Suppose while performing double-slit experiment, the space between the slits and the screen is filled with water. How does the interference pattern change?

The above is one of the questions which was asked in our Engineering Physics class.

My logic was, due to the fact that refraction will take place, the waves will lean closer to the normal, but I am unable to decide what kind of effect this change of medium will have on the wavefront which we typically use to depict Young's double-slit experiment.

I can only conclude that the distance between the fringes will end up reducing. Although, I'd appreciate it if someone with a more in-depth understanding of optics would answer it.

  • 1
    $\begingroup$ Consider spelling out acronyms. $\endgroup$
    – Qmechanic
    Dec 9, 2017 at 12:17
  • $\begingroup$ Consider that "due to the fact that refraction will take place, the waves will lean closer to the normal" is an expectation from ray optics which also holds that rays are straight unless they interact with a changing medium, and yet the 'rays' in the double slit experiment are draw as bending at slits without a medium. Or, to be less cryptic, the double slit experiment depends on wave optics, and you can't naively rely on ray optics: you have to understand how ray optics develop out of wave optics, or just plow ahead in the wave formalism. $\endgroup$ Dec 9, 2017 at 18:21
  • $\begingroup$ So, to move forward you shouldn't be asking yourself about rays, you should be asking how light waves in water differ from those in air or vacuum. $\endgroup$ Dec 9, 2017 at 18:22

1 Answer 1


When the light waves pass through water, it's wavelength becomes $\frac{\lambda}{\mu}$. Due to this, the condition for formation of maxima becomes $$y = n\frac{D\lambda}{\mu d}=\frac{y_0}{\mu}$$ Where $y_0$ is the position of the maxima without any water. So what effectively happens is that the fringes will move closer to the central maximum (since $\mu>1$)


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