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. 257Laws Reflection Refraction Huygens
8 MCQs9-step worked example
Source: NCERT OpticsOfficial key: NTA-verifiedLast updated: 8 Oct 2026
Try this first
In Huygens' construction, the shape of a wavefront at a later time is obtained by drawing
- A.the normal to the old wavefront at its centre
- B.the ray that leaves the source along the axis
- C.a circle around the source with radius equal to the wavelength
- D.a common tangent to the secondary wavelets from every point of the old wavefront
Tap to see the answer
Answer: D. Each point of the wavefront sends out secondary wavelets with the speed of the wave, and a common tangent (envelope) to them is the new wavefront (NCERT Class 12 Physics Chapter 10, page 257).
A is wrong: A is wrong because the normal to a wavefront is a ray, which shows the direction of energy flow, not the position of the wavefront at a later time.
B is wrong: B is wrong because a single ray from the source says nothing about where the whole surface of constant phase has moved.
C is wrong: C is wrong because a circle of radius equal to the wavelength has no link to the elapsed time. The wavelets have radius vt after a time t.
Laws Reflection Refraction Huygens, explained for NEET
A plane wavefront passes from air into glass and bends towards the normal. Did the wave speed up or slow down, and what did the older particle model of light predict? The usual slip is to recall Snell's law as sin i / sin r = v₂ / v₁, or to say the frequency changes at the boundary.
Huygens' principle is a geometrical construction. Each point of a wavefront (a surface of constant phase) is the source of secondary wavelets that spread out with the speed of the wave. A common tangent to those wavelets gives the wavefront at a later time (NCERT Class 12 Physics Chapter 10, page 257).
Refraction. A plane wavefront AB meets the boundary at angle i. While the edge B travels BC = v₁t in medium 1, the wavelet from A travels AE = v₂t in medium 2. The two right triangles give sin i = v₁t/AC and sin r = v₂t/AC, so sin i / sin r = v₁/v₂ (page 259). Bending towards the normal (r < i) therefore means v₂ < v₁. The wave theory predicts a slower speed in the denser medium, the opposite of the corpuscular model, and experiment later favoured the wave theory (page 259). With n = c/v this becomes n₁ sin i = n₂ sin r, Snell's law (page 259). In a denser medium the speed and wavelength both fall, but the frequency stays the same (page 260).
Rarer medium. For v₂ > v₁ the wave bends away from the normal, and sin i_c = n₂/n₁. Beyond the critical angle there is no refracted wave, only total internal reflection (page 260).
Reflection. With AE = BC = vt, the triangles EAC and BAC are congruent, so i = r (pages 260 and 261).
Watch-out: the speed ratio is v₁/v₂, with the incident medium on top. Check any answer by asking: is the wave slower in the second medium, and does the ray bend towards the normal?
How do you solve a Laws Reflection Refraction Huygens question? A worked example
- 1
Given
A ray of light falls on a glass surface of refractive index √3 at an angle of incidence of 60° from the normal (exact angle).
- 2
Required
The angle between the reflected ray and the refracted ray.
- 3
Concept
Reflection gives r_reflected = i. Refraction follows n₁ sin i = n₂ sin r, the law obtained from the Huygens construction (NCERT Class 12 Physics Chapter 10, pages 259 to 261).
- 4
Formula
n₁ sin i = n₂ sin r; reflected angle = i; the angle between the rays = 180° − i − r.
- 5
Substitution
1 × sin 60° = √3 × sin r, so sin r = sin 60°/√3 = (√3/2)/√3.
- 6
Calculation
sin r = 1/2, so r = 30°. The reflected ray lies at 60° to the normal on the incident side and the refracted ray at 30° on the other side, so the angle between them is 180° − 60° − 30° = 90°. The 1, 2, 3 and the angles 30° and 60° are exact and do not limit the significant figures.
- 7
Final answer
The reflected and refracted rays are at 90° to each other.
- 8
Common trap
Taking the angle between the rays as 60° (the angle of reflection) or 30° (the angle of refraction) and forgetting that both are measured from the normal, so the angle between the rays is 180° minus their sum.
- 9
Similar NEET-style question
Light is incident at 45° on a glass slab of refractive index √2. Find the angle between the reflected and refracted rays. (Answer: sin r = sin 45°/√2 = (1/√2)/√2 = 1/2, so r = 30°; the angle between the rays is 180° − 45° − 30° = 105°.)
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Can you answer these Laws Reflection Refraction Huygens 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
In Huygens' construction, the shape of a wavefront at a later time is obtained by drawing
Show answer and why every option is right or wrong
Answer: D. Each point of the wavefront sends out secondary wavelets with the speed of the wave, and a common tangent (envelope) to them is the new wavefront (NCERT Class 12 Physics Chapter 10, page 257).
Why A is wrong: A is wrong because the normal to a wavefront is a ray, which shows the direction of energy flow, not the position of the wavefront at a later time.
Why B is wrong: B is wrong because a single ray from the source says nothing about where the whole surface of constant phase has moved.
Why C is wrong: C is wrong because a circle of radius equal to the wavelength has no link to the elapsed time. The wavelets have radius vt after a time t.
MCQ 2Easy RecallPractice
A wavefront is defined as
Show answer and why every option is right or wrong
Answer: B. A wavefront is the locus of points that oscillate in phase, a surface of constant phase; the energy travels perpendicular to it (NCERT Class 12 Physics Chapter 10, page 257).
Why A is wrong: A is wrong because the line along which energy travels is the ray, which is perpendicular to the wavefront.
Why C is wrong: C is wrong because a ray is the path of energy propagation, and it cuts every wavefront at right angles.
Why D is wrong: D is wrong because the points on a wavefront share the same phase and, for a spherical wave, the same amplitude. The amplitude is not zero.
MCQ 3Direct ApplicationPractice
Light travels in medium 1 at 2.0 × 10⁸ m s⁻¹ and enters medium 2, where it travels at 1.5 × 10⁸ m s⁻¹. The angle of incidence satisfies sin i = 0.60. What is sin r?
Show answer and why every option is right or wrong
Answer: C. The construction gives sin i / sin r = v₁/v₂. So sin r = sin i × v₂/v₁ = 0.60 × (1.5/2.0) = 0.45 (NCERT Class 12 Physics Chapter 10, page 259).
Why A is wrong: A is wrong because 0.80 = 0.60 × (2.0/1.5) uses the inverted speed ratio v₁/v₂. The wave slows down, so the ray bends towards the normal and sin r must be smaller than sin i.
Why B is wrong: B is wrong because sin r = sin i would mean no change of speed and no bending.
Why D is wrong: D is wrong because 0.75 is only the speed ratio v₂/v₁; it was not multiplied by sin i.
MCQ 4Easy RecallPractice
A refracted ray bends towards the normal (r < i). What does the wave theory conclude about the speed of light in the second medium?
Show answer and why every option is right or wrong
Answer: A. From sin i / sin r = v₁/v₂, r < i gives v₂ < v₁. This prediction is opposite to that of the corpuscular model, and later experiments showed the wave theory to be correct (NCERT Class 12 Physics Chapter 10, page 259).
Why B is wrong: B is wrong because a larger speed in the second medium would bend the ray away from the normal. It is the corpuscular model that predicted a larger speed in the denser medium.
Why C is wrong: C is wrong because equal speeds give sin i = sin r, so there would be no bending.
Why D is wrong: D is wrong because the speed ratio follows from the angles alone, v₁/v₂ = sin i / sin r; the frequency does not enter.
MCQ 5CalculationPractice
Light of wavelength 6.0 × 10² nm in air (n = 1.0, c = 3.0 × 10⁸ m s⁻¹) enters glass of refractive index 1.5. What are its wavelength and frequency in glass?
Show answer and why every option is right or wrong
Answer: D. The wavelength ratio equals the speed ratio, λ₁/λ₂ = v₁/v₂ = n₂/n₁, so λ₂ = 6.0 × 10² nm / 1.5 = 4.0 × 10² nm. The frequency does not change at the boundary: ν = c/λ₁ = 3.0 × 10⁸ / 6.0 × 10⁻⁷ = 5.0 × 10¹⁴ Hz (NCERT Class 12 Physics Chapter 10, pages 259 and 260).
Why A is wrong: A is wrong because 9.0 × 10² nm multiplies the wavelength by n. In a denser medium the wavelength and the speed both decrease.
Why B is wrong: B is wrong because the wavelength is right, but 7.5 × 10¹⁴ Hz is c/λ₂, using the vacuum speed with the glass wavelength. The frequency stays 5.0 × 10¹⁴ Hz.
Why C is wrong: C is wrong because the frequency is right but the wavelength is unchanged from air. The wavelength decreases with the speed on entering a denser medium.
MCQ 6Direct ApplicationPractice
A ray in glass of refractive index 1.5 meets a glass-air boundary. The critical angle for this boundary is
Show answer and why every option is right or wrong
Answer: A. At the critical angle the refracted ray grazes the surface, r = 90°, and sin i_c = n₂/n₁ = 1.0/1.5 = 2/3, so i_c = sin⁻¹(2/3), about 42° (NCERT Class 12 Physics Chapter 10, page 260).
Why B is wrong: B is wrong because the defining relation is for the sine of the critical angle, sin i_c = n₂/n₁, not its cosine.
Why C is wrong: C is wrong because the relation involves the sine, not the tangent: sin i_c = n₂/n₁ follows from r = 90°.
Why D is wrong: D is wrong because 1/2 would be n₂/n₁ for an index of 2.0, or the difference n₁ − n₂ = 0.5. The ratio here is 1.0/1.5.
MCQ 7CalculationPractice
A plane wavefront makes an angle of 35° with a plane mirror. What angle does the reflected wavefront make with the mirror, and what is the angle between the incident and reflected rays?
Show answer and why every option is right or wrong
Answer: B. The congruent triangles EAC and BAC give i = r, so the reflected wavefront also makes 35° with the mirror (NCERT Class 12 Physics Chapter 10, pages 260 and 261). The ray is perpendicular to its wavefront, so each ray makes 35° with the normal, and the angle between the incident and reflected rays is 35° + 35° = 70°.
Why A is wrong: A is wrong because 35° is the angle each ray makes with the normal. The angle between the two rays is the sum of the two, 70°.
Why C is wrong: C is wrong because 55° is the complement. The angle of incidence equals the wavefront's angle with the mirror, 35°, not 90° − 35°.
Why D is wrong: D is wrong because the reflected wavefront makes the same 35° as the incident one (i = r), not twice that angle; only the angle between the rays is 70°.
MCQ 8Direct ApplicationPractice
Two transparent media A and B are separated by a plane boundary. The speeds of light in those media are 1.5 × 10⁸ m s⁻¹ and 2.0 × 10⁸ m s⁻¹, respectively. The critical angle for a ray of light for these two media is
Show answer and why every option is right or wrong
Answer: C. Light is slower in A, so A is the denser medium and total internal reflection happens for a ray going from A to B. With n = c/v, sin i_c = n_B/n_A = v_A/v_B = 1.5/2.0 = 0.750, so i_c = sin⁻¹(0.750) (NCERT Class 12 Physics Chapter 10, pages 259 and 260). This is an NTA question from 2022.
Why A is wrong: A is wrong because the critical angle is fixed by its sine, not its tangent: sin i_c = n_B/n_A = v_A/v_B = 1.5/2.0 = 0.750.
Why B is wrong: B is wrong because sin i_c = 0.500 does not follow from these speeds. The ratio of the speeds is 1.5/2.0 = 0.750.
Why D is wrong: D is wrong because a tangent does not give the critical angle, and 0.500 is not the ratio of these speeds; sin i_c = 0.750.
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What to remember before solving Laws Reflection Refraction Huygens questions
9 NCERT lines
From the above equation, we get the important result that if r < i (i.e., if the ray bends toward the normal), the speed of the light wave in the second medium (v2) will be less then the speed of the light wave in the first medium (v1). This prediction is opposite to the prediction from the corpuscular model of light and as later experiments showed, the prediction of the wave theory is correct.
-- NCERT Class 12 Physics, Ch. 10, p. 259In terms of the refractive indices, Eq. (10.3) can be written as n1 sin i = n2 sin r (10.6) This is the Snell’s law of refraction.
-- NCERT Class 12 Physics, Ch. 10, p. 259Key Fact
On refraction into a denser medium, wavelength and speed decrease but the frequency stays the same
The above equation implies that when a wave gets refracted into a denser medium (v1 > v2) the wavelength and the speed of propagation decrease but the frequency n (= v/l) remains the same.
-- NCERT Class 12 Physics, Ch. 10, p. 260We now consider refraction of a plane wave at a rarer medium, i.e., v2 > v1. Proceeding in an exactly similar manner we can construct a refracted wavefront as shown in Fig. 10.5. The angle of refraction will now be greater than angle of incidence; however, we will still have n1 sin i = n2 sin r . We define an angle ic by the following equation 2 1 sin c n i n = (10.8) Thus, if i = ic then sin r = 1 and r = 90°. Obviously, for i > ic, there can not be any refracted wave. The angle ic is known as the critical angle and for all angles of incidence greater than the critical angle, we will not have any refracted wave and the wave will undergo what is known as total internal reflection.
-- NCERT Class 12 Physics, Ch. 10, p. 260We next consider a plane wave AB incident at an angle i on a reflecting surface MN. If v represents the speed of the wave in the medium and if t represents the time taken by the wavefront to advance from the point B to C then the distance BC = vt In order to construct the reflected wavefront we draw a sphere of radius vt from the point A as shown in Fig. 10.6. Let CE represent the tangent plane drawn from the point C to this sphere. Obviously AE = BC = vt
-- NCERT Class 12 Physics, Ch. 10, p. 260If we now consider the triangles EAC and BAC we will find that they are congruent and therefore, the angles i and r (as shown in Fig. 10.6) would be equal. This is the law of reflection.
-- NCERT Class 12 Physics, Ch. 10, p. 261Key Fact
Prisms, lenses and mirrors from wavefronts: slower speed in glass delays part of the wavefront
Once we have the laws of reflection and refraction, the behaviour of prisms, lenses, and mirrors can be understood. These phenomena were discussed in detail in Chapter 9 on the basis of rectilinear propagation of light. Here we just describe the behaviour of the wavefronts as they undergo reflection or refraction. In Fig. 10.7(a) we consider a plane wave passing through a thin prism. Clearly, since the speed of light waves is less in glass, the lower portion of the incoming wavefront (which travels through the greatest thickness of glass) will get delayed resulting in a tilt in the emerging wavefront as shown in the figure. In Fig. 10.7(b) we consider a plane wave incident on a thin convex lens; the central part of the incident plane wave traverses the thickest portion of the lens and is delayed the most. The emerging wavefront has a depression at the centre and therefore the wavefront becomes spherical and converges to the point F which is known as the focus. In Fig. 10.7(c) a plane wave is incident on a concave mirror and on reflection we have a spherical wave converging to the focal point F.
-- NCERT Class 12 Physics, Ch. 10, p. 261From the above discussion it follows that the total time taken from a point on the object to the corresponding point on the image is the same measured along any ray. For example, when a convex lens focusses light to form a real image, although the ray going through the centre traverses a shorter path, but because of the slower speed in glass, the time taken is the same as for rays travelling near the edge of the lens.
-- NCERT Class 12 Physics, Ch. 10, p. 261Laws Reflection Refraction Huygens: NEET previous year questions (PYQs) with answers
21 questions in Optics
No question in our NEET 2020–2025 set targets this topic directly.
More in Optics: 9 exam traps and mistakes · 11 formulas · 7 question patterns from its other lessons.
Sources
NCERT refs: Class 12 Physics Chapter 10, p.257
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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