Filtering white light to a single frequency of coherent light.

White light, whether from an incandescent bulb or from a star, is characterized by multiple frequencies and multiple orientations of the perpendicular fields that make up the presence of photons as they propagate through space, if I understand correctly. Polarizing light reduces the intensity by filtering out many of the orthogonal orientations, and I think the result is coherent light of multiple frequencies. My question is, do I have that right, and how do we filter out all except a given frequency or frequencies, which I understand are used in various experiments like studying the photoelectric effect?

 

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You can use an arc light if you are trying to get ultraviolet light.

 

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polarized light is light that vibrates in the same plane. it isn't coherent because the resultant polarized light will not have the same phase.
i don't think i can explain it without a graphic.
the best way to explain this is to liken the polarizing "filter" to a diffraction grating.
if the grating is oriented to 45 degrees, then only light that can pass through the grating will also be 45 degrees.
passing this filtered light through a filter oriented at 90 degrees will block almost all of the light.
polarized light composes all light that vibrates in the same plane.
passing the above polarized light through a red filter will produce a red polarized light, but it will not be coherent because the waves aren't in phase with one another.

coherent light is multiple light waves of the same frequency and phase.

 

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Leopold has given part of the answer. If you plane-polarise white light you do not have coherent light. As he says, coherence, which is what you get in a laser, requires that all the photons are in phase with one another. But you do need a laser for that. Clearly, this can only be done for monochromatic light, i.e. light that is all the same frequency, since if you have several frequencies they may be in phase at one point in space, but will thereafter get out of phase, due the differences in wavelength.

To select a narrow band of frequencies from white light, you have to use the dispersion properties of light, either refraction through a prism or diffraction through, or reflection from, a grating. This, as I expect you know, causes different frequencies to be deflected through different angles, allowing you to select the frequency range you want.

But if you use a laser, you generate monochromatic light (from specific atomic excitations which occur at a characteristic frequency), so no filtering is needed.

What you say about the orientation in space of the electric and magnetic vectors in plane-polarised light is quite right. And, since frequency does not not come into it, you can indeed have a beam of plane-polarised white light.

Addendum: So far as I know, there is not a practical way to filter light, whether white or monochromatic, to select by phase, so as to make it coherent through filtering.

 

A laser itself is a chromatic filter of sorts, and if it uses a Brewster window or similar technology to reduce optical path losses until sufficient intensity is built up to escape the laser cavity, then the laser light will be linearly polarized, and also spatially and temporally coherent. This assures that photons in close proximity are "in step", and so there is minimal spreading due to interference of adjacent beam paths, as there otherwise would be from a non-coherent source of photon energy. The amount of spatial and temporal coherence depends in part on the geometry of the cavity and the particulars about how the optical or electrical pumping of the laser emission atoms is achieved.

Laser light also obeys the inverse square law once it leaves the cavity. Several reflective bounces within the cavity before the beam leaves assures that the beam will be tightly collimated, and will not decohere (begin spreading) for a considerable distance.

An optical laser or one with a cavity designed for optical pumping/amplification is also an amplifier for photons within a very tightly tuned range of wavelengths. This is to say, once energy saturation has been achieve within the cavity, only a tiny additional amount of energy of that particular wavelength is necessary in order to produce a cascade effect that is synchronized with the modulating signal. One photon can control the cascade of a great many. This is exactly what an amplifier does.

For most of the other threads on sciforums which I have started, I frequently make use of the properties of what a laser pulse (or pulses) are doing in order to illustrate what photon energy is doing in a vacuum as viewed from other inertial FoRs. I do it so often, I was expecting someone to point out that even a laser decoheres with increasing distance, and some physics I have read suggests that the inverse square law itself for ordinary incoherent light derives of the uncertainty principle. This makes some sense in terms of the more or less random quantum processes typically used to produce a wider spectrum of incoherent light by heating substances to incandescence. Fortunately, the device known as a laser cavity is a real game changer in terms of our being able to more tightly control the transitions of a certain metastable energy levels of laser emitting atoms. Without a device like that, most of my own thought experiments fall flat.

 

I think you are confusing passband, phase and polarization.

Whiteness refers to the passband, that it spans the entire visible spectrum. A light source is "white" because its frequency content is uniformly distributed in the visible band. It is "coherent" if every zero crossing of every wavefront, measured at some reference, occurs at the same time. Implied in this is that it's emitting a pure sine wave, and this lends itself to the gain realized in lasers. Polarization refers to the orientation of the field with respect to the direction of propagation.

The term "filter" usu. refers to a spectral envelope. A green filter has a hump in the envelope around the wavelengths we call "green", for example.

All optical devices will tend to diffract light and therefore randomize the phase to some extent.

The polarizing filter is often described as a picket fence. Only the waves oriented parallel to the pickets can pass between them; the rest are reflected or absorbed.

That loss of signal, whether due to the troughs on either side of the green hump, or whether due to rejection by the picket fence, amounts to a loss of power. All filters are by definition lossy, although ideally the filter could be so perfectly made and so perfectly matched to the source that it could hypothetically be lossless.

It's not clear to me what this has to do with the discovery of the photoelectric effect. It certainly has a lot to do with understanding spectrometry, interferometry, optics and instrumentation. Designers wrestle with reducing losses as a matter of daily practice.

What is your question concerning discovery of the photoelectric effect? The main point there has to do with energy exchange. The measurements made are over bulk effects. The instruments report ensemble averages. You seem to be concerned with the effects of optical losses on measuring luminosity or something. All of that is the subject of instrument design. But that's why we always state measurements within an error band. Two or three digits of precision are often sufficient in measuring power, and the scales may be optimized, such as using log scales, to extend the dynamic range indefinitely with minimal digits of precision.

 

I had taken the OP to mean the simple photoelectric effect finding that, as one scans light of increasing frequency, one finds photoelectrons start to be ejected only once the threshold frequency has been reached. For this, one obviously needs a source that allows frequencies to be scanned.

But the OP had got this confused with polarisation.

 

Any references for that? Seems to me just three general conditions guarantee the inverse square law:
The embedding space is isotropic, homogeneous, flat, non-absorbing, and 3D.
The fapp point source in question has a time-steady angular distribution output.
Conservation of energy.
All three above are purely classical constraints.

 

Thank you for the responses. I wasn't fully and correctly remembering a source diagram showing the process of getting coherent light from white light. I was searching with polarization instead of spatial filtering. There is a version of the diagram in this link:
http://electron6.phys.utk.edu/optics421/modules/m5/Coherence.htm

As the diagram shows, it was spatial filtering, like a pin hole, that reduces the intensity and at the same time increasing the spatial coherence, followed by a spectral filter, that can give us light that is one frequency, and coherent.

That would be the answer to the part of the OP about getting coherent light from white light. I wanted to set the stage for some questions I have about the photoelectric effect using one frequency of coherent light. I know I could just google my little heart out, but I'm looking for people who know a little more than a layman enthusiast can get out of the available material; an educated perspective from the scientific community.

The experiments that demonstrate the photoelectric effect show that the intensity of the light doesn't make the electrons jump any more energetically, and I take that to mean that it is the energy of the individual photons that determines what will be measured in terms of the distant the electrons will jump.

Am I correct in my understanding about that, and is it correct to say that the energy of photons at a single frequency is Planck's constant times the frequency?

 

The diagram in the post above gives the sequence of getting coherent light of a single frequency from white light in a repeatable and observable process. The size of the electron jump from a metal plate via the observable photoelectric effect shows that a photon of a given frequency, when it strikes a metal plate, can force an atom to eject an electron. The magnitude of the electron jump from the plate is a measure of the photon's energy, correct?

Is the photoelectric effect touted as evidence that photons are particles instead of waves, or is the photoelectric effect a part of the evidence leading to the description of photons as having wave and particle duality?

 

http://www.colorado.edu/physics/2000/quantumzone/photoelectric.html

 

http://en.wikipedia.org/wiki/Photoelectric_effect

http://www.nobelprize.org/nobel_prizes/physics/laureates/1921/

 

Yes, familiar links, but helpful to review to distinguish between intensity and frequency. I think it is right to say that intensity of light on metal refers to the number of photons in a given surface area over a given time? Also, given that a single photon has a given frequency from the instant that it is emitted until it strikes the metal, if that frequency is high enough, then a single photon can cause the effect of an electron being ejected from the plate.

Therefore, am I right to conclude that the frequency of a single photon can be determined by measuring the jump of the ejected electron.

 

The wave particle duality question is long obsolete in physics. Light is comprised of photons, which are quantum mechanical particles that have properties somewhat resembling particles and waves and is adequately described with one theory.

http://en.wikipedia.org/wiki/Wave–particle_duality#Treatment_in_modern_quantum_mechanics

 

No, intensity is power per unit area (or angle):
http://en.wikipedia.org/wiki/Luminous_intensity

That is true if the source and target are stationary and the gravitational field they are in is weak....

Yes.

 

Thanks for the interesting reference to QFT; certainly an area for more review on my part.

Thanks for the links and confirmation of the relationship between the magnitude of the "jump", and the ability to measure the frequency of a photon.

 

Most of the early errors have been corrected
After re-post of some I wrote on the photoelectric effect I give a not yet mentions cheap way to get well filtered beam of quite pure colored light.

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The KE of the electron ejected is always less than that of the photon by the metal's "work function."

Yes, filtering existing light must use dispersion (different frequencies travel in transparent materials at different speed. I can explain why, but have elsewhere already too.) If you fill a jar with some fine transparent dust (ground glass?) of one material and then pour in a clear liquid, typically they may have no or only one wave length where their index of refraction is exactly the same. - That wavelength will pass thru the mix, but all other will scatter out. Every time a photon comes to a interface with a discontinuous change in the index some will be refleced by the surface. For example both surfaces of a common glass window reflect about 4% so the light coming to you is only ~92% as intense as struck the first surface.

White light that passed thru many hundreds of interfaces between the solid transparent "dust" (powdered glass?) and the clear liquid will be quite a pure color - all else scattered out of the beam.

BTW, the "purest color" you will ever see is made by nature - the green of the "northern lights." Those photons have so many cycles that the are at least 2 meters long (I forget correct number).* Fourier Analysis shows that a pure color / frequency must have a huge number of cycles. Such well defined energy must have great uncertainity in when it was emitted. This green line does as to first order QM it is not even an allowed radiative transition. Why for years no one could figure out it is a "forbidden transition" of Oxygen. Man can not make it in the lab as huge high vacuum bottle would be needed.

* Some Sodium light I measured and have several posts telling how, were ~30cm long.

 

Thanks for that discussion of optical frequency filtering, and the related history about Planck's constant and the photoelectric effect.

 

I have a follow up question: In a double slit experiment, if the screen is replaced by a metal plate, and a laser of adequate frequency is pointing at the slits, would the pattern of ejected electrons from the plate show the same interference pattern as is shown when the photons strike the photosensitive screen?

 

yes.