Can polarization affect diffraction?

[fargoth]

I know the scalar theory of light, which doesn't include polarization effects.

I was taught it is quite accurate when the subject is bigger then the wavelength.
But today, someone told me that by using half plates and quarter plates in an optical setup with polarized light he can transfer energy from the zero frequency to higher frequencies.

According to what I know, this is impossible, but i don't know everything - so i ask you - do you know if this is possible?

 

[Claude Bile]

But today, someone told me that by using half plates and quarter plates in an optical setup with polarized light he can transfer energy from the zero frequency to higher frequencies.

I'm not 100% sure what you mean by this. Are you referring to spatial frequencies?

Most diffraction theories deal with a single propagating wave interacting with a set of obstacles. If the size of the obstacles is large, say 10 times bigger than the wavelength then polarisation effects are minimal because big objects "look" much the same to the wave no matter how the wave is polarised, and thus exhibit a very similar "response" (or resultant wave if you like) - as opposed to say a very thin wire which "looks" like a continuous sheet in one direction, and very discontinuous in the other (perpendicular) direction.

The picture is complicated when you have two or more waves. The relative polarisation between each wave suddenly becomes important is not as easily neglected as in the case with a single incident wave. This is because, in short, the interference pattern between two waves is polarisation dependent. If you take a classic double slit experiment and rotate the polarisation of one slit by 90 degrees the interference pattern will wash out. You cannot reasonably neglect polarisation when you have two or more waves in the general case.

Claude.

 

[fargoth]

The system we're talking about has only one beam of laser (I think that's what you meant by one wave, right?).

And yes, when i was talking about the zero frequency, i meant the spatial frequency, specifically i was referring to the light which remained in it's initial direction after going through the obstacle.

As for the size of the obstacle - it's minimal size is one and a half wavelengths...

 

[Andy Resnick]

Polarization will have a huge effect! It's easy to see why- diffraction gratings are spatially-oriented features (ruled lines, pressed patterns, etc), and so the scattering will definitely depend on the polarization.

Some solid-state devices, like acousto-optic modulators, also show a polarization dependence on the diffraction efficiency. Again, it's because there is a spatial organization to how the device works: strain wave patterns, in this case.

 

[fargoth]

lets say I use a spatial light modulator with a linear polarizer, liquid crystal and another linear polarizer - the light leaving this device will already be linearly polarized - will the pattern shown on the SLM change anything?

I know it won't matter much right after the SLM, because the pixel size is too big - but if i use a concentrating lens, will the pattern shown on the SLM change the polarization after the image got small enough?

if so, how can half plate followed by a quarter plate transfer energy from the zero order to other diffraction orders?

 

[Andy Resnick]

My understanding is that a SLM is not much more than a spatially-varying retarder. So, if you rotate the first polarizer with respect to the liquid crystal element, the output will vary because the (say) fast axis of the liquid crystal is no longer in a fixed relation with the polarization direction, resulting in altered amounts of retardance.

I don't understand your second paragraph.

For the third paragraph, if I am following you, you are rotating the linear polarization with a half wave plate, the light exiting the quarter wave plate will then change from circular-elliptical-linear-elliptical-circular. So it's clear that the liquid crystal element will act on the light differently as the polarization state of the input varies. Presumably as the half-wave plate rotates, the diffracted energy oscillates between zero- and first-order?

 

[fargoth]

SLM may be just a spatially-varying retarder, but in this case it's made of a linear polarizer, liquid crystal, and another linear polarizer in the same position as the first one - when no voltage is applied to the liquid crystal the refraction index for both polarization axis is the same - and so the linear polarization remains the same and the light passes through the pixel.
And when voltage is applied to the liquid crystal, the polarization is rotated in 90 degrees - and no light passes through the pixel.

The input of the system is just a plane wave in a circular polarization... so the SLM is the object of the system.

The size of the pixel is much bigger then 10 wavelengths, and the light leaving the SLM is linearly polarized according to the polarizers in it's two ends.

After the SLM, the light is focused with a convex lens, my question in the second paragraph was:
will the pattern of blocking and passing pixels in the SLM change in some way the polarization after the image of the SLM got small so that the pixel size of the image is about two wavelengths?

why does rotating the linear polarization in 90 degrees, and then making it circular\elliptic makes the zero order weaker and transfers it's energy to other orders?

 

[Claude Bile]

I know it won't matter much right after the SLM, because the pixel size is too big - but if i use a concentrating lens, will the pattern shown on the SLM change the polarization after the image got small enough?

No, any effect the SLM has on polarisation will occur at the device, not after you focus it down.

A word of caution though, you cannot assume the polarisation to remain invariant in any case for a strongly focused beam.

With regard to light coupling out of the 0th order, this occurs invariably because there is a spatial phase modulation occurring in conjunction with the spatial amplitude modulation.

Claude.

 

[fargoth]

No, any effect the SLM has on polarisation will occur at the device, not after you focus it down.

A word of caution though, you cannot assume the polarisation to remain invariant in any case for a strongly focused beam.

With regard to light coupling out of the 0th order, this occurs invariably because there is a spatial phase modulation occurring in conjunction with the spatial amplitude modulation.

Claude.

I didn't understand the port about strongly focused beam - if all the light is linearly polarized - why would it change when i focus it, if the spatial and phase modulation doesn't affect it's polarization?
and if it does affect the polarization, it must happen after the wave went out of the SLM - because right after the SLM there was a linear polarizer.

Doesn't the spatial phase modulation affect both polarizations the same way?
Can you show me some math to make it clearer? or point me to a good book on the subject?

 

[Andy Resnick]

Strongly focused beams (usually taken to mean numerical apertures greater than 0.5) no longer obey the paraxial approximation (sin u ~u). Detailed calculations of the electromagnetic field within such a focal volume show all kinds of bizarre effects: longitudinal waves, changes to polarization, etc. etc.

A SLM, and most all diffraction elements, work by interference to create a desired optical pattern in the far-field. By changing the phase lags in the individual pixels of a SLM (by changing the polarization input state, for example), the far-field diffraction pattern must change.

Without knowing any details about the SLM, I really can't say much more than that. I suspect there is coupling into higher diffraction orders as well, maybe you just aren't seeing it.

Edit: sorry, a book reference. Again, without knowing your comfort level it's hard to recommend something. The Handbook of Optics has a nice chapter on polarimetry by Chipman, one of my past professors. My go-to book is Azzam and Bashare, "ellipsometry and polarized light", but there's nothing in that specifically about SLMs. If someone has a good text or review article on the design and use of SLMs, I'd be very interested to know what it is.

 

[fargoth]

...a book reference. Again, without knowing your comfort level it's hard to recommend something.

I'm in my first year of my M.Sc. in Physics.
during my B.Sc I had one introductory course of optics (most of Eugene Hecht OPTICS), I had two labs in optics (built a laser using mirrors and a NeHe tube) and recorded a hologram with a CCD and reconstructed it's image using Fresnel numerically. (of course i also had the more basic labs like finding brewster's angle, measuring polarization from scattering, experimenting with Newton's rings and interference from slits etc.)

In my M.Sc i did the Linear Optics course, it focused on Fourier optics, and super resolution, but we had also looked at holograms, SLMs, Zernike’s phase-contrast microscope, filters etc.

I'd be happy if anyone could recommend me a book, because I find it hard to see how the polarization affect the resulting image without the proper math to back it up.

 

[Claude Bile]

Doesn't the spatial phase modulation affect both polarizations the same way?

No, especially where there are birefringent media involved. Only one polarisation component will "see" the fast/slow axis, and thus be shifted relative to the non-phase shifted portion of the wave.

Can you show me some math to make it clearer? or point me to a good book on the subject?

Typically apertures are characterised based on their aperture function $a(x)$, which is typically defined as follows.

$$E(x) = a(x)E_0(x)$$

Where E_0 is the field immediately before the aperture and E(x) is the field at the aperture. The aperture function, $a(x)$ is complex-valued in the general case. Like any complex-valued function, it can be expressed in polar form, as an amplitude and phase.

The function $a(x)$ can also be expressed as a sum of spatial frequencies via a Fourier transform. In the absence of spatial phase modulation (i.e. if an aperture modulates amplitude only) there will always be a DC term present in the FT of $a(x)$. The DC term corresponds to the 0th diffracted order.

To eliminate the 0th order, there need to be both positive and negative phase contributions to the diffracted field (as opposed to uniform positive or uniform negative phase contributions) hence the phase of the diffracted wave must be spatially modulated in order for this to happen.

My suspicion is that the Liquid crystal is modulating the phase of the wave in addition to the amplitude and changing the polarisation is allowing varying proportions of the wave to "see" the birefringence in the liquid crystal and thus cause light to progressively couple out of the 0th order beam as the polarisation is changed.