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The human eye detects light via a family of proteins called opsins. Different forms of photopsins are sensitive to different wavebands, which is what gives us color vision. Rhodopsin is sensitive mainly to greenish-blue light, and provides us with monochromatic night vision. Rhodopsin works by changing its conformation when it absorbs a photon; that change of conformation allows ions to flow through the rod cell's membrane and generate a signal. The signal from each rod cell is processed through the retina and passed to the visual cortex, where a representation of the visual field is constructed.
Human rhodopsin has a quantum efficiency (QE) of about 25% (there's a 25% chance a single photon will be absorbed and produce the rod-cell signal). By comparison, cat rhodopsin is more than 90% quantum efficient. 25% QE is sufficiently high to be observable -- a source of single photons can be seen by a dark-adpated person with normal vision.
Human rhodopsin has a quantum efficiency (QE) of about 25% (there's a 25% chance a single photon will be absorbed and produce the rod-cell signal). By comparison, cat rhodopsin is more than 90% quantum efficient. 25% QE is sufficiently high to be observable -- a source of single photons can be seen by a dark-adpated person with normal vision.
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This is a lab we did when I was an undergraduate, more than 20 years ago. I haven't done the setup myself since then, so I'm just going to describe it; if I have the opportunity run it again, then I can take pictures and publish this as an I'ble.
If someone else decides to tackle it, please feel free to contact me and I'll make this a collaboration.
If someone else decides to tackle it, please feel free to contact me and I'll make this a collaboration.
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For demonstrate that you have a single photon emitter, Could you do a HBT(Intensity interferometry) experiment for probe it?
If you use filter only decrease density but we have in this case a poissonian distribution. This implicates that you could have in one second(using you disertation) an indeterminated quantity of photon (mean could be 1 by second but no each second 1 photon).
Please don't say that this is way for observe single photons because you're promoting ignorance.
This demonstration, in fact, was adapted from an actual undergraduate physics lab experiment I did at UCLA when I was there (1984-1988). We first did measurements using a small photomultiplier tube and scaler, to verify that the single-photon rate was about 1 Hz. After that, we did the dark-adapted observation, using a "clicker" to count the flashes we thought we observed.
Using neutral-density filters reduces the intensity of the beam, while still preserving the Poissonian emission distribution, exactly as you say, but your conclusion is not correct.
The probability distribution means that on average, the time interval between successive individual photons will be longer than human POV resolution. For an individual observation, there is some probability (which you can calculate) that two photons reached your eye in a time interval shorter than human optical resolution (POV).
The design of the experiment is to get the Poisson interval to be long enough that the probability of a two-photon event is small. This is not "ignorance," it is mathematics.
Facts:
laser: coherent emitter with poissonian distribution.
Laser diode: have fluctuations. Because this you can't calculate density in each time.
Improper definition of this demonstration. If anybody do this, he are doing a LOW DENSITY EMITTER SYSTEM, never you've a single emitter. If you want a single emitter you can use a Single Quantum Dot structure. Actually in science, work with single emitters is considered a top research and you need a very expensive lab for work properly.
Be carefully: NEVER use your eyes for see laser emition. NEVER NEVER. And please DON'T PROMOTE THAT SOMEBODY DO THIS.
If you want to build a single emitter you need a sub-poissonian distribution. You can check this in many articles that works with Single emitters.
This diode have fluctuations if you are so crazy for see this flashes, you're detecting this fluctuations. not photons.
For other part, Please think in this: If we can detect single photons with our eyes, then when we are outside in a sunny day, why we don't become blind?
I propose to you that, erase this demonstration or rename properly. Because you're promoting ignorance.
You are correct that laser diodes (in fact any laser) have fluctuations such that you can only calculate a probability for single-photon detection, with no guarantees. QDs or trapped ions (e.g., in a MOT or Penning structure) can provide guaranteed single photon emission, which is required for true research-level work.
You are highly overstating the danger of laser emission. It is not significantly different than any other high-intensity light. The primary danger is from infrared lasers because the human blink reflex is only triggered by visible light. Consequently, the eye may be exposed to even a low-power IR laser for a relatively long time, resulting in significant thermal damage.
For visible light lasers, the danger is specifically graded according to intensity (power output). Below one milliwatt output (i.e. a full intensity green laser pointer) the hazard is considred "negligible." With the filtering described here, the visible power on target is femtowatts.
Finally, the human eye does detect single photons (if it didn't, we couldn't see anything at all). Under normal conditions, there are many, many single photon detections all happening at once, each within a different specific rhodopsin molecule. It is the temporal correlation of all those detections which forms the "scene" we see.
With both slits open, a many-hour exposure will get you a pattern of dots on the film in the form of bands (interference fringes). Covering one slit will get you a different pattern.
If you do build this, please take lots of pictures. I can add you to this I'ble as a collaborator, so you can attach pictures and narrative, or I can cross-link to your own I'ble if you decide to write one.
Wikipedia tells me that an ND4 results in 25% transmittance (http://en.wikipedia.org/wiki/Neutral_density_filter). Thus;
1st ND4
(2.8*10^15)*0.25 = 7*10^14
2nd ND4
(7*10^14)*0.25 = 1.75*10^14
3rd ND4
(1.75*10*14)*0.25 = 4.375*10^13
4th ND4
(4.375*10^13) = 1.09375*10^13
I'm pretty sure I must be missing something here, but I bought a dirt cheap laser pointer (1mW ~532nm) and shone it through my 10 stop ND filter (ND1024) and plenty of photons streamed through!
Thanks in advance for any assistance.
If you'd rather write up your own I'ble about your implementation, I completely understand; it just never hurts to ask :-)
ND1024
ND8
ND8
ND8
ND2
ND4
...and as expected it let through too much light (~669921875 photons/s if my calculations are correct). i added one piece of bin bag and nothing came through at all... I'm a bit stuck. Might go down the welder's goggle sroute as lenses seem very cheap on eBay. ND1024 photographic filters are not cheap! The other five filters mentioned above are all grads, obvioulsy using the darker end of the grad.
When I looked up ND filters at Thor Labs, they used a notation where the number indicates a quantity called "optical density", such that ND1 corresponds to 10% transmission, ND2 to 1% transmission and so on (i.e., transmission = 10-ND). See, for example http://www.thorlabs.us/NewGroupPage9.cfm?ObjectGroup_ID=266. Hence my statement that ND4 would correspond to 10-4 transmission, and a stack of four of them to 10-16.
The Wikipedia page you cite has the same definition, under Mechanism. However, the table on that page (ND Filter Types) describes exactly what you say above.
I don't know which notation is standard, but what I called "ND4" in my I'ble would be about the same as "ND8192" (0.012% transmission) in the Wikipedia table.
Ah, ha! Take a look at the caption for the photographic comparison on that Wikipedia page.
In their own table, that "1000x ND-Filter" would correspond to the same ND1024 you have, with about 0.1% trasmission. But in the caption, they label it as "ND3.0" (optical density 3.0), corresponding to what I wrote in the I'ble.
So maybe what I should have written is really "ND4.0" (optical density 4 == 0.0001 transmission). What do you think?
As far as your article is concerned, I don't think it is incorrect to refer to ND3.0 & ND4.0 but I'm not sure it is the clearest method. I would be inclined to write 'filter with optical density of 3.0' rather than 'ND3.0'.
I see that Thor do a filter with optical density of 6.0 for £14.04. By my calculations, if I had two of those added on to the above which I already have then I'd be in the right ballpark (altogether would be reduction of 10^-16.6 I think). Then I could remove some of the lower rated filters to get near the magic 1 photon per/second. Do you agree?
One more thing - sorry to be nitpicky! The note you have added implies that the type of NDx notation used in photography (ie. Cokin) refers to f-stops. As far as I can tell from the, admittedly misleading in places, Wikipedia article is that it actually refers to lens areas opening as a fraction of the complete lens where the resulting opening equals 1/x where x is taken from NDx.
The ND6.0 filter provides a reduction of 10-6, so two of them would get you 10-12. After that, all you need is another factor of 10-4 or so, which your can just about reach with ND1024 + ND8.
It sounds like dropping 30 quid will get you into the single photon game; very cool :-)
....................But having the photons stream in at parallel is crucial if one wishes to try a double slit-experiment, isn't it?
................A very long tube indeed.
Thanks for posting this niftyness!
In place of ND filters, it is also possible to use a stack of trash bags (black Hefty bags), but since they aren't calibrated, figuring out how many you need, and what the effective photon rate is becomes much more problematic.
Using a single-wavelength laser allows you to convert the beam power directly into an effective photon rate, using E=hf. A "green Christmas light" is a continuum source with a colored filter, and there's no good way to even guess the photon rate.
The Wikipedia entry for Planck's constant doesn't show the calculation either.
I think it goes like this. Frequency of the green light = speed of light / wavelength of the green light. 300 km/sec / 532 nm = 564 Tz
Planck's constant x frequency of green light = energy in Joules per particle
6.6 x 10-34 Joule sec X 564 Tz = 372 x 10 - 19 Joule/particle
Now the energy of the laser is 1 miliwatt or .001 watts which is .001 Joule per second.
Divide the energy of the laser by the energy in each particle to get the number of particles in the laser beam. .001 Joule/sec / 372 x 10 -19 = 2.69 x 10**16 particles
Is the picture a photon? May I have permission to use it in my paper?
I think i read that photons don''t have mass- how are they made?
1) The photon is nothing more than the quantization of the electromagnetic field. There's nothing "deeper" unless you consider classical EM "deeper" than quantum electrodynamics (QED).
2) The picture I used in my introduction is just a laser beam pointed at a camera lens. The rings and horizontal "disk" are diffraction effects.
3) Free photons don't have mass. That just means that their energy and momentum are equal (in natural units where c = 1).
4) Photons are "made" by any sort of electromagnetic transition: an electron changing energy levels in an atom, an oscillating electric current, a moving magnet.
If you have a moving magnet, then you get light. If you want to ask how that light interacts with atoms, then you need quantum mechanics, and QED is the most accurately verified scientific theory we know.
Classically, you get brighter (more intense) light when the amplitude of the oscillating fields is larger (intensity is amplitude squared).
Quantum mechanically, you get brighter (more intense) light when there are more photons of a given frequency. Each individual photon carries a fixed energy (frequency * h-bar), so more of them mean more energy.