possible to block light with light?

[Fayd]

forgive my idiocy with regard to physics (for some reason, my college physics course was less advanced than my high school physics course)

is it possible for one EM energy wave to change the course of another EM energy wave?

i'm not sure where i'm going with this...

 

[foges]

Umm, not that i know of. Light is its own antiparticle though. So light can cancel itself out, like with the double slit experiment. Also gravity bends light. But i dont think one photon of light can change the momentum of another like with two bowling balls hitting eachother.

 

[f95toli]

First of all, the photon does not have an antiparticle as such meaning you can't annhilate photons (also, where would the energy go?)

But to answer the first question: yes, sort of. Although the effects is extremely weak and can only be observed at very high energies; i.e. it can't be used for visible light or radiowaves.

 

[QuixoticOne]

No flashdark for you.

 

[Net]

btw, light is both a photon and a wave.

here are some principals you should be aware of.

These wavefronts interfere either constructively or destructively depending on how the peaks and valleys of the waves are related. If a peak falls on a valley consistently (called destructive interference), then the waves cancel and no light exists at that point.

http://www.physics.smu.edu/~sc...l/diffraction/lab.html

Scroll down to that quote on the page. The wave approach the diffraction grating ( see the gaps? the wave goes through those gaps and the direction of the wave changes.) imagine a sinusodial wave alone those lines. now see the red line shows the difference in length?

call the top line x1 and the longer one (with the red line) x2. then to get constructive interference you do x2 - x1 = n(landa) where n is an integer. to get destructive interference you simple do x2 - x1 = ( n + [1/2] ) * landa
landa = velocity / frequency

http://hyperphysics.phy-astr.g...yopt/imgpho/htuber.jpg

Why don't you see all the colors of the rainbow? (that is hydrogen seen through a diffraction grating btw). Why is there black? Because some of the waves destructively interfere canceling each other out.

So can you block light with light? not really. its not actually "blocking." the waves are canceling out. the trough of one wave coincides with the peak of another and it makes the average amplitude of the wave 0 at that point. So at that point there will be no light.

 

[Born2bwire]

Also should note you cannot expect to achieve perfect deconstructive interference. You will notice that in the slit diffraction experiments that photons will still land in the regions of deconstructive interference. The inherent finite sizes of potentials and the fact that the wave equation describes something akin to a probability distribution function prevents perfect cancellation.

 

[silverpig]

Originally posted by: Born2bwire
Also should note you cannot expect to achieve perfect deconstructive interference. You will notice that in the slit diffraction experiments that photons will still land in the regions of deconstructive interference. The inherent finite sizes of potentials and the fact that the wave equation describes something akin to a probability distribution function prevents perfect cancellation.

The banding is caused by geometry...

 

[Born2bwire]

Originally posted by: silverpig

Originally posted by: Born2bwire
Also should note you cannot expect to achieve perfect deconstructive interference. You will notice that in the slit diffraction experiments that photons will still land in the regions of deconstructive interference. The inherent finite sizes of potentials and the fact that the wave equation describes something akin to a probability distribution function prevents perfect cancellation.

The banding is caused by geometry...

Yes... that is the primary cause...

 

[Comdrpopnfresh]

If one stretches the meaning of "em energy wave"- then yes. When an electron is moving in free space, and it's directional velocity is changed by coulombic repulsion of ,say, a nuclei; it results in an increase in acceleration, and the subsequent release of an x-ray. that kinda radiation is called bremsstrahlung

 

[spikespiegal]

I believe the process of film based holography answers a lot of these questions.

 

[Roguestar]

Originally posted by: f95toli
First of all, the photon does not have an antiparticle as such meaning you can't annhilate photons (also, where would the energy go?)

But to answer the first question: yes, sort of. Although the effects is extremely weak and can only be observed at very high energies; i.e. it can't be used for visible light or radiowaves.

The photon is its own antiparticle .

 

[MrDudeMan]

Originally posted by: silverpig

Originally posted by: Born2bwire
Also should note you cannot expect to achieve perfect deconstructive interference. You will notice that in the slit diffraction experiments that photons will still land in the regions of deconstructive interference. The inherent finite sizes of potentials and the fact that the wave equation describes something akin to a probability distribution function prevents perfect cancellation.

The banding is caused by geometry...

It's normally distributed though, so there has to be some number of photons even in the deconstructive areas.

 

[Eeezee]

Originally posted by: f95toli
First of all, the photon does not have an antiparticle as such meaning you can't annhilate photons (also, where would the energy go?)

But to answer the first question: yes, sort of. Although the effects is extremely weak and can only be observed at very high energies; i.e. it can't be used for visible light or radiowaves.

Into an electron-positron pair. You can collide an electron-positron pair to create two photons as well.

Here is a Feynman diagram demonstrating the reaction (time can go either forward or backwards on this graph, so you can either collide an electron and a positron to create two photons or you can collide two photons to get an electron and a positron, note that the electron on the middle branch is virtual).

http://www.shef.ac.uk/physics/people/cbooth/eegg.gif

In any case, the photon IS its own antiparticle. This is allowed because it doesn't have a charge; a photon under time reversal is still a photon (but an electron under time reversal becomes a positron).

Edit: And also, OP, you can scatter photons off of each other. However, you can't think of this as classical billiard ball scattering due to the quantum nature of particles like the photon. You're not shooting single photons at single photons and watching them bounce off of each other. You're shooting clumps of photons often, and most of them won't even interact.

Furthermore, even if you do successfully manage to shoot a single photon and a single photon, there is a non-zero probability that they just won't interact at all. Thus, you can't simply bounce a laser beam off of another laser beam.

 

[Eeezee]

Originally posted by: MrDudeMan

Originally posted by: silverpig

Originally posted by: Born2bwire
Also should note you cannot expect to achieve perfect deconstructive interference. You will notice that in the slit diffraction experiments that photons will still land in the regions of deconstructive interference. The inherent finite sizes of potentials and the fact that the wave equation describes something akin to a probability distribution function prevents perfect cancellation.

The banding is caused by geometry...

It's normally distributed though, so there has to be some number of photons even in the deconstructive areas.

Agreed, we like to teach that the interference is perfectly destructive but in reality there will always be some photons hitting those regions, due to uncertainty, but not enough to detect easily.