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For the medium let’s use water. When light passes through water its wavelength decreases. It’s frequency stays constant. It changes its direction upon entering the water unless it enters the water orthogonal to the surface. It exits the water at the same angle it enters. The current explanation for this is the absorption and reemission of photons by the atoms in the water. I can understand how this explanation works for the frequency remaining the same, but how does it explain the decrease in wavelength? As far as the change In direction goes that’s too much for one question. I’ll just accept the marching band explanation we are all familiar with for now. It’s the wavelength question that I would like to know.

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One of the Maxwell equations in the vacuum has the magnetic and electric constants:

$$\nabla \times \mathbf B = \mu_0 \epsilon_0 \frac{\partial \mathbf E}{\partial t}$$

so that in the wave equation, derived from the above and the other three:

$$\frac{\partial^2 \mathbf E}{\partial t^2} = \frac{1}{\mu_0 \epsilon_0}\frac{\partial^2 \mathbf E}{\partial x^2}$$

the wave speed is determined by then.

In the water, that constants are different: $\mu_w$ and $\epsilon_w$, and the speed is smaller.

For a plane wave of the form: $\mathbf E = E(k_w(x - vt))$, where $v = (\mu_w \epsilon_w)^{-1/2}$

if the frequency $\omega = k_wv$ is the same as in the vacuum, then $k_wv = kc$ => $$k_w = \frac{c}{v}k = \left(\frac {\mu_w \epsilon_w}{\mu_0 \epsilon_0}\right)^{1/2}k$$

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