Every point on a wavefront is a source of secondary wavelets which propagate forward at the same speed; the new wavefront is the envelope of these wavelets.
-- NCERT Class 12 Physics, Ch. 10, p. 257Wavefront Huygens Principle
8 MCQs9-step worked example
Source: NCERT OpticsOfficial key: NTA-verifiedLast updated: 27 Sep 2026
Wavefront Huygens Principle, explained for NEET
The confusion that costs marks here is treating a wavefront as if it were a ray. A ray is a direction; a wavefront is a surface. They are perpendicular to each other, and questions exploit students who blur the two.
NCERT Class 12 Physics Chapter 10 states the principle on page 256: every point on a given wavefront acts as a source of secondary disturbance, and the wavelets emanating from these points spread out in all directions with the speed of the wave. The new wavefront at a later instant is the forward-going envelope (the common tangent surface) of these secondary wavelets.
Three things follow, and each is testable:
Wavefront shape depends on distance from the source. A point source in an isotropic medium produces spherical wavefronts. A line source produces cylindrical wavefronts. Very far from a point source, a small patch of the sphere is effectively flat — a plane wavefront. This is why sunlight reaching Earth is treated as plane.
The envelope is forward-going only. The construction generates secondary wavelets in all directions, but only the forward envelope is taken as the new wavefront. This is the step students skip when asked why there is no backward wave.
Wavelet radius is v·t, not λ. Each secondary wavelet drawn after time t has radius vt in a medium where the wave speed is v. In a denser medium v is smaller, so the wavelets are smaller — this is the geometric reason refraction bends the wavefront, though the quantitative law belongs to the refraction topic and is not re-derived here.
Watch-out: "wavefront" means a surface of constant phase, not constant amplitude and not maximum displacement. A crest surface is one wavefront; a trough surface is equally a wavefront. Any option defining it by amplitude is wrong by definition.
Can you answer these Wavefront Huygens Principle MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
A wavefront is best defined as a surface on which all points of the wave have
Show answer and why every option is right or wrong
Answer: C. A wavefront is a surface of constant phase; this is the definition on which Huygens' construction in NCERT Class 12 Physics Chapter 10, page 256, rests.
Why A is wrong: A is wrong because equal amplitude is not the defining property — amplitude falls off with distance from a point source even along a single spherical wavefront, yet that surface is still one wavefront.
Why B is wrong: B is wrong because it describes only the crest surface. A trough surface, where displacement is most negative, is equally a wavefront — the two differ in phase value, not in status.
Why D is wrong: D is wrong because wavelength is a property of the wave as a whole in a given medium, not something that distinguishes one surface from another within that wave.
MCQ 2Easy RecallPractice
According to Huygens' principle, each point on a wavefront behaves as
Show answer and why every option is right or wrong
Answer: B. NCERT Class 12 Physics Chapter 10, page 256, states that every point on a wavefront acts as a source of secondary disturbance, and the wavelets spread out with the wave's speed.
Why A is wrong: A is wrong because the construction involves no absorption or time delay at the points; the secondary wavelets start at the same instant the primary wavefront reaches those points.
Why C is wrong: C is wrong because it describes a standing-wave node. Huygens' points are sources of propagating disturbance, not stationary points of zero displacement.
Why D is wrong: D is wrong because the wavelets spread in all directions. Restricting emission to the forward direction is an assumption added afterwards when taking the envelope, not a property of the individual wavelets.
MCQ 3Easy RecallPractice
In Huygens' construction, the wavefront at a later instant is obtained as
Show answer and why every option is right or wrong
Answer: D. The new wavefront is the forward envelope — the common tangent surface — of all the secondary wavelets, as constructed in NCERT Class 12 Physics Chapter 10, page 256.
Why A is wrong: A is wrong because it describes superposition of amplitudes, which is a separate calculation. Huygens' construction is geometric: it locates the surface, not the resultant displacement on it.
Why B is wrong: B is wrong because the centres of the secondary wavelets are the points of the OLD wavefront. Joining them reproduces the original surface, giving no propagation at all.
Why C is wrong: C is wrong because in a homogeneous medium every wavelet has the same radius vt, so no one wavelet is largest; and even where speeds differ, a single wavelet is not the wavefront.
MCQ 4Concept TrapPractice
A point source of light is placed in a uniform isotropic medium. At a large distance from the source, a small region of the wavefront is treated as plane because
Show answer and why every option is right or wrong
Answer: A. The wavefront remains spherical, but over a small patch its curvature is negligible when the radius is very large — this is the plane-wavefront approximation used in NCERT Class 12 Physics Chapter 10, page 256.
Why B is wrong: B is wrong because wavelength is fixed by the source frequency and the medium's wave speed; it does not grow with distance.
Why C is wrong: C is wrong because wave speed in a uniform isotropic medium is a property of the medium and does not depend on distance from the source.
Why D is wrong: D is wrong because there is no distance at which the shape genuinely changes. The sphere stays a sphere; 'plane' is an approximation valid over a limited region, not a transition.
MCQ 5Direct ApplicationPractice
For a wave travelling outward from a point source in a uniform medium, the relationship between a ray and the wavefront at any point is that the ray is
Show answer and why every option is right or wrong
Answer: B. A ray marks the direction of propagation, and propagation is along the outward normal to the surface of constant phase — so the ray is perpendicular to the wavefront, consistent with the geometry in NCERT Class 12 Physics Chapter 10, page 256.
Why A is wrong: A is wrong because a tangential direction lies within the surface of constant phase; moving along it does not advance the wave at all.
Why C is wrong: C is wrong because it states the tangential relation in different words, and in addition a wavefront is a surface while a ray is a line — they cannot be parallel in the required sense.
Why D is wrong: D is wrong because a ray is a one-dimensional direction of energy transport and a wavefront is a two-dimensional surface of constant phase. Conflating them is the central confusion this topic tests.
MCQ 6Direct ApplicationPractice
A plane wavefront travels through a homogeneous medium in which the wave speed is v. Applying Huygens' construction, the secondary wavelets drawn after a time interval t are spheres of radius
Show answer and why every option is right or wrong
Answer: A. Each secondary wavelet spreads at the wave speed, so after time t its radius is the distance travelled, vt — the construction step described in NCERT Class 12 Physics Chapter 10, page 256.
Why B is wrong: B is wrong because v/t has units of acceleration, not length, and cannot be a radius.
Why C is wrong: C is wrong because t/v has units of time squared per length; it is dimensionally impossible as a radius.
Why D is wrong: D is wrong because it fixes the radius at λ regardless of elapsed time. The wavelet grows continuously with t; λ is merely the distance covered in one period.
MCQ 7Direct ApplicationPractice
A long straight filament emits light uniformly along its length in a uniform isotropic medium. Close to the filament, the wavefronts produced are
Show answer and why every option is right or wrong
Answer: C. Each point on the line emits secondary wavelets whose common tangent surface at equal distance from the line is a coaxial cylinder — the line-source case that accompanies the point-source and plane cases in NCERT Class 12 Physics Chapter 10, page 256.
Why A is wrong: A is wrong because a plane wavefront requires the source to be effectively at infinity. Close to the filament the surfaces of constant phase are curved around it.
Why B is wrong: B is wrong because a single sphere centred on the midpoint would place points near the filament's ends at different phases from points at the same perpendicular distance near the centre; the source is a line, not a point.
Why D is wrong: D is wrong because a conical surface contains points at varying perpendicular distance from the line, so the disturbance from the filament does not arrive at them in the same phase.
MCQ 8CalculationPractice
In Huygens' construction the secondary wavelets are generated in all directions, yet the wavefront is taken to advance only forwards. The step in the construction that removes the backward-travelling surface is that
Show answer and why every option is right or wrong
Answer: D. Two separate steps are involved: the wavelets do spread in all directions, and then the construction keeps only the forward envelope as the new wavefront — the selection described in NCERT Class 12 Physics Chapter 10, page 256.
Why A is wrong: A is wrong because the wavelets are spheres of uniform radius vt; the construction assigns no direction-dependent radius.
Why B is wrong: B is wrong because the wave speed in a homogeneous medium is a single scalar, the same in every direction. A backward envelope can in fact be drawn geometrically — it is discarded by the construction rule, not by any difference in speed.
Why C is wrong: C is wrong because it misplaces the restriction. The emission is genuinely all-directional in the principle as stated; the forward-only condition is imposed when the envelope is selected, one step later.
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How do you solve a Wavefront Huygens Principle question? A worked example
- 1
Given.
A point source emits light in a uniform medium in which the wave speed is v = 2.00 × 10⁸ m/s. At an instant taken as t = 0, the spherical wavefront from the source has radius r₀ = 4.00 × 10⁻¹ m. The construction is to be applied over an interval t = 1.00 × 10⁻⁹ s.
- 2
Required.
The radius of the secondary wavelets drawn on that wavefront, and the radius of the new wavefront obtained as their forward envelope.
- 3
Concept.
Every point on the r₀ wavefront acts as a source of secondary wavelets. Because the medium is uniform and isotropic, every wavelet has the same radius after the same interval. The forward envelope of a family of equal spheres, centred on a sphere of radius r₀, is a concentric sphere of radius r₀ + (wavelet radius).
- 4
Formula.
Wavelet radius = v t. New wavefront radius r = r₀ + v t. (No dossier formula applies — this is the construction rule itself, not one of the chapter's algebraic formulas.)
- 5
Substitution.
Wavelet radius = (2.00 × 10⁸ m/s)(1.00 × 10⁻⁹ s); r = 4.00 × 10⁻¹ m + that quantity.
- 6
Calculation.
Wavelet radius = 2.00 × 10⁻¹ m. Then r = 4.00 × 10⁻¹ m + 2.00 × 10⁻¹ m = 6.00 × 10⁻¹ m. Every quantity here is a measured value carrying three significant figures; no exact constants (counting numbers or mathematical constants) enter this calculation, so none needs to be excluded from the significant-figure count.
- 7
Final answer.
Secondary wavelets of radius 2.00 × 10⁻¹ m; new wavefront a concentric sphere of radius 6.00 × 10⁻¹ m.
- 8
Common trap.
Reading the wavelet radius as one wavelength rather than vt. Wavelength is the distance covered in one period; the interval given here is arbitrary and has no obligation to equal a period. A second trap is reporting 2.00 × 10⁻¹ m as the new wavefront radius — that is the increment, not the radius. The envelope is concentric with the original surface, so the increment adds to r₀.
- 9
Similar NEET-style question.
A plane wavefront travels through a medium of wave speed 1.50 × 10⁸ m/s. Using Huygens' construction over an interval of 2.00 × 10⁻⁹ s, state the radius of the secondary wavelets and the shape and displacement of the new wavefront. *(Answer: wavelets of radius 3.00 × 10⁻¹ m; the envelope of equal spheres centred on a plane is a parallel plane, displaced 3.00 × 10⁻¹ m forward.)*
What to remember before solving Wavefront Huygens Principle questions
More in Optics: 9 exam traps and mistakes · 11 formulas · 7 question patterns from its other lessons.
Wavefront Huygens Principle questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
Sources
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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