Constructive Interference of Electromagnetic Waves

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The electric and magnetic parts of an electromagnetic wave have the same wavelength and the same spatial relationship, so if the electric parts line up the magnetic ones do, too. There is only one wave, not three. If they are traveling in the same direction (as in a laser) they will stay in phase. Constructive interference works fine-that is what makes lasers work and interference patterns in light.

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Pi_Co
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Pi_Co

Updated on August 03, 2020

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  • Pi_Co
    Pi_Co about 2 years

    So I was wondering if Electromagnetic wave has the same property of interference as normal waves. I understand that both the electric and magnetic parts of the wave would have to be in the same position at the same time. To negotiate the fact that only one part of the wave would match up at one time due to the fact that light can't go faster than it's self I had the idea for three waves all intersecting at the same point. All of these waves would be of different wavelengths so that they do not interfering with each other before the main point. If I knew of a good easy way off making a fbd I would but i am rather new to really doing physics Ioutside of a high school classroom. So I was wondering if constructive interference worked on electromagnetic waves and if it does whether or not is decrease's the wavelength (increasing the energy).

  • Pi_Co
    Pi_Co about 8 years
    What I meant was to Have three seperate waves intersecting on one point of the main wave. Resulting in one wave of a different wavelength. Thank you for the clarification and i apologize for not stating my question clearly.
  • Ross Millikan
    Ross Millikan about 8 years
    If you have three EM waves, you can sum the amplitudes and will get constructive/destructive interference. Two of different wavelengths give you the product of the sum and difference frequencies. Adding in a third will give a more complex spectrum, but it will be dominated by the sums and differences of the three basic frequencies.
  • Pi_Co
    Pi_Co about 8 years
    By doing that in a specific way to get one of these nine combinations to be whiten the visible spectrum thus creating an almost infinitesimally small points that creates a wave that is visible to the human eye in 3-d space? Of course you would need a material (like fog) to excite in this tiny area to emit the light.
  • Ross Millikan
    Ross Millikan about 8 years
    If you are interested in only one combination, you can only look at the two waves that contribute to it. It would be hard to make the difference frequency be in the visible unless one of the waves were already visible-UV monochromatic waves are hard to come by. You could imagine adding two $9.4 - 10.6 \mu m\ CO_2$ frequencies to get one in the visible spectrum. The size of the fringe will depend on the frequency difference. Whether you can make it bright enough to be visible I have no idea.
  • Pi_Co
    Pi_Co about 8 years
    Not to argue but wouldn't that end up with a wave with the wavelength of 21μm which is way over the visible spectrum. Or if contrary to my idea does the wavelength get bigger when combining these in constructive interference for electromagnetic waves.
  • Ross Millikan
    Ross Millikan about 8 years
    No, you add the frequencies, so the wavelength is roughly half. But I am off a factor $10$. That gets $5 \mu m$, not $0.5 \mu m$ You need a much closer infrared laser than $CO_2$
  • Pi_Co
    Pi_Co about 8 years
    Ok your still confusing me because your using a a distance to describe the frequency of a light wave but ok.