Exercises — Huygens' principle — wavefront propagation

Same phase. A wavefront is the surface joining all points of the wave that are oscillating in step (same point in their up-and-down cycle — same phase, as defined above) at that instant.

Plane (flat). Close to the source the wavefronts are spheres. Look at a tiny patch of a huge sphere — like standing on the Earth, it looks flat. So far away the curvature is negligible and the wavefront is effectively a plane. See the parent for the sphere→plane picture.

$r=vt$ Because the wavelet travels at the medium speed $v$ for a time $t$, and distance = speed $\times$ time.

$n=\frac{c}{v}=\frac{3.0\times 10^8}{2.0\times 10^8}=1.5$ Why divide? $n$ measures how many times slower light is inside the medium than in vacuum. $n=1.5$ means light is $1.5\times$ slower in the glass.

$r=vt=(2.0\times 10^8)(5.0\times 10^{-9})=1.0\text{m}$

From the Huygens refraction result in the parent note (medium 1 = air, medium 2 = glass), $\frac{\sin i}{\sin r_{\text{ref}}}=\frac{v_1}{v_2}=\frac{n_2}{n_1}=\frac{1.5}{1}=1.5$ So $\sin r_{\text{ref}}=\frac{\sin 30^{\circ }}{1.5}=\frac{0.5}{1.5}=0.3333$ $r_{\text{ref}}=\arcsin (0.3333)=19.47^{\circ }$ What it looks like (figure s01): end $A$ of the wavefront enters the slow medium first, so it lags; the far end $B$ is still speeding through air. The wavefront tilts and bends toward the normal — exactly what a smaller angle ($19.5^{\circ }<30^{\circ }$) means.

(b) first — frequency is unchanged. Wavefronts join continuously at the boundary; crests cannot pile up or vanish, so the number of crests passing per second ($f$) is conserved. (a) Since $\lambda =v/f$ and $f$ is fixed, $\lambda \propto v$. Speed drops by the factor $n$: $v_2=v_1/n$. So ${\lambda }_2=\frac{{\lambda }_1}{n}=\frac{600}{1.5}=400\text{nm}$ The wavelength shrinks in the slower medium; the frequency stays the same.

Points $A$, $C$ lie on the mirror; $B$ is the lagging end of the incoming front; $E$ is where $A$'s wavelet has reached. Look at triangles $ABC$ and $AEC$ (figure s02):

• Both are right-angled ($AB\perp BC$ on the incoming front; $AE\perp$ its own reflected front $CE$).
• They share the hypotenuse $AC$.
• $BC=AE=vt$ (equal radii, because the same speed $v$ acted for the same time $t$).

Two right triangles with equal hypotenuse and one equal leg are congruent (RHS congruence). Therefore $\angle BAC=\angle ECA.$ These two equal angles are exactly the angle of incidence and the angle of reflection measured off the surface, hence off the normal: $\boxed{\angle i=\angle r}$ Why equal radii is the whole engine: if $v$ or $t$ differed, $BC\ne AE$, the triangles would not be congruent, and the reflected front would tilt by a different angle. Equal speed forces the symmetry.

Glass is denser than water, so light speeds up entering water ⇒ it bends away from the normal ⇒ expect $r_{\text{ref}}>20^{\circ }$. $n_1\sin i=n_2\sin r_{\text{ref}}$ $\sin r_{\text{ref}}=\frac{n_1}{n_2}\sin i=\frac{1.5}{1.33}\sin 20^{\circ }=(1.1278)(0.3420)=0.3858$ $r_{\text{ref}}=\arcsin (0.3858)=22.69^{\circ }$ Indeed $22.7^{\circ }>20^{\circ }$ — bent away from the normal, as predicted.

(a) At the critical angle, $r_{\text{ref}}=90^{\circ }$ so $\sin r_{\text{ref}}=1$: $n_1\sin {\theta }_c=n_2\sin 90^{\circ }=n_2$ $\sin {\theta }_c=\frac{n_2}{n_1}=\frac{1.0}{1.5}=0.6667$ ${\theta }_c=\arcsin (0.6667)=41.81^{\circ }$ (b) Huygens picture (figure s03): in air the wavelet from $A$ spreads faster ($v_2>v_1$), so its radius $AD=v_{2}t$ is larger than the gap $BC=v_{1}t$ that end $B$ still has to cover. To draw the refracted front you must lay a tangent from $C$ to the big air-wavelet. As $i$ grows, the required tangent tilts until, at ${\theta }_c$, the tangent lies flat along the surface ($r_{\text{ref}}=90^{\circ }$). Beyond that, $AD$ is so big that no tangent into medium 2 exists — the geometry has no solution — so all energy stays behind: total internal reflection.

Huygens' form of Snell's law relates the speeds directly: $\frac{\sin i}{\sin r_{\text{ref}}}=\frac{v_1}{v_2}$ $v_2=v_1\frac{\sin r_{\text{ref}}}{\sin i}=(2.4\times 10^8)\frac{\sin 30^{\circ }}{\sin 45^{\circ }}=(2.4\times 10^8)\frac{0.5}{0.7071}$ $v_2=1.697\times 10^8\text{m/s}$ The wave slowed down (smaller refraction angle ⇒ denser medium), consistent with $v_2<v_1$.

The Fresnel–Kirchhoff obliquity factor is $K(\theta )=\frac{1}{2}(1+\cos \theta ).$

• Forward: $\theta =0^{\circ }\Rightarrow \cos 0=1\Rightarrow K=\frac{1}{2}(1+1)=1$ (full amplitude).
• Backward: $\theta =180^{\circ }\Rightarrow \cos 180^{\circ }=-1\Rightarrow K=\frac{1}{2}(1-1)=0$ (zero amplitude).

So the wavelet contributes fully forward and nothing backward. That is why Huygens could discard the backward envelope — it isn't a fudge, the amplitude there genuinely vanishes. See Fresnel–Kirchhoff Diffraction for the full derivation.

A crest is a specific phase point of the wavefront. Because the field is continuous across the boundary, whatever oscillation happens on the medium-1 side of the surface is instantly matched on the medium-2 side — the boundary cannot store, create, or destroy crests. So if a crest arrives at the interface, one crest departs into medium 2 at the same instant. If two consecutive crests arrive $T$ apart, two consecutive crests must depart $T$ apart: $f_2=\frac{1}{T}=f_1.$ The speed and wavelength adjust ($\lambda =v/f$), but the rate of crest crossings is conserved. This is the wave-continuity argument behind phase continuity.

Wavelength in the medium: ${\lambda }_{\text{glass}}=\frac{{\lambda }_{\text{vac}}}{n}=\frac{589}{1.52}=387.5\text{nm}.$ Number of wavelengths across a $3.0\mu \text{m}=3000\text{nm}$ slab: $N=\frac{3000}{387.5}=\mathrm{7.742.}$ Not a whole number — so the wave exits the slab shifted in phase relative to a wave that skipped the glass. This leftover fraction ($0.742$ of a wavelength) is exactly the optical-path idea that powers thin-film interference.

L1.1 same phase · L1.2 plane (flat) · L1.3 $r=vt$ L2.1 $n=1.5$ · L2.2 $r=1.0\text{ m}$ · L2.3 $r_{\text{ref}}=19.47^{\circ }$ L3.1 ${\lambda }_2=400\text{ nm}$, $f$ unchanged · L3.2 $△ABC\cong △AEC$ (RHS) $\Rightarrow \angle i=\angle r$ · L3.3 $r_{\text{ref}}=22.69^{\circ }$ (bends away) L4.1 ${\theta }_c=41.81^{\circ }$ · L4.2 $v_2=1.697\times 10^8\text{ m/s}$ L5.1 $K(0^{\circ })=1,K(180^{\circ })=0$ · L5.2 $f_2=1/T=f_1$ (crest count conserved) · L5.3 ${\lambda }_{\text{glass}}=387.5\text{ nm},N=7.742$ Every numeric line above is machine-checked in the verification block attached to this note.