Will two lasers deliver exactly double the heating?

[Bullington]

If you shine two lasers at the same spot that are the same frequency, and; exactly out of phase; would there still be energy at the one spot they are pointed at?

 

[Bullington]

If you shine two lasers of equal frequencies on one spot, will the energy absorbed at the one spot be double?

 

[Simon Bridge]

Where two EM fields exactly cancel, there is zero EM energy and no photons are detected.
This is how interference patterns work.

In the laser example - to be exact, for ideal lasers, you'd need the beams to be coincident.
This would mean there is no beam.

 

[Paul Colby]

Lasers are not perfect. They have a phase or coherence length which can be quite significant number of meters. This translates into a coherence time which is on the order of microseconds. So two lasers that are not phase locked would go in and out of phase in the MHz range or (more likely) much higher. I wouldn't expect to see anything by eye but a very fast photo diode might see the fluctuations. If any true laser expert is listening, please comment with more accurate numbers.

 

[Bullington]

Ah, I see, so I will rephrase the question; if we shine two lasers with equal frequency, at the same spot and they don't interfere with each others path, what could the energy at that one spot be?

 

[Khashishi]

If they don't interfere with each other's path, then the energy on the spot will be double (over a time average).

 

[pixel]

Ah, I see, so I will rephrase the question; if we shine two lasers with equal frequency, at the same spot and they don't interfere with each others path, what could the energy at that one spot be?

Just curious why you would think, without interference, that it would be anything other than double?

 

[Bullington]

Just curious why you would think, without interference, that it would be anything other than double?

Well, I would think it wouldn't be double, maybe if the lasers interfered constructively then I would say the energy would increase, but two lasers on the same spot without any interference, that doubles? This is a new topic for me so I'd like to look into it a bit more, is there any reference for the energy being double?

 

[Paul Colby]

The simplest way to get two phase locked lasers with identical frequencies is to use a single laser and some beam splitters. The beams may be recombined to show the interference patten phenomena your asking about. Because the phase coherence length is real large with a decent laser, this is real simple.

 

[mrspeedybob]

Even if the 2 beams do interfere with each other destructively, the energy absorbed by the target should still be the sum of the 2 beams.
If it less, then where did the non-absorbed energy go?

 

[Paul Colby]

If it less, then where did the non-absorbed energy go?

Elsewhere is the short answer. The best one might do to exactly overlap two destructively canceling beams is likely feeding two ports of a 4 port beam splitter. In this case all the power would exit one port while none would go out the other (canceling one). Beam splitters conserve energy so it all balances when the numbers are worked.

 

[Bullington]

Even if the 2 beams do interfere with each other destructively, the energy absorbed by the target should still be the sum of the 2 beams.
If it less, then where did the non-absorbed energy go?

Ah, this is interesting,

I would like to know where I can find out more about this; I believe there is energy, but I can't justify.

 

[Bullington]

Lets say we do the slit experiment, is there a way to test the energy of the dark spots? I'm getting conflicting information,.

Someone said that even if the two lasers are destructive there energy will be double. Is there any documentation on this? Maybe a good introductory book?

 

[Simon Bridge]

You can test the power per unit area using a light detector. This test is done many times a year and the density falls off in the dark areas. I don't see a contradictory reply in the linked thread.

Note: in the regeon of ideal total destructive interference, there is zero probability of detecting photons.
Therefore there is zero em energy arriving there.

A detector will collect photons from nearbe areas which may not have total destructive interference.
It is also lokely tgat a real laser system departs from the idealized geometry you are using in your thoughts on this matter.

 

[Abhishek Sethi]

Hi,
For superposition of beams of EM waves, the energy will be re-distributed in the space. So, a dark spot can have 0 energy.