Huygens' principle — wavefront propagation: Worked examples

This page is the drill ground for Huygens' principle — wavefront propagation. We take the construction rules you already met and push them into every corner — every sign of the refractive index change, the flat ( $90°$ ) grazing case, the "critical" case where refraction dies, and a couple of exam-style twists. Nothing here contradicts the parent; it makes the parent's rules do work.

Before we start, six symbols you will see everywhere. If any feels unfamiliar, we build it fully so no notation sneaks up on you.

The scenario matrix

Every problem this topic can throw is one of these cells. The worked examples below are tagged with the cell they hit, and together they fill the whole grid.

Cell	Situation	What is special about it
C1	Plane wave in one uniform medium	No bending — pure straight propagation, radius $r = v Δ t$
C2	Reflection, general $i$	$∠ i = ∠ r$ , congruent triangles
C3	Refraction rarer → denser ( $v_{2} < v_{1}$ )	Bends toward normal, $r < i$
C4	Refraction denser → rarer ( $v_{2} > v_{1}$ )	Bends away from normal, $r > i$
C5	Degenerate: $i = 0$ (normal incidence)	No bend at all; wavefront just slows/speeds
C6	Limiting: critical angle, $r \to 90°$	Refracted wave grazes the surface; beyond it → total internal reflection
C7	Wavelength / frequency across boundary	$f$ fixed, $λ$ and $v$ change
C8	Real-world word problem	Translate a physical story into $i, r, v$
C9	Grazing incidence $i \to 90°$	Wave skims the surface — the limiting geometry
C10	Exam twist: what Huygens predicts vs a trap	Backward wavelet / obliquity, ray⊥wavefront

Prerequisite links you may want open: Snell's Law and Refractive Index, Laws of Reflection, Phase and Path Difference.

Example 1 — Plane wave marches forward (Cell C1)

Forecast: guess the distance before reading — is it a few centimetres, or a few metres?

Figure — Huygens' principle — wavefront propagation

Reading the figure: the left cyan line is the starting wavefront W1; the faint white semicircles are the secondary wavelets, each of radius $r = v Δ t$ , emitted from five points along W1; the right cyan line W2 is their common forward tangent; the amber arrow marks the distance $v Δ t$ the wave has advanced.

Each point on the wavefront emits a wavelet of radius $r = v Δ t$ . Why this step? This is the whole Huygens recipe — every point is a secondary source, and in time $Δ t$ its little sphere grows to radius $v Δ t$ . $r = (2.25 \times 1 0^{8}) (2.0 \times 1 0^{- 9}) = 0.45 m$
The new wavefront is the common tangent of all these equal circles. Equal circles whose centres sit on a straight line have a straight tangent line, parallel to the first, a distance $0.45 m$ ahead. Why this step? Same $v$ , same start-time ⇒ same radius ⇒ the envelope cannot tilt. A flat wave stays flat.
Shape: still a plane, moved $0.45 m$ forward.

Example 2 — Reflection at a general angle (Cell C2)

Forecast: will $r$ be bigger, smaller, or equal to $35°$ ?

Reading the figure: the white horizontal line is the mirror; $A$ and $C$ are the two points where the wavefront meets the mirror ( $A$ first, $C$ later). $B$ is the far end of the incident wavefront (the cyan line $B C$ ); $E$ is the tip of the wavelet $A$ has thrown back into the medium (the amber line $E C$ is the reflected wavefront). The dashed line at $A$ is the normal.

Set $B C = A E = v t$ . Why this step? Both distances are covered in the same time $t$ at the same speed $v$ (same medium), so they are equal — this is the key fact.
Both triangles have a right angle, so $A C$ is the hypotenuse of each. The incident wavefront $B C$ is, by definition, perpendicular to the incident rays' travel; end $B$ arrives at $C$ moving straight into the surface line $A C$ , so $∠ B C A = 90°$ . Likewise the reflected wavefront $A E$ is perpendicular to the reflected travel, giving $∠ E A C = 90°$ . The side opposite each right angle is $A C$ , hence $A C$ is the shared hypotenuse of both $△ A B C$ and $△ A E C$ . Why this step? You cannot call two triangles "right-angled with hypotenuse $A C$ " until you show where the right angles sit — the RHS congruence test in the next step depends on it.
Apply RHS congruence. The two triangles share hypotenuse $A C$ , both are right-angled (Step 2), and $B C = A E$ (Step 1). By RHS (right-angle–hypotenuse–side), $△ A B C ≅ △ A E C$ , so $∠ B A C = ∠ E C A$ , which translates to $∠ i = ∠ r$ . Thus $r = 35°$ . Why this step? Congruent triangles force the incidence and reflection geometry to match; the wavefront angle equals the ray-from-normal angle (each is " $90°$ minus the surface angle"), so the equality carries over.

Example 3 — Refraction rarer → denser, bends toward normal (Cell C3)

Forecast: the wave slows down — do you expect the ray to bend toward or away from the normal?

Reading the figure: the white horizontal line is the air–glass boundary; the cyan arrow above is the incident ray at $i = 60°$ ; the amber arrow below is the refracted ray at $r = 35.3°$ ; the dashed vertical line is the normal. In the underlying wavefront picture, $B C = v_{1} t$ (air side) and $A D = v_{2} t$ (glass side) share the common hypotenuse $A C$ along the boundary — $D$ is simply the tip of $A$ 's wavelet inside the glass.

Write Huygens' Snell relation $\frac{sin i}{sin r} = \frac{v _{1}}{v _{2}}$ . Why this step? In the derivation $B C = v_{1} t$ and $A D = v_{2} t$ over the shared hypotenuse $A C$ , so $sin i = B C / A C$ and $sin r = A D / A C$ put the sines in the ratio of the speeds.
Solve for $sin r$ : $sin r = sin i \cdot \frac{v _{2}}{v _{1}} = sin 60° \cdot \frac{2.0}{3.0} = 0.8660 \times 0.6667 = 0.5774$ Why this step? We want $r$ , so isolate $sin r$ ; the slower medium ( $v_{2} < v_{1}$ ) shrinks $sin r$ .
Take arcsin: $r = arcsin (0.5774) = 35.3°$ . Why this step? $arcsin$ answers "which angle has this sine?" — it undoes the sine to recover the angle itself.

Example 4 — Refraction denser → rarer, bends away (Cell C4)

Forecast: the wave speeds up — this time, toward or away from the normal?

Same relation $sin r = sin i \cdot \frac{v _{2}}{v _{1}}$ . Why this step? Snell's law does not care which medium is "first"; only the speed ratio matters.
Plug in: $sin r = sin 30° \cdot \frac{3.0}{2.0} = 0.5 \times 1.5 = 0.75$ Why this step? Now $v_{2} > v_{1}$ , so $sin r > sin i$ — the factor is bigger than 1.
Arcsin: $r = arcsin (0.75) = 48.6°$ . Why this step? Undo the sine to read off the actual bend angle.

Example 5 — Normal incidence, the degenerate case (Cell C5)

Forecast: does anything bend when you hit dead-on?

Use $sin r = sin i \cdot \frac{v _{2}}{v _{1}}$ with $sin 0° = 0$ . $sin r = 0 \times \frac{v _{2}}{v _{1}} = 0 \Rightarrow r = 0°$ Why this step? Zero times anything is zero, so the formula automatically gives no bend.
Geometrically: ends $A$ and $B$ of the wavefront reach the surface at the same instant (the wavefront is parallel to the surface), so there is no "one end arrives first" to create a tilt. Why this step? The tilt in refraction comes entirely from one end lagging; remove the lag, remove the bend.
What still changes: speed $v$ and wavelength $λ = v / f$ . Frequency $f$ stays fixed. Why this step? We must report what the boundary does do even when it doesn't bend — the wave still slows and its crests crowd closer, so $λ$ shrinks while the "beat rate" $f$ is untouched.

Example 6 — Critical angle, the limiting case (Cell C6)

Forecast: as we tilt the beam more and more, the refracted ray flattens toward the surface — at what angle does it "run out of room"?

Reading the figure: the white horizontal line is the glass–air boundary (glass below, air above); the cyan arrow rising from below is the incident ray at $i_{c} = 41.8°$ ; the amber arrow lying flat along the surface is the refracted ray at $r = 90°$ — it grazes rather than escapes; the dashed vertical line is the normal.

Set the limiting condition $r = 90°$ , so $sin r = 1$ . Why this step? $90°$ is the largest an angle from the normal can be while still pointing along the surface — beyond it there is no forward direction left for a refracted wave.
Snell's law $n_{1} sin i_{c} = n_{2} sin 90°$ gives $sin i_{c} = \frac{n _{2}}{n _{1}} = \frac{1.0}{1.5} = 0.6667 \Rightarrow i_{c} = arcsin (0.6667) = 41.8°$ Why this step? Index form $n_{1} sin i = n_{2} sin r$ is the cleanest here; setting $r = 90°$ isolates $i_{c}$ .
For $i > i_{c}$ : the equation $sin r = (n_{1} / n_{2}) sin i > 1$ has no real solution. Physically the refracted wavelet cannot form; all energy reflects — total internal reflection. Why this step? A sine can never exceed 1; the algebra itself signals the refracted wave has vanished, so we must interpret it physically rather than force a fake angle.

Example 7 — Frequency fixed, wavelength changes (Cell C7)

Forecast: does the light look more red or more blue inside the glass? (Trick — colour is set by frequency.)

Frequency in air $f = c / λ_{1}$ . $f = \frac{3.0 \times 1 0 ^{8}}{600 \times 1 0 ^{- 9}} = 5.0 \times 1 0^{14} Hz$ Why this step? Frequency is the "crests per second," and it is what stays continuous across the boundary (wavefronts can't pile up or vanish).
Speed in glass $v_{2} = c / n_{2} = 2.0 \times 1 0^{8} m/s$ , then $λ_{2} = v_{2} / f$ . $λ_{2} = \frac{2.0 \times 1 0 ^{8}}{5.0 \times 1 0 ^{14}} = 4.0 \times 1 0^{- 7} m = 400 nm$ Why this step? With $f$ fixed and $v$ smaller, $λ = v / f$ must shrink.
Shortcut check: $λ_{2} = λ_{1} \cdot v_{2} / v_{1} = 600 \times (2/3) = 400 nm$ ✓. Why this step? An independent route to the same $λ_{2}$ guards against an arithmetic slip in Step 2.

Example 8 — Real-world word problem (Cell C8)

Forecast: is this the same critical-angle idea as Example 6, just with water instead of glass?

Recognise it is total internal reflection going water→air ( $n_{1} > n_{2}$ ). Set $r = 90°$ . Why this step? "Surface acts like a mirror" = no light escapes upward = we are at or past the critical angle.
Compute $sin i_{c} = n_{2} / n_{1} = 1.00/1.33 = 0.7519$ , so $i_{c} = arcsin (0.7519) = 48.8°$ Why this step? Identical machinery to Example 6 — Huygens gives one rule for every medium pair.
Interpret: looking up within $48.8°$ of straight-up, she sees the sky; beyond it, mirror. Why this step? The physics question asked for the viewing behaviour, so we must translate the abstract $i_{c}$ back into what the swimmer's eye actually sees — otherwise the number is meaningless.

Example 9 — Grazing incidence, the $i \to 90°$ limit (Cell C9)

Forecast: if the beam barely skims in at $90°$ , does the refracted ray go steeply into the glass or stay close to the surface?

Set the limiting incidence $i = 90°$ , so $sin i = 1$ . Why this step? $90°$ from the normal means the ray lies along the surface — the most extreme incidence physically possible. It probes the ceiling of refraction.
Snell's law $n_{1} sin 90° = n_{2} sin r$ gives $sin r = \frac{n _{1}}{n _{2}} sin 90° = \frac{1.0}{1.5} = 0.6667 \Rightarrow r = arcsin (0.6667) = 41.8°$ Why this step? With $sin i$ at its maximum of $1$ , $sin r$ hits its maximum $n_{1} / n_{2}$ , so $r$ can never exceed this value — every real ray from air lands inside the cone of half-angle $41.8°$ .
Interpret: all light entering the glass from air is squeezed into a cone of half-angle $41.8°$ around the normal; nothing refracts steeper. Why this step? This is the mirror-image of Example 6 — the grazing-in limit ( $r = 41.8°$ ) equals the critical angle for going back out, which is exactly why refraction is reversible.

Example 10 — Exam twist: the backward-wavelet trap (Cell C10)

Forecast: guess $K$ straight forward ( $θ = 0°$ ) and straight backward ( $θ = 180°$ ) before computing.

Forward: $θ = 0°$ , $cos 0° = 1$ , so $K = \frac{1}{2} (1 + 1) = 1$ — full strength. Why this step? The obliquity factor is the exact rule that replaces Huygens' hand-waved "ignore backward," so we test it in the direction the wave should be strongest.
Backward: $θ = 180°$ , $cos 180° = - 1$ , so $K = \frac{1}{2} (1 + (- 1)) = 0$ Why this step? Setting the backward direction and evaluating shows the amplitude genuinely vanishes, not by decree but by the Fresnel–Kirchhoff factor.
Sideways sanity: $θ = 90°$ , $cos 90° = 0$ , $K = \frac{1}{2}$ — half strength, but these sideways wavelets are out of phase along the envelope, so they cancel too. Why this step? We must rule out the middle case too — a non-zero $K$ at $90°$ could look like a surviving sideways blur, so we check that interference (not amplitude) is what kills it, closing every loophole in the student's argument.

Connections

Huygens' principle — wavefront propagation (parent)
Snell's Law and Refractive Index (Examples 3, 4, 6, 8, 9)
Laws of Reflection (Example 2)
Phase and Path Difference (Example 7)
Fresnel–Kirchhoff Diffraction (Example 10, obliquity factor)
Wave Optics — Interference
Young's Double Slit Experiment
Diffraction