The sodium D-lines are a pair of narrow, closely spaced, approximately equal intensity spectral lines with a mean wavelength of approximately 589 nm, with wavelength difference 0.60 nm. A Michelson interferometer is set up to study the D-lines from a sodium lamp. High contrast fringes are seen for zero pathlength difference between the two arms of the interferometer. The fringes disappear when the pathlength difference is increased to 0.29 mm.

Question: If the spectral lines have approximately Gaussian shapes, with a width of 50 pm (taken between the points of the line shape where the intensity falls to e −1/2 of the peak intensity), what is the maximum fringe contrast (visibility) seen for a pathlength difference of around 4 mm?

What is the "line shape" and what is it plotted as a function of? Considering that the width of the peaks in the line shape is given in units pm, I suppose it must be some sort of length. However, I am not sure what length, because the one on the screen is not general enough (it depends on the distance etc.).

Fringe visibility: $$ v = \frac{I_{\max}-I_{\min}}{I_{\max}+I_{\min}} $$

  • $\begingroup$ Can you sketch and describe your experimental setup, so I can answer more fully? See my answer here for discussion of coherence on visibility: from the formulas there, the $1/e$ coherence length for a 50pm wide source is about 1.5mm, therefore the visibility of fringes when this source acts on a path difference of 4mm is of the order of $\exp(-(4/1.5)^2)\approx 0.001$. Only visible in high SNR conditions. $\endgroup$ Dec 15, 2017 at 9:48
  • $\begingroup$ @WetSavannaAnimalakaRodVance I am afraid it will be hard to supply more information about the set-up, other than that it is a Michelson-Morley interferometer. This is a theoretical question I have been given, not an actual experiment I have performed. $\endgroup$
    – Jhonny
    Dec 17, 2017 at 10:57


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