PhysicsISC Class 12 · Wave Optics

Huygens' Principle: Wavefronts, Secondary Wavelets and the Laws of Reflection and Refraction

Updated 15 October 2026 · Fact-checked

A wavefront is the locus of points vibrating in the same phase. Huygens' principle says every point on a wavefront acts as a source of secondary wavelets moving at the wave speed, and the new wavefront is the forward envelope of these wavelets. Applied at a boundary, it gives $\frac{\sin i}{\sin r} = \frac{v_1}{v_2}$, which is Snell's law.

Understand Huygens' Principle and Wavefronts

Wavefronts. A wavefront is a surface on which every point oscillates in the same phase. The direction of energy flow, the ray, is always perpendicular to the wavefront.

Point source: spherical wavefronts (concentric spheres, not "circles"; the ISC 2024 examiners accepted only the word spherical).
Line source (e.g. a slit lit from behind): cylindrical wavefronts.
Very distant source (the Sun, a star) or light from a point source after a convex lens placed at its focus: plane wavefronts.

Huygens' principle.
1. Every point on a wavefront is a source of secondary wavelets, which spread out in the forward direction with the speed of the wave in that medium.
2. The new position of the wavefront after time t is the surface touching all these wavelets (the forward envelope).
(The backward wave is ignored.)

Proof of the law of refraction (ISC 2024 Q13(i), ISC 2025 Q14). A plane wavefront AB meets the boundary XY between medium 1 (speed $v_{1}$ , rarer) and medium 2 (speed $v_{2}$ , denser) obliquely; end A touches the surface first. Let t be the time for B to reach $B_{1}$ on the surface: $B B_{1} = v_{1} t$ . In the same time the wavelet from A travels $A A_{1} = v_{2} t$ in medium 2. The refracted wavefront is $A_{1} B_{1}$ , the tangent from $B_{1}$ to the wavelet of radius $v_{2} t$ centred on A.
In right triangles $A B B_{1}$ and $A A_{1} B_{1}$ (common hypotenuse $A B_{1}$ ), the angle between the incident wavefront and the surface equals i, and that between the refracted wavefront and the surface equals r:
$sin i = \frac{B B _{1}}{A B _{1}} = \frac{v _{1} t}{A B _{1}}, sin r = \frac{A A _{1}}{A B _{1}} = \frac{v _{2} t}{A B _{1}} .$
Dividing:
$\frac{s i n i}{s i n r} = \frac{v _{1}}{v _{2}} = constant =_{1} n_{2} .$
With medium 1 as vacuum, $n = \frac{c}{v}$ . Since $v_{2} < v_{1}$ in a denser medium, r < i: the ray bends towards the normal.

Proof of the law of reflection. Same construction, but the wavelet from A returns into medium 1 at speed $v_{1}$ : $A A_{1} = v_{1} t = B B_{1}$ . Triangles $A B B_{1}$ and $B_{1} A_{1} A$ are congruent (right angles, common hypotenuse $A B_{1}$ , equal sides), so i = r. Note that the ISC examiners penalised candidates who proved reflection when refraction was asked.

Frequency, speed and wavelength. At the boundary each wavelet is set in motion by the incoming wave, so the frequency does not change (it is fixed by the source). Since v = fλ, $\frac{λ _{2}}{λ _{1}} = \frac{v _{2}}{v _{1}}$ : in a medium of index n, $v = \frac{c}{n}$ and $λ = \frac{λ _{0}}{n}$ .

Denser to rarer. If $v_{2} > v_{1}$ , then r > i and the ray bends away from the normal. When $sin i = \frac{v _{1}}{v _{2}}$ , r = 90°: this is the critical angle. For larger i there is no refracted wavefront: total internal reflection. (Refraction through prisms and lenses using Huygens' theory is not required.)

Huygens' Principle and Wavefronts: formulas and key facts

Snell's law from wavefronts: $\frac{s i n i}{s i n r} = \frac{v _{1}}{v _{2}} =_{1} n_{2}$; Huygens construction.
Absolute refractive index: $n = \frac{c}{v}$; Medium 1 vacuum.
Wavelength in a medium: $λ = \frac{λ _{0}}{n}$; Frequency unchanged.
Construction lengths: $B B_{1} = v_{1} t, A A_{1} = v_{2} t$; Common hypotenuse AB₁.
Critical angle: $sin c = \frac{v _{1}}{v _{2}} = \frac{1}{n}$; Denser to rarer.

How to solve Huygens' Principle and Wavefronts questions

Proof questions need a correct diagram first; numericals use n = c/v and Snell's law.

1Draw the boundary, mark the rarer and denser media, the incident wavefront AB with ray arrows, and the normal.
2Mark $B B_{1} = v_{1} t$ in medium 1 and the wavelet of radius $A A_{1} = v_{2} t$ in medium 2; draw the tangent $A_{1} B_{1}$ .
3Mark i between AB and the surface, r between $A_{1} B_{1}$ and the surface.
4Write $sin i = \frac{B B _{1}}{A B _{1}}$ and $sin r = \frac{A A _{1}}{A B _{1}}$ , divide, and state that the ratio is constant.
5Numericals: speed $v = \frac{c}{n}$ ; wavelength $\frac{λ _{0}}{n}$ ; frequency unchanged; angles from $n_{1} sin i = n_{2} sin r$ .

Huygens' Principle and Wavefronts in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Waves turning towards a beach

Suppose straight ocean waves of wavelength 20 m travel at 4 m/s in deep water and reach a shallow shelf, where they slow to 3 m/s. They meet the shelf edge at 50°.

Find the new direction, wavelength and frequency.

Try it first, then show the solutionSolution

Huygens: $sin r = \frac{v _{2}}{v _{1}} sin i = 0.75 \times sin 50° \approx 0.575$ , r ≈ 35.1°.
Frequency unchanged: $f = \frac{4}{20} = 0.2$ Hz; new wavelength $= \frac{3}{0.2} = 15$ m.

Answer: About 35° (turned towards the normal); 15 m; 0.2 Hz. That is why waves arrive almost parallel to a beach.

Memory hook: Slow side turns the wavefront round, like a car wheel hitting sand.

Why sunlight has plane wavefronts

Imagine a window 1 m wide receiving sunlight; the Sun is $1.5 \times 1 0^{11}$ m away.

How far does the spherical wavefront from the Sun bulge across the width of the window?

Try it first, then show the solutionSolution

Over half-width 0.5 m, the bulge (sagitta) of a sphere of radius R is about $\frac{( 0.5 ) ^{2}}{2 R}$ .
$\frac{0.25}{3 \times 1 0 ^{11}} \approx 8.3 \times 1 0^{- 13}$ m, far smaller than one wavelength.

Answer: About $8 \times 1 0^{- 13}$ m: for all practical purposes the wavefront is plane.

Memory hook: Far enough away, every sphere looks flat.

Tricks to remember Huygens' Principle and Wavefronts

Short memory aids. Say each one aloud once, then use it on a question the same day.

Same t, different speeds
$B B_{1} = v_{1} t$ , $A A_{1} = v_{2} t$ ; share the hypotenuse $A B_{1}$ .
Use it when: The refraction proof.
Point, line, far
Spherical, cylindrical, plane.
Use it when: Types of wavefronts.
Frequency is the boss
f is set by the source; v and λ change together.
Use it when: Speed and wavelength in media.
Rays cross wavefronts at right angles
Rays ⊥ wavefronts, always.
Use it when: Drawing diagrams.

Huygens' Principle and Wavefronts: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Wavefront from a line source · 1 mark · MCQBoard level

The wavefront produced by a narrow illuminated slit (a line source) is:
(a) spherical
(b) cylindrical
(c) plane
(d) elliptical

Try it first, then show the solutionSolution

Points equidistant from a line form a cylinder around it.

Answer: (b).

Example 2: Speed, wavelength and frequency in water · 2 marks · Short answerBoard level

Light of wavelength 589 nm in air enters water of refractive index 4/3. Find its speed, wavelength and frequency in water.

Try it first, then show the solutionSolution

$v = \frac{3 \times 1 0 ^{8}}{4/3} = 2.25 \times 1 0^{8}$ m/s; $λ = \frac{589}{4/3} \approx 442$ nm. (1)
$f = \frac{3 \times 1 0 ^{8}}{589 \times 1 0 ^{- 9}} \approx 5.09 \times 1 0^{14}$ Hz, the same in air and water. (1)

Answer: $2.25 \times 1 0^{8}$ m/s; ≈ 442 nm; ≈ $5.09 \times 1 0^{14}$ Hz.

Example 3: Proof of Snell's law · 3 marks · Long answerHarder board level

A plane wavefront travels obliquely from a rarer medium (speed $v_{1}$ ) into a denser medium (speed $v_{2}$ ). Draw the incident and refracted wavefronts and use Huygens' principle to prove that $\frac{s i n i}{s i n r}$ is constant.

Try it first, then show the solutionSolution

Diagram: boundary XY, rarer above, denser below; incident wavefront AB with A on the surface; $B B_{1} = v_{1} t$ ; wavelet of radius $A A_{1} = v_{2} t$ about A; refracted wavefront $A_{1} B_{1}$ tangent to it; i and r marked; ray arrows. (1)
In right triangles $A B B_{1}$ and $A A_{1} B_{1}$ : $sin i = \frac{B B _{1}}{A B _{1}} = \frac{v _{1} t}{A B _{1}}$ , $sin r = \frac{A A _{1}}{A B _{1}} = \frac{v _{2} t}{A B _{1}}$ . (1)
$\frac{s i n i}{s i n r} = \frac{v _{1}}{v _{2}}$ , a constant for the pair of media (the refractive index of medium 2 relative to 1). (1)

Answer: $\frac{s i n i}{s i n r} = \frac{v _{1}}{v _{2}} =$ constant.

Example 4: Measuring the construction · 3 marks · Short answerHarder board level

A plane wavefront in air strikes a medium in which light travels at $2 \times 1 0^{8}$ m/s, at an angle of incidence of 45°. In the Huygens construction $A B_{1}$ = 4.0 cm. Find the refractive index, the angle of refraction, and the lengths $B B_{1}$ and $A A_{1}$ .

Try it first, then show the solutionSolution

$n = \frac{3 \times 1 0 ^{8}}{2 \times 1 0 ^{8}} = 1.5$ ; $sin r = \frac{s i n 45°}{1.5} \approx 0.471$ , r ≈ 28.1°. (1)
$B B_{1} = A B_{1} sin i = 4.0 \times 0.707 \approx 2.83$ cm. (1)
$A A_{1} = A B_{1} sin r \approx 4.0 \times 0.471 \approx 1.89$ cm; check $\frac{2.83}{1.89} \approx 1.5$ . (1)

Answer: n = 1.5; r ≈ 28.1°; ≈ 2.83 cm and ≈ 1.89 cm.

Example 5: From glass into air · 3 marks · Short answerHOTS / competency

Light inside glass (n = 1.5) meets the glass-air surface at 30°. Using the wavefront picture, find the angle of refraction, say which way the ray bends, and find the angle beyond which no refracted wavefront exists.

Try it first, then show the solutionSolution

Now $v_{2} > v_{1}$ , so the wavelet in air is larger than the distance travelled in glass: $sin r = 1.5 sin 30° = 0.75$ , r ≈ 48.6°. (1)
r > i: the ray bends away from the normal. (1)
No refracted wavefront when $sin r$ would exceed 1: $sin c = \frac{1}{1.5}$ , c ≈ 41.8°. (1)

Answer: ≈ 48.6°, away from the normal; beyond ≈ 41.8° total internal reflection.

Example 6: Proof of the law of reflection · 4 marks · Case-basedHOTS / competency

Suppose a plane wavefront AB strikes a plane mirror obliquely, with A touching the mirror first.
(i) In time t, how far does B travel to reach the mirror at $B_{1}$ ? (1 mark)
(ii) How large is the wavelet from A in the same time, and why? (1 mark)
(iii) Identify the reflected wavefront. (1 mark)
(iv) Prove that i = r. (1 mark)

Try it first, then show the solutionSolution

(i) $B B_{1} = v t$ .
(ii) Radius $A A_{1} = v t$ : the wavelet travels back in the same medium at the same speed.
(iii) $A_{1} B_{1}$ , the tangent from $B_{1}$ to the wavelet centred on A.
(iv) Right triangles $A B B_{1}$ and $B_{1} A_{1} A$ share hypotenuse $A B_{1}$ and have $B B_{1} = A A_{1}$ , so they are congruent: the angles the wavefronts make with the mirror are equal, i = r.

Answer: (i) vt; (ii) vt, same medium; (iii) $A_{1} B_{1}$ ; (iv) congruent triangles give i = r.

Where students lose marks in Huygens' Principle and Wavefronts

The slips examiners see most often, and what to write instead.

Mistake: Proving reflection when refraction is asked
Reported by the ISC examiners in both 2024 Q13(i) and 2025 Q14.
Do this: For refraction, the wavelet in medium 2 has radius $v_{2} t$ , not $v_{1} t$ .
Usually costs: 2 to 3 marks.
Mistake: Unlabelled diagram
Both years: no ray arrows, i and r not marked, media not named, wavefronts missing.
Do this: Label AB, $A_{1} B_{1}$ , $B B_{1}$ , $A A_{1}$ , i, r, rarer and denser.
Usually costs: 1 mark.
Mistake: Wrong sines
Candidates wrote $sin i = \frac{A A _{1}}{A B _{1}}$ .
Wrong working
$sin i = \frac{A A _{1}}{A B _{1}}$ , $sin r = \frac{B B _{1}}{A B _{1}}$ .
Right working
$sin i = \frac{B B _{1}}{A B _{1}}$ , $sin r = \frac{A A _{1}}{A B _{1}}$ .
Do this: i belongs with the incident medium, so with $B B_{1} = v_{1} t$ .
Usually costs: 1 mark.
Mistake: "Circular" wavefronts
In 2024 Q1(B)(v) "circular" was not accepted for a point source.
Do this: Write "spherical".
Usually costs: 1 mark.

Exam tips

Spend time on the diagram: it carries a full mark in the 3-mark proof.
Finish the proof by saying "= constant = refractive index".
State the principle in two parts: secondary wavelets, forward envelope.
Say explicitly that frequency is unchanged on refraction.

Practice questions on Huygens' Principle and Wavefronts

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
A ray of light is always:
(a) parallel to the wavefront
(b) perpendicular to the wavefront
(c) at 45° to the wavefront
(d) unrelated to the wavefront
Hint
Energy flows normal to equal-phase surfaces.
Answer
(b).
Q2 · 1 mark · Short answerBoard level
Define a wavefront.
Hint
Phase.
Answer
The locus (surface) of all points of the medium vibrating in the same phase.
Q3 · 2 marks · Short answerHarder board level
Light of wavelength 500 nm in air enters glass of refractive index 1.6. Find its wavelength and speed in the glass.
Hint
Divide by n.
Answer
312.5 nm; $1.875 \times 1 0^{8}$ m/s.
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The frequency of light does not change when it passes from air into glass.
Reason (R): Frequency is fixed by the source, since each secondary wavelet at the boundary is driven by the incoming wave.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
Which quantity is set by the source?
Answer
(a).
Q5 · 3 marks · Long answerHarder board level
Using Huygens' principle and a labelled diagram, prove the law of reflection.
Hint
Same speed both ways.
Answer
Incident wavefront AB; $B B_{1} = v t$ ; wavelet about A of radius $A A_{1} = v t$ ; reflected wavefront $A_{1} B_{1}$ tangent to it. Triangles $A B B_{1}$ and $B_{1} A_{1} A$ are congruent (right angle, common hypotenuse, equal sides), so i = r.
Q6 · 2 marks · Short answerHOTS / competency
A ray in air is incident at 60° and refracted at 35°. Find the refractive index and the speed of light in the medium.
Hint
$\frac{s i n 60°}{s i n 35°}$ .
Answer
n ≈ 1.51; v ≈ $1.99 \times 1 0^{8}$ m/s.
Q7 · 2 marks · Short answerHOTS / competency
Light goes from water (n = 1.33) into glass (n = 1.5) at an angle of incidence of 40°. Find the angle of refraction using the ratio of speeds.
Hint
$\frac{v _{1}}{v _{2}} = \frac{1.5}{1.33}$ .
Answer
$sin r = \frac{1.33 s i n 40°}{1.5} \approx 0.570$ , r ≈ 34.8°.
Q8 · 4 marks · Case-basedHOTS / competency
Case study: Suppose light of wavelength 600 nm (in air) enters a diamond of refractive index 2.42.
(i) Find its speed in diamond. (1 mark)
(ii) Find its wavelength in diamond. (1 mark)
(iii) Find the critical angle for diamond-air. (1 mark)
(iv) Use (iii) to explain why a cut diamond sparkles. (1 mark)
Hint
c = sin⁻¹(1/n).
Answer
(i) ≈ $1.24 \times 1 0^{8}$ m/s. (ii) ≈ 248 nm. (iii) ≈ 24.4°. (iv) The small critical angle means most light inside strikes faces at more than 24.4° and is totally reflected many times before leaving through a few faces, so it emerges in bright flashes.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Wavefront from a point source?Show answer
Spherical.
2.Huygens' principle in one line?Show answer
Every point on a wavefront sends out secondary wavelets; their forward envelope is the new wavefront.
3.Ratio of sines in terms of speeds?Show answer
$\frac{s i n i}{s i n r} = \frac{v _{1}}{v _{2}}$ .
4.n = 1.5: speed of light?Show answer
$2 \times 1 0^{8}$ m/s.
5.Which quantity is unchanged on refraction?Show answer
Frequency.

Huygens' Principle and Wavefronts: frequently asked questions

Why is the backward wavelet ignored?

Huygens simply assumed the wavelets act only forward; a full wave theory shows the backward amplitude is zero, so the assumption is justified.

Why does light bend towards the normal in glass?

Light is slower in glass, so the part of the wavefront that enters first lags behind; the wavefront swings round and the ray (perpendicular to it) turns towards the normal.

Does Huygens' principle explain diffraction?

Yes, qualitatively: wavelets from points of an opening spread into the geometrical shadow and interfere.

What does the ISC 2027 syllabus need here?

Wavefronts and their shapes, Huygens' principle, and proofs of the laws of reflection and refraction using it. Refraction through a prism or lens by Huygens' theory is not required.

How was this asked recently?

ISC 2024 Q13(i) and ISC 2025 Q14 (3 marks each) asked the Huygens proof of Snell's law with a labelled diagram; ISC 2024 Q1(B)(v) asked the wavefront shape for a point source.