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One of the conditions stated for interference to occur was source should be monochromatic.

why this condition needs to be satisfied for interference to occur?

Is it because all mathematical equations use wavelength and single color means single wavelength?

Shrini Kulkarni
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  • Monochromatic light isn't actually necessary for interference; see white-light interferometry. More generally, I think that some form of statistical coherence is all that is needed for interference effects to occur; Emil Wolf has a good book on the statistical interpretation of interference effects called "Introduction to the theory of coherence and polarization of light". But monochromatic interference is simpler and more widely taught than broadband interference AFAIK. – DumpsterDoofus Feb 04 '14 at 17:59
  • @DumpsterDoofus Why not turn that into an answer? – joshphysics Feb 04 '14 at 18:22
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    From your questions it looks like you are just starting out on this topic, but you can answer this question for yourself. Try assuming light of two well defined frequencies (bichromatic?) and looking at the resulting patterns. The math is more tedious that for the monochromatic case, but should be within your reach. Having done that you will know instead of having been told. – dmckee --- ex-moderator kitten Feb 04 '14 at 18:38

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As @DumpusterDoofus and @dmckee mentioned, monochromatic light ($ \lambda = \lambda \pm d\lambda$) is not necessary for interference to occur. Interference can occur on light regardless of its spectral content, polarization state, etc. We are interested in interference as an experimental observation tool, and for our purposes, we like to work with monochromatic light because it is just easier.

sammy gerbil
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mcodesmart
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