In my book it says we can use the boundary conditions of electromagnetic waves to derive the refraction law of light. How to derive it?
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2 Answers
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If you have a traveling wave, incident at an angle $\theta$ to the normal to the boundary, then you can express the E-field of the wave as $$\vec{E}(\vec{r},t) = E_0 \exp[i (\vec{k}\cdot\vec{r} - \omega t)] \hat{u}\ , $$ where $\hat{u}\cdot \vec{k} = 0$.
The form of the transmitted and reflected wave can be written in the same format, but with (potentially) different values for $\vec{k}$ and $\omega$.
From there, it is easier to use an example - suppose the boundary is the plane $y=0$ and the normal to the boundary is the y-axis and the incident wave travels in the $x-y$ plane and has $\vec{k} = k(\sin\theta\ \hat{i} + \cos\theta\ \hat{j})$ and $\hat{u} = \hat{k}$.
At the boundary, where $\vec{r} = x\hat{i}$ then the general form of the incident E-field written above becomes $$\vec{E}(\vec{r},t) = E_0 \exp[i (kx\sin\theta - \omega t)] \hat{k}\ , $$ with similar expressions for the reflected and transmitted E-fields, but written with different values of $k$, $\theta$ and $\omega$.
But the continuity condition that says the tangential component of the E-field must be the same either side of the boundary means that $$\vec{E}_i + \vec{E}_r = \vec{E}_t\ , $$ and if this is hold true for any value of $x$ or $t$ along the boundary, then the spatial and temporal parts of the arguments of the exponential functions for the incident, reflected and transmitted waves have to be the same.
This leads to the equality $$ k_i \sin\theta_i = k_t \sin \theta_t\ ,$$ where the subscripts refer to incident and transmitted waves, and $$\omega_i = \omega_t\ .$$ Now, because the speed of light, $c/n = \omega/k$, where $n$ is the refractive index, then if $\omega$ is constant, one obtains $$ n_i \sin\theta_i = n_t \sin\theta_t\ . $$
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Of course you start with a planar boundary btw vacuum ($\epsilon_0$) and and linear isotropic medium ($\epsilon$) and then consider the continuity of the normal and tangential components of the appropriate fields.
Aside: ofc the frequency on either side have to match. That question has come up here before, and there is no way to satisfy boundary conditions with different frequencies on either side.
Looking at Maxwell's equation in media, you should be able to se that the normal component of $\vec D$ is matched at the boundary, and the tangential component of $\vec E$ is, too. (Unless I don't recall correctly...then reverse it).
Anyway, that introduces a incident angle dependent projection of $\vec D$ and $\vec E$ that depends on $\epsilon/\epsilon_0$ ($\propto n^2$, iirc), that is solved at Snell's angle.
