in solving the problem