PhysicsCBSE Class 12 · Wave Optics

Huygens Principle and Wavefronts Class 12: Proving Reflection and Refraction

Updated 15 October 2026 · Fact-checked

A wavefront is a surface of points vibrating in the same phase; rays are perpendicular to it. Huygens principle says each point of a wavefront emits secondary wavelets, and the new wavefront is their forward envelope. With a plane wavefront it proves the laws of reflection (i = r) and refraction (sin i/sin r = v₁/v₂).

Understand Huygens Principle and Wavefronts

Drop a pebble in a pond and circular ripples spread out. Every point on a ripple can be thought of as starting its own tiny ripple, and the next ripple is where all these tiny ripples join up. Huygens used this simple idea to explain how light travels, reflects and refracts.

Wavefronts. A wavefront is the locus of all points that vibrate in the same phase. Its shape depends on the source:

A point source gives spherical wavefronts.
A line source (a slit) gives cylindrical wavefronts.
Far from any source, a small part of a spherical wavefront is effectively plane (light from the Sun or a star).

The direction of travel (a ray) is always perpendicular to the wavefront. The energy travels outwards with the wavefront at the wave speed.

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 forward envelope (common tangent surface) of these wavelets. (Backward wavelets are ignored; a full wave treatment shows they cancel.)

Refraction of a plane wave (Snell's law). A plane wavefront AB meets the plane surface between medium 1 (speed $v_{1}$ ) and medium 2 (speed $v_{2}$ ) at angle i. Point A touches the surface first. While the other edge B travels a distance $B C = v_{1} t$ in medium 1 to reach the surface at C, the wavelet from A travels $A E = v_{2} t$ into medium 2. The refracted wavefront is the tangent CE from C to the wavelet. From the right triangles ABC and AEC (common hypotenuse AC): $sin i = \frac{B C}{A C} = \frac{v _{1} t}{A C}$ and $sin r = \frac{A E}{A C} = \frac{v _{2} t}{A C}$ . So $\frac{s i n i}{s i n r} = \frac{v _{1}}{v _{2}}$ . Since $n = \frac{c}{v}$ , this gives $n_{1} sin i = n_{2} sin r$ : Snell's law. If $v_{2} < v_{1}$ the ray bends towards the normal.

Reflection of a plane wave. The same construction with the wavelet from A spreading back into medium 1 at the same speed v: $A E = B C = v t$ . Triangles ABC and AEC are congruent (same hypotenuse AC, equal sides), so the angle of reflection equals the angle of incidence: i = r.

Consequences.

On refraction the frequency stays the same (it is set by the source), while speed and wavelength change: $λ_{2} = \frac{λ _{1} v _{2}}{v _{1}}$ , i.e. $λ_{m e d i u m} = \frac{λ _{0}}{n}$ .
Going from denser to rarer ( $v_{2} > v_{1}$ ), the refracted ray bends away from the normal; if $sin i > \frac{v _{1}}{v _{2}}$ , no refracted wavefront can be drawn (the wavelet does not reach far enough): total internal reflection.
A convex lens turns a plane wavefront into a converging spherical one; a concave mirror does the same on reflection; a thin prism tilts a plane wavefront.
The wave theory correctly predicts that light is slower in water than in air, which the old corpuscular theory got wrong.

Huygens Principle and Wavefronts: formulas and key facts

Snell's law from Huygens: $\frac{s i n i}{s i n r} = \frac{v _{1}}{v _{2}} = \frac{n _{2}}{n _{1}}$; From triangles sharing the hypotenuse AC.
Law of reflection: $i = r$; Wavelet travels the same distance in the same medium.
Refractive index and speed: $n = \frac{c}{v}$; Speed is less in a denser medium.
Wavelength in a medium: $λ_{m e d i u m} = \frac{λ _{0}}{n}, ν unchanged$; Frequency is fixed by the source.
Total internal reflection (wave picture): $sin i > \frac{v _{1}}{v _{2}}$; No refracted wavefront possible (v₂ > v₁).

How to solve Huygens Principle and Wavefronts questions

Use this method for wavefront constructions and related numericals.

1Identify the wavefront shape from the source: point (spherical), line (cylindrical), distant (plane).
2For a construction, draw the incident plane wavefront AB touching the surface at A, the extra path BC = v₁t, and the wavelet of radius v₂t (refraction) or v₁t (reflection) centred at A.
3Draw the tangent from C to the wavelet: that is the new wavefront; mark i and r between the wavefronts and the surface.
4Use the two right triangles with common hypotenuse AC to get sin i and sin r.
5In numericals, use $\frac{s i n i}{s i n r} = \frac{v _{1}}{v _{2}}$ , $v = \frac{c}{n}$ and $λ = \frac{λ _{0}}{n}$ , keeping the frequency unchanged.
6State clearly which medium is faster; that decides whether the ray bends towards or away from the normal.

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.

Change of wavelength in water

Suppose the yellow light from a sodium street lamp (wavelength 589 nm in air) shines into a pond. Take the refractive index of water as 1.33.

(a) Find the speed of the light in water. (b) Find its wavelength in water. (c) Does its colour change? Explain using the wavefront picture.

Try it first, then show the solutionSolution

$v = \frac{c}{n} = \frac{3 \times 1 0 ^{8}}{1.33} \approx 2.26 \times 1 0^{8}$ m/s.
$λ = \frac{589}{1.33} \approx 443$ nm.
No. The secondary wavelets at the surface are driven by the incoming wave, so they oscillate at the same frequency; colour is set by frequency, so it stays yellow even though the wavelength is shorter.

Answer: (a) about 2.26 × 10⁸ m/s; (b) about 443 nm; (c) no, the frequency (and colour) is unchanged.

Memory hook: Into water, light slows and squeezes its wavelength but keeps its beat.

Why wavefronts bend

Imagine a marching band in straight rows crossing at an angle from a smooth road (where they walk at 1.2 m/s) onto a muddy field (where they walk at 0.8 m/s). Each row is like a wavefront.

(a) Explain why the rows change direction at the edge. (b) If the rows meet the edge at 40°, find the new angle using the wavefront rule. (c) Which way do they turn, towards or away from the normal?

Try it first, then show the solutionSolution

The end of a row that reaches the mud first slows down while the rest still moves fast, so the row swings round, just as a wavefront does at a boundary.
$sin r = sin 40° \times \frac{0.8}{1.2} \approx 0.643 \times 0.667 \approx 0.429$ , so r ≈ 25.4°.
Towards the normal, since they move more slowly in the second region.

Answer: (a) the end that enters first slows down first; (b) about 25.4°; (c) towards the normal.

Memory hook: Slow side first, the wavefront swings towards the normal.

Tricks to remember Huygens Principle and Wavefronts

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

Point-sphere, line-cylinder, far-plane
Source shape decides the wavefront shape.
Use it when: Naming wavefronts.
Rays stand on wavefronts
A ray is always at 90° to the wavefront.
Use it when: Drawing constructions.
BC in medium 1, AE in medium 2
BC = v₁t (the late edge catching up), AE = v₂t (the wavelet in the new medium).
Use it when: Huygens construction for refraction.
Frequency is fixed
Wavelength and speed change at a boundary; frequency never does.
Use it when: Refraction numericals.

Huygens Principle and Wavefronts: solved examples

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

Example 1: Light entering glass · 2 marks · Very short answerBoard level

Light of wavelength 540 nm in air enters glass of refractive index 1.5. Find (a) its speed, (b) its wavelength and (c) its frequency in glass.

Try it first, then show the solutionSolution

(a) $v = \frac{3 \times 1 0 ^{8}}{1.5} = 2 \times 1 0^{8}$ m/s. (½)
(b) $λ = \frac{540}{1.5} = 360$ nm. (½)
(c) $ν = \frac{c}{λ _{0}} = \frac{3 \times 1 0 ^{8}}{5.4 \times 1 0 ^{- 7}} \approx 5.56 \times 1 0^{14}$ Hz, unchanged in glass. (1)

Answer: (a) 2 × 10⁸ m/s; (b) 360 nm; (c) about 5.56 × 10¹⁴ Hz.

Example 2: Proof of Snell's law using Huygens principle · 5 marks · Long answerBoard level

(a) State Huygens principle. (1)
(b) Using a plane wavefront incident on a plane surface separating two media, draw the construction and prove Snell's law of refraction for light going into a denser medium. (4)

Try it first, then show the solutionSolution

(a) Every point of a wavefront is a source of secondary wavelets moving forward at the wave speed; the new wavefront is the forward tangent envelope of these wavelets. (1)
(b) Plane wavefront AB meets surface PP' at A, at angle i. In time t, B moves to C on the surface: BC = v₁t. (1)
In the same time, the wavelet from A spreads into medium 2 with radius AE = v₂t. The tangent CE from C to this wavelet is the refracted wavefront, at angle r to the surface. (1)
In right triangles ABC and AEC: $sin i = \frac{v _{1} t}{A C}$ , $sin r = \frac{v _{2} t}{A C}$ . (1)
So $\frac{s i n i}{s i n r} = \frac{v _{1}}{v _{2}} = \frac{c / n _{1}}{c / n _{2}} = \frac{n _{2}}{n _{1}}$ , i.e. $n_{1} sin i = n_{2} sin r$ . Since v₂ < v₁, r < i. (1)

Answer: Huygens principle as stated; sin i/sin r = v₁/v₂, which gives n₁ sin i = n₂ sin r.

Example 3: Proof of the law of reflection · 3 marks · Short answerHarder board level

Using Huygens principle, show that a plane wavefront reflected from a plane surface obeys the law of reflection.

Try it first, then show the solutionSolution

Wavefront AB meets the surface at A at angle i. In time t, B travels BC = vt to the surface. (1)
Meanwhile the wavelet from A spreads back into the same medium with radius AE = vt. The tangent CE is the reflected wavefront, making angle r with the surface. (1)
Right triangles ABC and AEC share hypotenuse AC and have BC = AE, so they are congruent: i = r. (1)

Answer: The congruent triangles give i = r.

Example 4: Angle of refraction from speeds · 2 marks · Very short answerHarder board level

Light travelling at 3 × 10⁸ m/s strikes a medium in which its speed is 2.25 × 10⁸ m/s, at an angle of incidence of 60°. Find the angle of refraction and the refractive index of the medium.

Try it first, then show the solutionSolution

$sin r = sin 60° \times \frac{2.25}{3} \approx 0.866 \times 0.75 \approx 0.650$ , so r ≈ 40.5°. (1)
$n = \frac{3}{2.25} \approx 1.33$ . (1)

Answer: r ≈ 40.5°; n ≈ 1.33.

Example 5: Numbers in the Huygens construction · 5 marks · Long answerHOTS / competency

A plane wavefront in air meets the flat surface of glass (n = 1.5) at 30°. In the construction, the length of surface AC between the first and last points of contact is 10 cm.
(a) Find BC and the time t for the wavefront edge to travel it. (2)
(b) Find the radius AE of the wavelet in glass at that time. (1)
(c) Find r from the construction and check it with Snell's law. (2)

Try it first, then show the solutionSolution

(a) BC = AC sin i = 10 × 0.5 = 5 cm. (1)
$t = \frac{0.05}{3 \times 1 0 ^{8}} \approx 1.67 \times 1 0^{- 10}$ s. (1)
(b) $A E = v_{2} t = 2 \times 1 0^{8} \times 1.67 \times 1 0^{- 10} \approx 0.0333$ m = 3.33 cm. (1)
(c) $sin r = \frac{A E}{A C} = \frac{3.33}{10} = 0.333$ , r ≈ 19.5°. (1)
Snell: $sin r = \frac{s i n 30°}{1.5} = 0.333$ . Same result. (1)

Answer: (a) 5 cm and about 1.67 × 10⁻¹⁰ s; (b) about 3.33 cm; (c) r ≈ 19.5°, agreeing with Snell's law.

Example 6: Shapes of wavefronts · 4 marks · Case-basedHarder board level

Read the passage and answer.

A student studies how wavefronts change shape.
(a) What is the shape of the wavefront from a distant star reaching a telescope? (1)
(b) What shape does a convex lens give to this wavefront after it passes through? (1)
(c) What is the shape of the wavefront from a narrow illuminated slit? (1)
(d) How does a thin prism change a plane wavefront? (1)

Try it first, then show the solutionSolution

(a) Plane. (1)
(b) A converging spherical wavefront, centred on the focus (the middle of the wavefront is delayed more, as it passes through the thicker glass). (1)
(c) Cylindrical. (1)
(d) It tilts it: the part passing through the thicker base is delayed more, so the emerging plane wavefront is turned towards the base. (1)

Answer: (a) plane; (b) converging spherical; (c) cylindrical; (d) tilted towards the base.

Where students lose marks in Huygens Principle and Wavefronts

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

Mistake: Drawing the wavelet in the second medium with the same radius as BC.
Forgetting the speed changes.
Do this: BC = v₁t; AE = v₂t. Draw AE shorter for a denser medium.
Usually costs: 1 to 2 marks of the construction.
Mistake: Marking i and r between the rays and the surface.
Confusing the angles.
Do this: In the construction, i and r are the angles between the wavefronts and the surface (equal to the ray angles with the normal).
Usually costs: 1 mark.
Mistake: Saying frequency changes on refraction.
Thinking all properties change.
Do this: Frequency stays the same; speed and wavelength change.
Usually costs: 1 mark.
Mistake: Not stating Huygens principle before using it.
Jumping into the proof.
Do this: State both parts of the principle in the first lines.
Usually costs: 1 mark in a 5-mark answer.

Exam tips

The Huygens proof of Snell's law (or of reflection) carried 3- and 5-mark questions in 2023 (55/1/1 Q21(a) and 55/5/1 Q26), 2024 (55/1/1 Q33(b) and 55/5/1 Q33(a)).
Draw the construction large and clear: incident wavefront, surface, BC, wavelet arc, refracted wavefront, angles.
Write v₁t and v₂t on the diagram; the proof then takes three lines.
Mention that the frequency stays the same and the wavelength changes; examiners can attach it as a follow-up part.
Doppler effect for light is not in the 2026-27 syllabus.

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
The wavefront of light from a very distant source is
(a) spherical (b) cylindrical (c) plane (d) elliptical
Hint
A tiny part of a huge sphere.
Answer
(c) plane.
Q2 · 1 mark · Very short answerBoard level
Define a wavefront. What is the angle between a ray and the wavefront?
Hint
Same phase.
Answer
A wavefront is the locus of all points vibrating in the same phase. A ray is perpendicular (90°) to the wavefront.
Q3 · 1 mark · Very short answerHarder board level
Light of wavelength 500 nm in vacuum enters a medium of refractive index 1.25. Find its wavelength and frequency in the medium.
Hint
λ/n; frequency unchanged.
Answer
λ = 400 nm; ν = 3 × 10⁸/5 × 10⁻⁷ = 6 × 10¹⁴ Hz (unchanged).
Q4 · 1 mark · Very short answerHarder board level
Light goes from medium 1 into medium 2 with i = 45° and r = 30°. Find the ratio of the speeds v₁ : v₂.
Hint
sin i/sin r = v₁/v₂.
Answer
v₁/v₂ = sin 45°/sin 30° = √2 ≈ 1.41.
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The frequency of light does not change when it passes from air into glass.
Reason (R): The secondary wavelets at the surface are set oscillating by the incident wave, so they oscillate at its frequency.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
What sets the frequency?
Answer
(a).
Q6 · 3 marks · Short answerHOTS / competency
Using the Huygens construction, explain why no refracted wavefront exists when light goes from glass to air at an angle of incidence greater than the critical angle.
Hint
Compare AE = v₂t with AC.
Answer
In air (v₂ > v₁) the wavelet from A has radius AE = v₂t, while BC = v₁t = AC sin i. A tangent from C to the wavelet exists only if AE < AC, i.e. sin i < v₁/v₂. If sin i > v₁/v₂ (i beyond the critical angle), the wavelet already reaches past C, no tangent can be drawn, and there is no refracted wavefront: the light is totally reflected.
Q7 · 2 marks · Very short answerHOTS / competency
Sketch (describe) the refracted wavefront and state how the wavelength changes when a plane wave passes from water into air at an angle.
Hint
Air is the faster medium.
Answer
The wavelet in air (radius v_air t) is larger than BC, so the refracted wavefront is tilted more steeply: the ray bends away from the normal (r > i). The wavelength increases in air (λ = λ_water × n_water), while the frequency stays the same.

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.Define wavefront.Show answer
The locus of points vibrating in the same phase.
2.State Huygens principle.Show answer
Each point of a wavefront emits secondary wavelets; the new wavefront is their forward envelope.
3.Wavefront from a line source?Show answer
Cylindrical.
4.Ratio sin i/sin r in terms of speeds?Show answer
v₁/v₂.
5.What is unchanged when light refracts?Show answer
Its frequency.
6.Angle between a ray and its wavefront?Show answer
90°.

Huygens Principle and Wavefronts: frequently asked questions

Why are backward wavelets ignored in Huygens principle?

Huygens simply assumed that only the forward envelope counts. A more complete wave theory shows that the backward wavelets cancel out, so no backward wave forms.

How is a wavefront related to a ray?

A ray is a line perpendicular to the wavefronts, showing the direction in which the wave and its energy travel.

Does the wave theory predict light is slower in water?

Yes. Huygens' construction gives sin i/sin r = v₁/v₂; since light bends towards the normal entering water, its speed in water must be less than in air, which experiments confirm.

What does a convex lens do to a plane wavefront?

The centre of the wavefront passes through more glass and is delayed more than the edges, so the wavefront becomes a converging spherical surface centred on the focus.

Is the Doppler effect for light in the syllabus?

No. For 2026-27 the wave optics chapter covers wavefronts, Huygens principle, interference and single-slit diffraction (qualitative); the Doppler effect and polarisation are not assessed.