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Just a little question. I was wondering why the waves of different colors (so different wave lengths) in white light do not interfere one another. Why interference is not taken place? I believe white light is not a result of interference since a prism can separate different waves in white light. Please explain me. Thanks.

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marked as duplicate by John Rennie, user36790, sammy gerbil, ACuriousMind, heather Aug 26 '16 at 22:36

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  • $\begingroup$ Newton and Young used white light from the sun and where able to see colorised fringes. The fringes where blurry. So even from white light one get fringes on an observation screen. $\endgroup$ – HolgerFiedler Aug 25 '16 at 17:17
  • $\begingroup$ That means it is taken place for some extent, even though not highly detectable? Thanks Holger. $\endgroup$ – Dinuka Aug 25 '16 at 17:21
  • $\begingroup$ Yes to your last comment. BTW, more information see physics.stackexchange.com/questions/212899/… $\endgroup$ – HolgerFiedler Aug 25 '16 at 18:24
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White light is composed of light of all colors. When you speak of interference, as in an interferometer, you are using coherent light of one wavelength. It is easy to see interference fringes from a LASER. But white light also interferes since it is composed of all wavelengths. It's just that the contrast of the fringes is overwhelmed by the light of the other wavelengths. When split and re-combined, each wavelength within the white light will interfere with itself at a different location. So the re-combined energy for each wavelength will be somewhere between fully constructive and fully destructive interference. You will not see the fringes of one particular wavelength because of the signal to noise problem. But white light interferometers exist. You can get White Light Fringes when the test and reference paths are exactly equal. This is how a white light interferometer works (using a Miru microscope objective by Nikon). When the microscope objective is looking at some surface that is rough, only that part of the surface whose overall optical path is EXACTLY equal to that of the reference arm will be dark. As the surface is scanned towards and away from that plane, other areas on that surface will become dark and you can build up a surface height map that is very accurate and precise. (I think I got the spelling of the Miru objective correct... I hope).

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  • $\begingroup$ Thanks Paul. This helps. Didn't have a broad knowledge about this. $\endgroup$ – Dinuka Aug 25 '16 at 20:04

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