Huygens' Proof of Reflection and Refraction

Huygens' principle reproduces the two basic laws of geometrical optics. The derivations use only the construction of wavelets and elementary trigonometry — no new physics is needed.

Reflection

Setup. A plane wavefront $AB$ in medium 1 strikes a flat reflecting surface at angle of incidence $i$. Let $C$ be the foot of the perpendicular dropped from $B$ to the surface; $A$ touches the surface first.

While the wavelet from $B$ travels to $C$ in time $t$ (distance $BC = vt$), the wavelet from $A$ expands into a hemisphere of radius $vt$ above the surface.

The reflected wavefront $CD$ is tangent to this hemisphere from $C$.

Geometry.

Triangles $ABC$ and $ADC$ share hypotenuse $AC$, with $AD = BC = vt$. They are right-angled (incident and reflected wavefronts are perpendicular to rays). Hence

$\sin i = \frac{BC}{AC}, \quad \sin r = \frac{AD}{AC}$

So $\sin i = \sin r$, i.e., $\boxed{i = r}$. The incident, reflected, and normal are coplanar.

Refraction (Snell's law)

Setup. Plane wavefront $AB$ in medium 1 (speed $v_1$) meets a flat interface; the refracted wavefront $CD$ moves through medium 2 (speed $v_2 < v_1$ for a denser medium 2).

In time $t$:

• Wavelet from $B$ reaches $C$ on the interface: $BC = v_1 t$.
• Wavelet from $A$ has entered medium 2 and travelled into it a distance $AD = v_2 t$.

Both right triangles share hypotenuse $AC$ along the interface.

$\sin i = \frac{BC}{AC} = \frac{v_1 t}{AC}, \quad \sin r = \frac{AD}{AC} = \frac{v_2 t}{AC}$

Divide:

$\frac{\sin i}{\sin r} = \frac{v_1}{v_2}$

Using $v = c/n$:

$\frac{\sin i}{\sin r} = \frac{n_2}{n_1}$

$\boxed{n_1 \sin i = n_2 \sin r}$

This is Snell's law, derived purely from the wave construction.

Worked Example

A plane wave in air ($n=1$) is incident at $60^\circ$ on water ($n = 4/3$). Find the refraction angle.

$\sin r = \frac{n_1}{n_2}\sin i = \frac{1}{4/3}\sin 60^\circ = \frac{3}{4}\cdot\frac{\sqrt 3}{2} \approx 0.6495$

$r \approx 40.5^\circ$

Common Confusions

• The line $AC$ along the interface is the same in both triangles — that's what makes the proof work.
• The angle of incidence is between the incident ray and the normal, equivalently between the incident wavefront and the interface.
• The speed used in the wavelet construction is the speed in that medium: $c/n$, not $c$.

Key Takeaways

• Huygens construction reproduces reflection ($i = r$) and refraction ($n_1\sin i = n_2 \sin r$).
• The argument uses only two right triangles sharing a hypotenuse along the interface.
• Refractive index ratio = inverse ratio of wave speeds.