Huygens' principle — wavefront propagation: Visual walkthrough

Step 1 — What is even moving? Draw the wavefront

WHAT. Before any bending, we need the thing that bends. A wavefront is a line (in 2D) joining all points of the wave that are "doing the same thing at the same moment" — all cresting together. Picture a straight rank of soldiers marching in step: the rank is the wavefront.

WHY. We track this equal-phase line, not individual light "particles", because the whole trick of Huygens is: every point on this line is a tiny new source. If we know where the line is now, we can build where it is next.

PICTURE. The blue line $A B$ is our wavefront. The pale-yellow arrow, drawn perpendicular to it, is the ray — the direction the whole rank marches. Remember: ray and wavefront are always at right angles (never parallel).

Step 2 — Let it march: distance = speed × time

WHAT. Give the wave a stopwatch. In a time $t$ the rank advances a distance. If the wave moves at speed $v_{1}$ (metres per second in medium 1), then in $t$ seconds it covers

$d = speed in medium 1 v_{1} \times elapsed time t$

WHY. Speed literally means "distance per unit time", so distance is just speed times time. This is the only physics input — everything after is geometry.

PICTURE. Two blue lines: the old wavefront and, a distance $v_{1} t$ ahead, the new one. They stay parallel because every point marched the same $v_{1} t$ . This is the parent's "plane stays plane".

Step 3 — Tilt the wavefront onto a boundary

WHAT. Now let the marching rank hit a slanted line $X Y$ — the surface between medium 1 (top, speed $v_{1}$ ) and medium 2 (bottom, slower, speed $v_{2}$ ). Because the wavefront is slanted, one end — call it $A$ — reaches the surface first, while the other end is still up in medium 1 at a point we call $B$ .

WHY. If the wavefront hit the surface perfectly flat-on, both ends would cross together and nothing interesting would happen. The tilt is what makes one end wait while the other crosses — and waiting is what makes light bend.

PICTURE. $A$ sits on the boundary. $B$ still hangs above it. The angle between the incoming wavefront and the surface (equivalently, between the incoming ray and the normal, the dashed perpendicular to $X Y$ ) is the angle of incidence $i$ .

Step 4 — The race: end $B$ finishes in medium 1 while end $A$ starts in medium 2

WHAT. Two things happen in the same time interval $t$ :

End $B$ , still in medium 1, races down to meet the surface at a point $C$ . It travels $B C = v_{1} t (v_{1} = fast top speed) .$
End $A$ , having already crossed, sends its Huygens wavelet forward into medium 2. In the same $t$ it spreads only $A D = v_{2} t (v_{2} = slow bottom speed, so A D < B C) .$

WHY. This is the heart of the whole thing. $A$ and $B$ get the same clock time $t$ , but $A$ is in the slow medium and $B$ in the fast one. Same time, different speeds ⟹ different distances. That mismatch is the bend.

PICTURE. $B C$ is a long blue segment (fast). $A D$ is a short pink arc of radius $v_{2} t$ — the secondary wavelet from $A$ (Huygens: every point is a source). Because $v_{2} < v_{1}$ , the arc is smaller than $B C$ .

Step 5 — Draw the new wavefront: the common tangent

WHAT. By time $t$ , end $B$ has just arrived at $C$ (so its wavelet has radius zero there), while $A$ 's wavelet has grown to radius $A D = v_{2} t$ . Every point between $A$ and $C$ crossed at some in-between moment, so its wavelet has an in-between radius. The new wavefront in medium 2 is the straight line from $C$ that is tangent to all these wavelets — it touches $A$ 's circle exactly at $D$ .

WHY. Huygens' rule: the next wavefront is the common tangent (envelope) of all the secondary wavelets. Off the tangent, wavelets are out of step and cancel; along it they reinforce.

PICTURE. The refracted wavefront $C D$ (blue) leans against the pink wavelet at $D$ , with a right angle at $D$ ( $A D ⊥ C D$ , because a tangent is perpendicular to the radius at the touch point).

Recall Why is

$A D$ perpendicular to $C D$ ? A tangent line touches a circle at exactly one point, and the radius to that point always meets the tangent at $9 0^{\circ}$ . Here $C D$ is tangent to $A$ 's wavelet at $D$ , so $A D ⊥ C D$ . ::: This right angle is what makes triangle $A C D$ a right triangle in Step 6. Which point is the refracted wavefront pinned to on the surface? ::: Point $C$ , where end $B$ has just arrived.

WHAT. Look at the two triangles that meet along the surface segment $A C$ :

$△ A B C$ : right angle at $B$ , hypotenuse $A C$ , and the side $B C = v_{1} t$ sits opposite the angle at $A$ , which equals $i$ .
$△ A D C$ : right angle at $D$ , hypotenuse $A C$ (the same segment), and side $A D = v_{2} t$ sits opposite the angle at $C$ , which equals the angle of refraction $r$ .

Now use the definition of sine — "opposite over hypotenuse":

$sin i = \frac{opposite}{hyp} = \frac{B C}{A C} = \frac{v _{1} t}{A C}, sin r = \frac{A D}{A C} = \frac{v _{2} t}{A C} .$

WHY sine, and why these two angles? Sine turns an angle into a ratio of sides, which is exactly the language a triangle speaks. We pick $i$ and $r$ because both angles ride on the same hypotenuse $A C$ — so when we divide, $A C$ cancels and the messy geometry disappears.

PICTURE. Both triangles drawn on the shared base $A C$ : yellow highlights the two "opposite" sides $B C$ and $A D$ ; the angle $i$ lives at $A$ , the angle $r$ lives at $C$ .

Step 7 — Divide, and Snell's law falls out

WHAT. Divide the two sine expressions. The common time $t$ and the common hypotenuse $A C$ both cancel:

$\frac{s i n i}{s i n r} = \frac{B C / A C}{A D / A C} = \frac{B C}{A D} = \frac{v _{1} t}{v _{2} t} = \frac{v _{1}}{v _{2}} .$

WHY. Everything that depended on how far or how long the wave travelled ( $t$ , $A C$ ) has vanished. What survives is only the ratio of the two speeds — a fixed property of the two media. A constant ratio is exactly what a "law" is.

Using $n = \frac{c}{v}$ (refractive index = light speed in vacuum ÷ speed in the medium):

PICTURE. The final annotated diagram: incident ray, bent refracted ray, and the boxed result — see Snell's Law and Refractive Index and Phase and Path Difference for where this feeds forward.

Step 8 — The edge cases (never leave the reader stranded)

WHAT & WHY & PICTURE, all three at once, for the situations the clean derivation glossed over:

Straight-on incidence ( $i = 0$ ). Wavefront parallel to surface, both ends cross together, no tilt develops. Then $sin i = 0 \Rightarrow sin r = 0 \Rightarrow r = 0$ . Light passes straight through, un-bent — but still slower inside.
Same medium ( $v_{1} = v_{2}$ ). The ratio is $1$ , so $sin i = sin r$ , $i = r$ : no bending. There was never a boundary worth bending at.
Slow → fast ( $v_{2} > v_{1}$ ). Now $A D > B C$ : the wavelet outruns end $B$ . So $sin r > sin i$ , $r > i$ : the ray bends away from the normal.
Critical angle & total internal reflection. Going slow→fast, keep increasing $i$ . The required $sin r = \frac{v _{2}}{v _{1}} sin i$ climbs toward $1$ . At the critical angle $i_{c}$ where $sin i_{c} = \frac{v _{1}}{v _{2}}$ , we get $r = 9 0^{\circ}$ — the refracted wavefront skims along the surface. Beyond $i_{c}$ , $sin r > 1$ is impossible: no tangent wavefront can be drawn in medium 2, so all the light reflects back. (See Laws of Reflection.)

The one-picture summary

Everything on one slate: the tilted incident wavefront $A B$ , the race ( $B C = v_{1} t$ long, $A D = v_{2} t$ short), the two right triangles sharing hypotenuse $A C$ , and the bent wavefront $C D$ — with the boxed law underneath.

Recall Feynman: the whole walkthrough in plain words

A line of friends marches in step (the wavefront). They reach a muddy field cut on a slant. The friend nearest the mud steps in first and slows down; his neighbours in mud slow too, but the friends still on firm ground keep sprinting. In the same few seconds, the sprinters cover a long stretch while the mud-walkers only shuffle a little. So the whole line pivots — it's now aimed more steeply into the mud. That pivot is the bend. Write "sprinter distance = $v_{1} t$ " and "mud distance = $v_{2} t$ " on the same slanted baseline; the two right triangles they form share the baseline, so dividing their sines cancels the baseline and the time, leaving just $v_{1} / v_{2}$ . Speed things up instead of slowing them (mud → firm ground) and the line pivots the other way. Slow them so hard that no forward line can even form, and the whole rank bounces back — total internal reflection.

Connections

Huygens' principle — wavefront propagation (parent)
Snell's Law and Refractive Index
Laws of Reflection
Phase and Path Difference
Wave Optics — Interference
Young's Double Slit Experiment
Diffraction