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Can anyone explain the relationship between the refractive index, the speed, wavelength and angle of a wave?

in my book is states that $$n = \frac{v_1}{v_2} = \frac{\sin θ_1}{\sin θ_2} = \frac{λ_1}{λ_2}$$ but it doesn’t state any explanation for the relationship.

Qmechanic
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dayum
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    See https://physics.stackexchange.com/q/97894/123208 https://physics.stackexchange.com/q/334145/123208 https://physics.stackexchange.com/q/466/123208 & links therein – PM 2Ring Mar 10 '24 at 03:08

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Well, if you draw phase fronts at the boundary, you should be able to convince yourself that the transverse component of the wave vector needs to be the same in both media:

$$ k_1\sin\theta_1 = k_2\sin\theta_2 $$

so

$$ \frac{\sin\theta_1}{\sin\theta_2} = \frac{k_2}{k_1} = \frac{\lambda_1}{\lambda_2} $$

(If you want a picture, put it in the question).

The dispersion relation in media is:

$$ \omega = vk = \frac{ck} n $$

Put that in the first formula and you get Snell's Law.

It's just boundary conditions. Phase on each side of the boundary has to match, at all ties.

JEB
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