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A monochromatic red ray of light, enters a beaker of water , $\gamma = \gamma ' / \mu $ , $\mu $ is the refractive constant.

The wavelength should reduce, (taking $\mu$ as $1.33$ ), so the light should become green, but we see it as red light. Why?

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  • $\begingroup$ I don't think it can reduce to that extent so as to change the colour. $\endgroup$ – Wrichik Basu Aug 16 '17 at 17:49
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This is an important question because it gets at the concept of color, and what that really means. Mitchell is correct that when you observe scattering from a laser beam sent through a beaker of water, you are really only seeing the photons which have left the water ($\mu\approx1.33$) and the beaker (glass $\mu\approx1.5$), traveled through the air ($\mu\approx1.00029$), entered your eye ($\mu\approx1.33$), and were absorbed by your retina. So which color do you see? And does it change if you repeat the experiment underwater or use oil ($\mu\approx1.5$) in the beaker instead?

The key to understanding why we always see red light in this experiment rather another color is due to the nature of your photoreceptors, which (like most light absorbers) are not concerned with the wavelength of the light. They instead respond to the frequency of the light, $\nu$ (or photon energy $E=h\nu$), which remains unchanged as light propagates from one (linear) medium to the next. The reason for this is that there are certain energies, determined by quantum mechanics, which are assigned to the electronic states in the absorbing material (your retina, in this case). In order to conserve energy, the photon needs to have the right energy to be absorbed, promoting the material from a low energy state to a high energy state. Wavelength has nothing to do with it. Indeed, when we say "color" in reference to light, we are often actually speaking of the frequency rather than wavelength. When we do refer to a wavelength as a color, it is always with an implicit assumed refractive index, usually $\mu=1$.

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  • $\begingroup$ But what about c= (frequency)(wavelength) ... $\endgroup$ – Sidharth Giri Sep 19 '17 at 6:33
  • $\begingroup$ @SidharthGiri Yes, you are right. But you need to use the wavelength of light in the medium, which is (wavelength)/(refractive index). This describes the fact that light travels slower in a dielectric compared to vacuum. But regardless of the refractive index, the frequency of light stays the same, and that is what's important for your perception of color. $\endgroup$ – Gilbert Sep 19 '17 at 12:42
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You are seeing that part of the ray that is coming out of the beaker, with its original wavelength. If the monochromatic radiation has reached your eye through the beaker then it means that it would have got out of the beaker.

The emergent radiation has the same wavelength as the incident radiation.

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