The following text is from Concepts of Physics by Dr. H.C.Verma, from the chapter on "Light Waves", for the number of fringes that will shift in a Young's double slit experiment (YDSE) when one of slits is covered by a transparent sheet of thickness $t$ made of a material of refractive index $\mu$:

When the light travels through a sheet of thickness $t$, the optical path travelled id $\mu t$, where $\mu$ is the refractive index. When one of the slits is covered by the sheet, air is replaced by the sheet and hence, the path changes by $(\mu-1)t$. >> One fringe shifts when the optical path changes by one wavelength. Thus the number of fringes shifted due to the introduction of the sheet is $\frac{(\mu-1)t}{\lambda}$. <<

Could anyone please explain what the author means by the statement marked by ">>" and "<<"? I understood the rest of the text. I thought when we introduce the transparent sheet, the entire interference pattern (all fringes) will be shifted on the screen. But according to the given statement, it seems only few fringes will shift their positions. If this is the case, how will the resultant interference pattern look like? Will there be huge regions of pure darkness and regions where shifted fringes coincide with existing fringes? (I don't think so)

There is no further explanation for this in my textbook. I think the Wikipedia article on Fringe Shift is some other phenomenon (reason: Michelson–Morley experiment is new to me and not introduced anywhere in my textbook).


1 Answer 1


The author means that the entire interference pattern shifts, and that the magnitude of the shift is by one fringe.

Similarly, if $(\mu-1)t=n\lambda$, the entire interference pattern will shift, and the magnitude of the shift will be by $n$ fringes.

  • $\begingroup$ Thank you for your answer. Could you please explain why does a shift of one fringe takes place when optical path changes by one wavelength? Is that because fringe width is given by $w=\frac{D\lambda}{d}$ where $D$ is the distance of the slit plane from the screen and $d$ is the distance between the slits? $\endgroup$
    – Vishnu
    Dec 9, 2019 at 16:04
  • $\begingroup$ The core reason is that if you shift the light that's coming from that slit by one wavelength, the net change is completely null ─ the two states of light are indistinguishable. From that it follows that the effect on the interference pattern must also be null, i.e. if there is a shift, then it must be such that the pattern isn't changed. Since a shift by one wavelength is the smallest shift on the wave such that the states are identical, the shift on the pattern must also be the smallest possible, i.e. it must be by one fringe only. $\endgroup$ Dec 9, 2019 at 16:41
  • $\begingroup$ Thank your for your reply. I didn't understand "From that it follows ... pattern isn't changed. " because, if we have no net change in the pattern, then why do we have such a formula (for number of fringes the entire pattern shifts)? Or in other words, if there is no change in the pattern, how will be able to tell there was a shift in pattern in the first place? I think, when we gradually increase the thickness of the sheet from $0$ to $t$, the pattern oscillates with amplitude equal to fringe width. So any shift by integral multiples of $\lambda$ results existing pattern ($0$ fringe shift). $\endgroup$
    – Vishnu
    Dec 10, 2019 at 12:14

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