Polarizing filters problem

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In summary, the problem involves two sheets of polaroid arranged as polarized and analyzer. 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. When unpolarized light of intensity I_0 is incident from the left, the intensity of the light emerging on the right can be calculated using the equations I=I_0\cos^2\phi or E=E_0\cos\phi. At \alpha=\frac{\pi}{2}, the intensity drops down to \frac{1}{2}I_0, and at \alpha=0, the intensity is \frac{1}{2}I_0\cos^2\
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bowma166
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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 [tex]\phi[/tex] about the direction of incidence and then rotated by an angle [itex]\alpha[/itex] about the vertical direction. If unpolarized light of intensity [itex]I_0[/itex] 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


[tex]I=I_0\cos^2\phi[/tex]
or
[tex]E=E_0\cos\phi[/tex]

The Attempt at a Solution


Well the intensity drops down to [itex]\frac{1}{2}I_0[/itex] 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 [itex]\phi[/itex], making the component of the light that can pass through smaller. It's then turned about the vertical axis by an angle [itex]\alpha[/itex], 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 [itex]\alpha=0[/itex], then rotate it away by [itex]\alpha[/itex] 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 [tex]\alpha=\frac{\pi}{2}[/tex], [tex]I=\frac{1}{2}I_0[/tex]

at [tex]\alpha=0[/tex], [tex]I=\frac{1}{2}I_0\cos^2\phi[/tex]

Thanks.
 
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  • #2
Anyone?
 
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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
 

1. What is the purpose of a polarizing filter?

A polarizing filter is used to reduce glare and reflections from non-metallic surfaces, such as water, glass, and foliage. It also enhances color saturation and contrast in photographs.

2. How does a polarizing filter work?

A polarizing filter is made of a special material that only allows light waves that are vibrating in a specific direction to pass through. This results in the filter blocking out certain types of light, such as glare and reflections, while allowing other light to pass through. This helps to reduce unwanted reflections and increase color saturation.

3. What is the difference between circular and linear polarizing filters?

Circular polarizing filters are used for DSLR cameras and have a special layer that helps to prevent the camera's autofocus and light meter from being affected by the filter. Linear polarizing filters are used for film cameras and do not have this special layer.

4. Can a polarizing filter be used with other filters?

Yes, a polarizing filter can be used with other filters, but it is important to pay attention to the order in which they are stacked. The polarizing filter should always be the first filter attached to the lens, with other filters stacked on top of it. This will ensure that the polarizing effect is not diminished by other filters.

5. Are there any drawbacks to using a polarizing filter?

One potential drawback of using a polarizing filter is that it can reduce the amount of light entering the camera, resulting in a darker image. This can be counteracted by adjusting the camera's exposure settings. Additionally, polarizing filters can also create a noticeable darkening effect in the sky, so it is important to adjust the angle of the filter to avoid this effect if desired.

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