= Solution
The <Huygens-Fresnel principle> constructs a new <wavefront> as the envelope of secondary wavelets emitted by points on an earlier <wavefront>. Let the <phase velocities> in the two media be $v_1=c/n_1$ and $v_2=c/n_2$, with <refractive indices> evaluated at the light's fixed <frequency>.
\Image[/past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2019-ia-physics-q3-wavelets.png]
{title=An original secondary-wavelet construction of refraction at a plane interface}
For <refraction from secondary wavelets>, let point $A$ of the incident <wavefront> reach the interface first. A second point $B$ reaches the interface at $C$ after time $\Delta t$, so $BC=v_1\Delta t$. During the same interval, the secondary wavelet centered on $A$ advances radius $AD=v_2\Delta t$ into the second medium. The refracted <wavefront> $CD$ is tangent to this wavelet, hence $AD\perp CD$; also $BC\perp AB$ because a ray is normal to its <wavefront> in these isotropic media.
Let $\theta_1,\theta_2$ be the incident and refracted ray angles from the interface normal. They also equal the corresponding acute angles between the <wavefronts> and the interface. The two right triangles have common hypotenuse $AC$, giving
$$
\sin\theta_1=\frac{BC}{AC}=\frac{v_1\Delta t}{AC},\qquad
\sin\theta_2=\frac{AD}{AC}=\frac{v_2\Delta t}{AC}.
$$
Their ratio eliminates both $AC$ and $\Delta t$:
$$
\frac{\sin\theta_1}{\sin\theta_2}=\frac{v_1}{v_2}=\frac{n_2}{n_1},
\qquad
\boxed{n_1\sin\theta_1=n_2\sin\theta_2.}
$$
This is <Snell's law>. In particular, entering a medium of larger <refractive index> reduces the ray's angle to the normal. The construction assumes a stationary interface and homogeneous isotropic media with a propagating refracted wave.
Back to article page
