Interference: Optical path difference and phase difference etc. AH

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C.Roger
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Hi. I'm very confused on working with destructive and constructive interference and how to calaculate Optical path Difference and phase difference in different situations. For instance why is it different for thin film interference and bloomed lenses. Sorry I dont think that makes much sense. As in how do you know for destructive interference OPD = lamda/2 that kind of thing. Thanks!
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lordaxil
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The basic principle is that if the phase of reflected wave is shifted by \pi radians relative to original then interference will be destructive. This can occur in a couple of different ways:

i) For thin films (say, oil on water), with a phase shift of \pi radians at air-oil interface on reflection and a  \lambda/2 optical path difference from oil layer.

ii) For bloomed lens (coating on glass) with a phase shift of \pi radians on reflection at BOTH air-coating and coating-lens interfaces and a  \lambda/4 OPD from coating.

Light gets a phase shift of \pi radians on reflection when it moves from a medium of lower to higher refractive index, but not the other way around.

There are some diagrams and a more quantitative analysis here:
http://physics.bu.edu/py106/notes/Thinfilm.html
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Croger16
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Hi, thank you for the reply. I was wondering if you could just explain a bit more to me what optical path difference and phase difference is?

Thanks
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lordaxil
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Sure - an optical path difference (OPD) is the product of the geometric distance through a medium (i.e. that measured by a ruler) with its refractive index in that medium (i.e. relative speed of light). The phase shift (or difference) resulting from the OPD will be equal to 2 \pi (\rm{OPD}/\lambda), where \lambda is the wavelength.
Light can also pick up a phase shift from being reflected if it moves from a medium of lower to higher refractive index. This occurs instantaneously at point of reflection and doesn't depend on OPD.
Hope that helps.
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