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I recently learnt the young's double slit experiment and its basic formulas, but i wondered what would (mathematically) happen if i put one of the 2 slits into a refractive medium such as water or glass. If a formula exists please show the derivation.

Note : The medium is only in between the slit and the screen , everything else is the same .

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  • $\begingroup$ clarify if the refractive medium covers the whole path length light takes via one slit or not. Where are the entry and exit boundaries for this medium? $\endgroup$ – JMLCarter Mar 30 '17 at 15:27
  • $\begingroup$ I have edited my question , hopefully its more clear than it was before . $\endgroup$ – Harshit Pandey Apr 1 '17 at 2:47
  • $\begingroup$ I have to assume the medium is thin, and resides near the slit. Otherwise light rays from the second slit to different fringes will pass through it...? $\endgroup$ – JMLCarter Apr 1 '17 at 3:23
  • $\begingroup$ Many slots are created on glass slides $\endgroup$ – Lambda Jun 10 '18 at 2:35
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The main effect of placing a refractive medium in one of the beam paths is that the phase of that beam will be delayed by an amount depending on the refractive index of the medium and the thickness of the medium. If the phase of one of the beams is delayed slightly, the interference fringes will shift laterally. Delay the phase by 1/2 cycle, and the fringes will shift by 1/2 of their spacing.

Note that the whole fringe pattern will not be shifted laterally. The fringe spacing (which is greatest in the middle and gets smaller toward the outside) will not be changed in the pattern; only the locations of the bright and dark fringes will change. In other words, for small shifts the amount of lateral shift is proportional to the fringe spacing. If the amount of shift is an integral number of full cycles, then the fringe pattern will return to its original form. That's not exactly true. If the total phase shift is greater than the coherence length (in cycles) of the light source, then the light from the two slits will not interfere to form fringes. Fringe contrast is a function of the mutual coherence of the two beams.

So, a two-slit interferometer with an adjustable-thickness refractive medium in one path can be used to measure the coherence function of a light source.

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The fringe pattern will be narrower.

Try thinking about [and specify] the frequency of Light here, then remember that $c = f\lambda$, the speed of light in water is $225,000,000$ so then $\lambda$ for light of some given frequency becomes $225000000/f$, and try combining that with the equations like $dsin(\theta) = n\lambda$

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  • $\begingroup$ no because the OP intends the projection screen isn't in the refractive medium, only the slit. I think the answer is about refractive effects especially the increase in path length. $\endgroup$ – JMLCarter Mar 30 '17 at 15:21
  • $\begingroup$ Oh, I didn't think that was particularly clear from the question. In that case it would just be an application Snell's law though surely? $\endgroup$ – user95137 Mar 30 '17 at 15:52
  • $\begingroup$ Maybe, just, that could be uninteresting though, I did post a comment asking for clarification. $\endgroup$ – JMLCarter Mar 30 '17 at 16:07
  • $\begingroup$ I think it's physically impossible to cover all the paths over the entire path length through one slit in one medium, but not the other. The path's for different fringes overlap. $\endgroup$ – JMLCarter Mar 30 '17 at 16:08
  • $\begingroup$ Oh jeez I didn't even see that part, either that's an edit or my brain just skipped a few pulses $\endgroup$ – user95137 Mar 30 '17 at 16:40
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In the conventional derivation that is done in schools, for the sake of simplicity the light coming from both the slits travels through the same medium. For sustained constructive interference the path difference of the two coherent sources must be an integral multiple if lambda ie. n lambda and for destructive interference equal to integral multiple of lambda/2. Now let the path travelled by light in a medium of thickness 't' and refractive index 'u' be 'x'. Here, x = u t

So for any specific arrangement of yours you simply have to find the path difference of the light from the two sources the same way you do otherwise, but take multiply the distance by 'u' for the corresponding ray.

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The fringes will be more blurred.

If you will perform such an experiment here is the explanation how to prove what happens. First at all we have to discover what happens if we use only one of the two slits. You will see that one slit still form an intensity distribution (fringes) on the observers screen.

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Airy disk

Observing single slits with different media in the slit or behind the slit one should see a change in the dimensions of the fringes and the distances between them. The used by you medium inside the slit is responsible for the interaction of the light with the slit and a medium behind the slit is responsible for the deflection due to the refraction index.

Since you are able to perform such an experiment - and if you are able I ask you to write about the results here - than you perhaps are able to perform an experiment where you will change the simple the distance between the slits to prove that the common intensity distribution behind the slits is simply the overlapping distribution from the single slits.

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