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Imagine the familiar setup of the double slit experiment using laser but it is submerged in a tank filled with water, would there be any interference pattern? I also read somewhere that tonic water contains quinine which glows in black light or UV-A, what if I replace the water in the tank with tonic water and switch the laser with the black light version. Now would there be any interference pattern in this scenario?

user6760
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If the experiment is done in water rather than air, there will still be an interference pattern.

Water will not destroy the interference pattern. A path difference between the rays will still be produced (assuming not a lot of scattering occurs). Perhaps the main difference will be that the wavelength of the light in the water will be refracted by an amount$^1$ $$\lambda ' = \frac{\lambda}{n}\tag1$$ where $n$ is the refractive index, which also means the fringes will be shifted (closer together).

If you switched the light to a UV-A source, since this light is of a wavelength $320\text{ nm}-400\text{ nm}$ (in air), will mean the refracted light in water ($n=1.33$) has a wavelength in the range $$\lambda '\approx 240\text{ nm} -300\text{ nm}$$ which is high energy and into the UV range. The refraction pattern will probably be not very (if at all) detectable. And it may also be that with tonic water (even with smaller wavelength light), there may be enough scattering (due to particulates and bubbles) so as to prevent the formation of a discernible interference pattern (c.f. what happens in the Tyndall effect).

$^1$ Refractive index, $n$ is defined as $$\boxed{n=\frac cv}\tag2$$ where $c$ is the speed of light (in vacuum), $v$ is the same for the medium, where $c=f\lambda$ and $v=f\lambda '$. Note that due to energy conservation, frequency $f$ does not change when travelling from air to water. Using equation (2) we can now write $v=\large\frac cn$ so that $$f\lambda ' = \frac{f\lambda}{n}$$ so that we obtain equation (1) $$\lambda ' = \frac{\lambda}{n}$$

joseph h
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    When the light leaves the water and travels agin through air the original wavelength is restored. – my2cts Jan 04 '24 at 22:18
  • @my2cts No. I thought about that, but the OP is talking about the entire setup being "submerged in a tank". That is, the source, slits and screen. Thanks for the suggestion though. – joseph h Jan 04 '24 at 22:27
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    I think @my2cts is right. The reason that the interference pattern is "visible" on the screen is that the photons scatter off of the screen and travel to our eye; so if our eye is out of the water, then their logic applies. And even if the eye is underwater too, it should still be visible; it's the energy of the photon that is directly relevant as to whether the pigments in our retina are excited, not the photon's wavelength. – Michael Seifert Jan 05 '24 at 21:37
  • @MichaelSeifert Again, no. The point of the question is to ask whether an interference pattern forms. Whether the observer is underwater or not is totally irrelevant. Whether we can see a pattern is also irrelevant. The idea that the question asks if we see an interference pattern renders the entire question as pointless, since yes, the light travels in air to our eyes. Think about it like this: rather than having a screen, assume we have some instruments that can detect the maxima of the pattern in the water. Will it register an interference pattern/shifted fringes? Thanks Michael. – joseph h Jan 05 '24 at 22:06
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    I'm not disputing that an interference pattern forms; I agree that it does. I'm questioning your description of it as "not very (if at all) visible" in your answer based on the wavelength of the light as measured in the water. But maybe I've misunderstood what you mean by that. – Michael Seifert Jan 05 '24 at 22:59
  • @MichaelSeifert I see Michael. That probably was not the best terminology on my part. I will edit this as soon as I'm free. Thanks again. – joseph h Jan 05 '24 at 23:15
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Not quite a double slit experiment, but still an experiment on light interference in a liquid (Optics Express Vol. 14, Issue 14, pp. 6434-6443 (2006)) :

Immersion interference lithography was used to pattern gratings with 22-nm half pitch. This ultrahigh resolution was made possible by using 157-nm light, a sapphire coupling prism with index 2.09, and a 30-nm-thick immersion fluid with index 1.82.

akhmeteli
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The only change would be that the fringe WIDTH will reduce to $$\lambda\cdot( \frac Dn) \cdot\text{slit width}$$

joseph h
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