Light and intensity

[chiddler]

If light #1 has intensity I and light #2 has equal intensity are both constructively shined upon the same spot, what is the intensity on that spot?



With constructive interference, amplitude doubles and, according to the intensity equation, amplitude is in a square relation with intensity.

So. Answer is 4I.

I understand how this works mathematically. But it seems like energy is being created. Intensity is power / area = energy / area * time. Each light by itself will have x energy/time and then it's quadrupled to 4x when they are together.

Where is this energy coming from?

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[MedPR]

If light #1 has intensity I and light #2 has equal intensity are both constructively shined upon the same spot, what is the intensity on that spot?



With constructive interference, amplitude doubles and, according to the intensity equation, amplitude is in a square relation with intensity.

So. Answer is 4I.

I understand how this works mathematically. But it seems like energy is being created. Intensity is power / area = energy / area * time. Each light by itself will have x energy/time and then it's quadrupled to 4x when they are together.

Where is this energy coming from?

Which intensity equation are you talking about? Sorry, can't help, but I am curious as well.

 

[chiddler]

Which intensity equation are you talking about? Sorry, can't help, but I am curious as well.

this ridiculously long eqution I = ½ρω²A²v. This equation is presented in EK's book.

 

[MedPR]

this ridiculously long eqution I = ½ρω²A²v. This equation is presented in EK's book.

Oh that stupid equation that I ignored. Where's milski when you need him?

 

[pfaction]

Dammit, I actually knew this.

Intensity of a Wave = Amplitude squared / Radius squared. I was going to say that since the amplitude is doubled, 2sq/1 is 4x.

 

[milski]

Yes, it's amplitude squared. The opposite paradox is two waves in destructive interference with total intensity of 0. The trick is that you cannot create a setup which has only constructive or only destructive interference - whatever you gain in some regions you are going to lose in others.

 

[chiddler]

I don't understand what the trick is. Can you please explain.

Oh and hey where does the destructed energy go? Is it dissipated as heat?

Probably not...a mechanical wave doesn't dissipate that energy as heat.

 

[milski]

You have to consider all the energy along the beam of light. Yes, in the region where there is constructive interference there will be more energy than you would expect but for each such region there will be regions with no energy at all. At the end, it all evens out. The only way to avoid that is to have the entire beams be completely coincidental, which makes them virtually a single beam.

 

[chiddler]

okie dokie.

thanks very much for the responses.

 

[milski]

okie dokie.

thanks very much for the responses.

It was a cool question. I spend a while with a big "WTF" over my head before I gave up and started researching for an answer. Supposedly someone even wrote a paper on that.

 

[chiddler]

It was a cool question. I spend a while with a big "WTF" over my head before I gave up and started researching for an answer. Supposedly someone even wrote a paper on that.

thank you. it is a bit refreshing to know that you bleed, too

 

[MedPR]

Yes, it's amplitude squared. The opposite paradox is two waves in destructive interference with total intensity of 0. The trick is that you cannot create a setup which has only constructive or only destructive interference - whatever you gain in some regions you are going to lose in others.

You have to consider all the energy along the beam of light. Yes, in the region where there is constructive interference there will be more energy than you would expect but for each such region there will be regions with no energy at all. At the end, it all evens out. The only way to avoid that is to have the entire beams be completely coincidental, which makes them virtually a single beam.

So you're saying that even though we think of completely destructive interference as regions of 0 energy, there isn't an actual loss of energy. The energy that isn't in the destructive regions must have gone somewhere else and that somewhere else ends up being everywhere non-destructive which includes constructive regions. So the constructive regions actually get more energy than we think they would?

 

[ipodtouch]

In the spirit that

1 Wave + 1 Wave = intensity x4

And if this multiplication is from energy received from destructive interference.
Can we conclude that for every constructive interference, there is also an accompanying destructive interference?

# constructive = # destructive



It seems difficult to understand how this would result in a beam that is "brighter," unless the light is reduced to a smaller total area.

 

[milski]

That's correct, energy is never lost. You have to be careful applying the formula that chiddler quoted. It's for the average intensity along a beam/media and assumes that the wave is a simple harmonic wave.

There is more energy at the regions with constructive interference the same way that there is no energy in the regions where there is destructive interference. You cannot make any simple conclusions about the sizes and the number of such areas since you also have to account for areas where you have only partial constructive/destructive interference. All these will depend strong on the geometry of your setup.

TL;DR: Energy is preserved, let the grad students worry about the details.

 

[milski]

In the spirit that

1 Wave + 1 Wave = intensity x4

And if this multiplication is from energy received from destructive interference.
Can we conclude that for every constructive interference, there is also an accompanying destructive interference?

# constructive = # destructive



It seems difficult to understand how this would result in a beam that is "brighter," unless the light is reduced to a smaller total area.

Not really, since you gain 2 extra units at fully constructive interference but lose only one for fully destructive. Leaving it as "it's complicated but it works out in a way that energy is not lost" is good enough for me.