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I have a problem with interferometer fringe shift. I am learning it from my physics book. In the book, it is written,

"As the L1 (Displacement of moveable mirror) is changed, the pattern of interference fringes is observed to shift. If M1 (Moveable Mirror) is displaced through distance equal to $\lambda /2$, a path difference of double of this displacement is produced, i.e equal to $\lambda$."

I want to know why, the path difference generated is double of displacement of mirror? Is it just an assumption? And tell me about that fringe shift.

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  • $\begingroup$ We have MathJax active on the site allowing you to write mathematical symbols and expressions in a LaTeX-math-mode equivalent language. I've done the Greek letters here for you. See physics.stackexchange.com/editing-help for a little help and the wide internet for more general help on LaTeX. $\endgroup$ Apr 28 '16 at 16:41
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The light is reflected from a mirror. If the mirror moves a distance $\frac \lambda 2$ then the incident ray of light has to travel an extra distance $\frac \lambda 2$ to reach the mirror and then the reflected ray of light has to move an extra $\frac \lambda 2$ a total extra distance of $\lambda$.

If the path length changes by one wavelength then there is a movement of one fringe.

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  • $\begingroup$ Doesn't fringe spacing depends upon the wavelength of light used and the distance of screen or observer? Then there should be different path difference for fringe shifting for different lights. It shouldn't be [lambda] for every light. $\endgroup$ Apr 28 '16 at 17:01
  • $\begingroup$ You see a set of interference fringes because of a difference in the path length between the two arms of the interferometer. $\endgroup$
    – Farcher
    Apr 28 '16 at 17:02
  • $\begingroup$ What are arms of interferometer? $\endgroup$ Apr 28 '16 at 17:03
  • $\begingroup$ You see a set of interference fringes because of a difference in the path length between the two arms of the interferometer. If the light source only has one wavelength then moving a mirror by $\frac \lambda 2$ will cause a one fringe displacement. If there are a number of wavelengths present then each of the fringe patterns will move different amounts and you will see them moving relative one another. So the fringe shift will be $2 \times \dfrac{\text{distance moved by mirror}}{\text{wavelength of light}}$ $\endgroup$
    – Farcher
    Apr 28 '16 at 17:08
  • $\begingroup$ The interferometer is in the form of a cross and I called them the arm. $\endgroup$
    – Farcher
    Apr 28 '16 at 17:09

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