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Light

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trennonix

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Hello,
we recently learned about light interference and Young's experiment and we also learned that a beam of light (the photons in that beam) can cause a discharge of a metal such as Zinc.
What we also learned was that two beams of lights can hit each other and pass without any change what so ever.

So i thought that on a dark line of young's experiment the photons would still be there even though we can't see light. I asked my teacher if putting a charged plate of zinc in a dark area would cause it to discharge.
She couldn't say; so here i am looking for an answer on this forum.

Excuse me if i didn't use scientific terms, but i'm french educated, so i get physics in french

Thanks
 

EN0

Member
Remember that light is particles and waves. Light is both wave and particle, but never both at the same time. You may wonder what does light wave? Nothing. It travels in a vacuum in a field it creates. A changing magnetic field creates a changing electrical field, which creates a magnetic field, and so on. In addition, when you look at a light beam, you're seeing pure, massless energy. When light travels through space, it has wave properties; when it interacts with matter, it behaves like a particle.

Also, high energy light knocks the electrons out of metal.
 

trennonix

New Member
so... when a charged plate of zinc is placed in a dark area created by young's experiment, would that cause it to discharge?

i never got that thing about why light can never be both particles and wave
 

Boron

New Member
It may be a little wacky to think about, but no. The photons do not exist at the dark spots (that is why they are dark) and subsequently the zink plate will not discharge due to the photoelectric effect.

When the waves cancel in those regions, nothing is left to hit the screen (or zink). It's a little mind boggling and often brings up questions of conservation of energy.. Quite frankly I'm actually sorta fuzzy on it, but I'm pretty sure what I wrote above is correct :).
 

trennonix

New Member
so you are saying that the photons simply dissapear, and then they are recreated :S
or the photoelectric effect depends on the sum of the phase of the photons?
just formulating these questions gives me a headache ;)
 

Boron

New Member
Well.. the problem with all this is that you cannot picture light as being just a particle (photons) or just a wave. As it was stated before its both, in some instances it behaves like particles other times waves.

What happens in this specific case is best understood by quoting wikipedia's article on the double slit experiment:

"Note that it is the probabilities of photons appearing at various points along the detection screen that add or cancel. So if there is a cancellation of waves at some point that does not mean that a photon disappears; it means that the probability of a photon's appearing at that point will disappear, and the probability that it will appear somewhere else increases."

Hope that helps :)
 

trennonix

New Member
by saying : "it is the probabilities of photons appearing at various points along the detection screen that add or cancel"

you mean that the the photons that were supposed to hit that dark area are pushed away?

i still can't get the idea how the protons are their but they won't cause light emission and they won't create the photoelectric effect.

(if i'm saying none sens please feel free to walk away :p )
 

Boron

New Member
Ah see the thing is that the photons are NOT there at all. So they do not hit the plate and do not cause the photoelectric effect.

As for being pushed away, I suppose you could kinda see it as such. The photons that should have impacted that section of the screen are instead impacting the peak points on the screen, making them brighter.

There was actually and experiment done where they managed to get a light source so dim that it was essentially one photon passing at a time and what you see is that the distribution of where these photons hit the screen over time follows the pattern of destructive and constructive interference. So the probability of where you would find a photon was given by the additive and cancelling nature of waves. Here's a video that demonstrates this in a somewhat entertaining way:

YouTube - The Infamous Double Slit Experiment

Really wonky to think about, but hey, that's quantum physics for you.
 

trennonix

New Member
WOW, that video almost fried my neurons !!!
when they did that experiment of sending one photon at a time, did they try to see through which slit it went, and if so, did it lose its wave properties like the electrons did in that mind bending video?
if you have any links to that experiment i would greatly appreciate you posting them.
 

Boron

New Member
Yhep, the fact that simply being there to measure what happens has an impact on what happens is a basic concept of quantum mechanics.

When photons are streaming through and I try to see which one they go through, the mear presence of my measuring will cause them to demonstrate something called the complimentarity principle. They will stop being both when observed.

When I actually fire a single photon at a time we STILL observe the interference pattern, which means that somehow that photon split up and went through both slits at the same time, probabilities cancelled and then the photon hit the screen.

**broken link removed**

That site shows some pictures of what the pattern would look like if photons were going one by one.
 

trennonix

New Member
man! you can't imagine the amount of questions buzzing through my head

after a photon splits up does it show as one bright spot even if the 2 "parts" hit each other at destructive phases (remember when we talked about 'probability' not 'destruction')?

now i'm not going to attempt to ask anything else since it would only lead to confusion as these stuff are way out of my league.
 
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