In the drivation of Snell's law for light as EM waves, we have the wave vector components parallel to the interface $k1\parallel$ = $k2\parallel$ as shown in the picture.
From $k_{1x} = k_{2x}$, we have $k_1sin\theta_1 = k_2sin\theta_2$. From the well known equations below, $$ \frac{k_2}{k_1}=\frac{\lambda_1}{\lambda_2}=\frac{v_1}{v_2}=\frac{n_2}{n_1}=\frac{p_2}{p_1} $$ where the relationships among wave vector $\vec{k}$, wavelength $\lambda$, velocity $\vec{v}$, refractive index n and momentum $\vec{p}$ are shown, we arrive at the Snell's law $n_1sin\theta_1 = n_2sin\theta_2$. The perpendicular $k\perp$ component in medium 2 can be expressed in terms of medium 1's as: $$ {k_{2y}}=\frac{n_2}{n_1}\frac{cos\theta_2}{cos\theta_1}{k_{1y}}=\sqrt{k_{1y}^2+((\frac{n_2}{n_1})^2-1){k_1}^2} $$ and ${k_{2y}} > {k_{1y}}$ for ${n_2}>{n_1}$ as expected.
Since $\vec{p}=(h/2\pi)\vec{k}$, there is a momentum gain of from $p_1$ to $p_2$ across the boundary and it has been discussed here and here. Since $F=\frac{dp}{dt}$ and $v_1$ and $v_2$ are constant speeds, there must be an impulsive photon force that is normal to the interface surface of the discontinuity. What is that normal force that pushes the photon across from 1 to 2 and why does the photon speed in medium 2 become slower? How about the law of conservation of momentum?
