(N/A) Let $PP'$ represent the surface separating medium-$1$ and medium-$2$. Let $v_1$ and $v_2$ be the speeds of light in medium-$1$ and medium-$2$ respectively.
$A$ plane wavefront $AB$ propagating in the direction $AA'$ is incident on the interface at an angle $i$. Let $\tau$ be the time taken by the wavefront to travel the distance $BC$. Thus,$BC = v_1 \tau$.
To determine the shape of the refracted wavefront,draw a sphere of radius $v_2 \tau$ from point $A$ in the second medium. Let $CE$ be the tangent plane drawn from point $C$ to this sphere. Then $AE = v_2 \tau$,and $CE$ represents the refracted wavefront.
In $\triangle ABC$,$\sin i = \frac{BC}{AC} = \frac{v_1 \tau}{AC}$.
In $\triangle AEC$,$\sin r = \frac{AE}{AC} = \frac{v_2 \tau}{AC}$.
Dividing the two equations:
$\frac{\sin i}{\sin r} = \frac{v_1 \tau / AC}{v_2 \tau / AC} = \frac{v_1}{v_2} = n_{21}$.
This is Snell's law of refraction. Since the incident ray,the refracted ray,and the normal all lie in the same plane,the laws of refraction are derived.