Quantum angle selectivity of polirizers filter

[beda pietanza]

given that light can be polirized at any plane while the polirized
filter has only a precise orientation, I assume there is a range of
angles of polirizzation that go though the polirized filter others are cut out.
how large is the angle selectivity of the polirized filters ?
how is this selectivity taken into account in the formalism ?

The same for electronic spin: how selective is the Gerlach apparatus on
the spin angle ?
Is the intensity of the magnet tuned up to the kinetic energy of the electron ?

thanks of any help,

best regards

beda pietanza

 

[Meir Achuz]

The intensity of the transmitted light is I=I_0 cos^2\theta. (Malus law)
There is also a cos^2 dependence to the up-down deflection in Stern-Gerlach.

 

[cesiumfrog]

I assume some polarisers (say birefringent materials) effectively make perfect quantum measurements (the outcome being one of two pure states). However, the stereotypical polariser (a series of parallel lines) seems potentially very inefficient because (even aside from whether it blocks half the beam that is correctly polarised) it would be expected to permit some light with polarisation deviating by a small angle (in the limit where the lines are widely spaced, the polariser will have no polarising effect on blue light). Do such polarisers really produce a mixture of quantum states?

 

[beda pietanza]

The intensity of the transmitted light is I=I_0 cos^2\theta. (Malus law)
There is also a cos^2 dependence to the up-down deflection in Stern-Gerlach.

OK, thanks,
the intensity of the transmitted light should depend also from the "angle selectivity" of the polirizing filter used and should also depend on the specific frequency of light, how these variable comes into play?

Same for Stern-Gerlach magnet how the speed of the electrons and the intensity of the magnet are taken into account ?

I=I_0 cos^2\theta doesn't containes all varables involved

best regards

beda pietanza

 

[Meir Achuz]

It doesn't need to. That law just comes from a vector dot product.

 

[beda pietanza]

It doesn't need to. That law just comes from a vector dot product.

I'm not a physicist nor a matematician, so please making a example: if we have solar light going through the filter , the various components of the spectrum, how are they attenuated?

If I use a different filter with different caratteristics how do you calculate the outcome?

best regards

beda pietanza

 

[Meir Achuz]

It only depends on cosine^2 of the angle between P and polarizer.
It's that simple.

 

[beda pietanza]

It only depends on cosine^2 of the angle between P and polarizer.
It's that simple.

I'am not convinced, I still think there are the "angle selectivity" and the “band width” of the filter that affect the intensity of the light coming out.

EI. using a filter box (containing 3 identical filters in a row) you don’t know the content of the filter box, if you still use your formula without knowing the characteristics of the filter box I suppose your results would be incorrect.

best regards

beda pietanza

 

[Meir Achuz]

Other effects, unrelated to polarization, can affect the intensity.

 

[DrChinese]

The term "angle selectivity" is not one usually associated with a filter. A polarization filter does have a range of wavelengths that it is rated for. Within that range, it is basically going to be the COS^2(theta) relationship which applies.

 

[Voltage]

Sorry if I should start a new thread on this: I raised a query about the polarization of light under Classical Physics, and Claude Bile suggested I ask it under QM.

Voltage, BeauGeste - The question of whether a single photon can possesses a linear polarisation pushes the edge of my knowledge to be honest. While a single photon must possesses angular momentum of +/- h-bar units, a photon can exist in a superposition of two states with the photon having a 50% probability of becoming either RH or LH polarised when a "measurement" is performed. You could perhaps call a photon in such a superposition of states a single photon with linear polarisation. You could similarly regard an unpolarised photon as being in a superposition of two orthogonal linearly polarised states. (I would recommend asking this question again in the QM forums, you will most likely get a more accurate answer there.)

I've read about the use of polarization filters in series, where two at 90 degrees cut out all the light whilst introducing a third intermediate filter let's some through. But I can't quite get the picture, even classically. Can you tell me whether a single photon can be considered to be linearly polarised? Or all photons always considered to have some element of circular polarization and hence are left or right cicular or elliptical? Sorry to be a little naive, my QM is weak.

 

[DrChinese]

I've read about the use of polarization filters in series, where two at 90 degrees cut out all the light whilst introducing a third intermediate filter let's some through. But I can't quite get the picture, even classically.

I personally see this situation as one which demonstrates the quantum nature of light. Specifically, you witness the effect of an observation (i.e. the observer) on the photon . The polarizer causes the photon's state function to collapse into one in which it is either aligned with the polarizer or orthogonal (90 degrees offset) to the polarizer.

The actual transmission function - cos^2(theta) - was discovered nearly 200 years ago by Malus. This was prior to the advent of quantum theory, which is why it is often characterized as being part of the classical picture.

 

[Voltage]

Thanks Dr Chinese. I've had a Google® and have printed a few things out to read offline, including Classical and Quantum Malus' Law by Krzysztof Wodkiewicz. If there's anything else you could suggest I'd be grateful.