# Snell's law and what determines color of light

White light is dispersed by a prism into the colors of the visible spectrum with wavelengths ranging from violet 380 to red 750 nanometers. By Snell’s law, the refractive index $$n_{21}=n_2/n_1=sin⁡θ_1/sin⁡θ_2 =v_1/v_2= λ_1/λ_2$$ . For example, the refractive index $$n_{21}$$ of water is 1.3 and glass is 1.5. Light moves slower in glass than in water because its wavelength has shortened more.

There is a discussion of whether the light color is determined by frequency or wavelength. We see rainbow in a glass prism because it reflects, part of the light is scattered, by glass back to air. So, its wavelength has not really changed because it changed from $$λ_1$$ in air to $$λ_2$$ in glass back to $$λ_1$$ in air. But if we could observe it inside the glass, what color would we see?

The wavelength spectrum inside, $$λ_2 (=λ_1/1.5)$$, is shifted towards blue [253 nm, 500 nm]. If the light color our retina and brain detect is based on wavelength (wave-like $$λ = \frac{h}{p}$$), we would not be able to see red, orange, and yellow. Even if the medium is water, we lose red and orange colors because it is shifted to [292 nm, 577 nm]. On the other hand, if light color is based on frequency ν (particle-like $$E=hν$$ or Photoelectric effect), the observer inside the medium would be able to see the same spectrum as the one outside because there is no frequency change. Is this argument plausible?

On one hand, you see light when light stimulates a rod or cone in the retina. I don't know the photochemistry. But I presume a photon promotes an electron to an excited state. This implies the photon must have a suitable energy. Since $$E = h\nu$$, it must have a suitable frequency.