Wave Optics – Huygen's Principle and Doppler's Effect of Light: NEET MCQs with Answers

13 free questions on Wave Optics – Huygen's Principle and Doppler's Effect of Light for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

Syllabus covered: Huygen's Principle and Doppler's Effect of Light: wavefront and Huygens' principle, Laws of reflection and refraction using Huygens principle.

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Question 1

Previous-year question
Which one of the following phenomena is NOT explained by Huygens' construction of wavefront? [AIPMT 1988]
  1. Refraction
  2. Reflection
  3. Diffraction
  4. Origin of spectra
Show answer
Answer: (D) Origin of spectra
Explanation: Huygens' principle treats every point on a wavefront as a source of secondary wavelets, and this successfully accounts for reflection, refraction and diffraction. The origin of spectra, however, arises from transitions between discrete atomic energy levels and is explained only by quantum theory, not by a wave construction.

Question 2

A plane wavefront AB meets a plane mirror obliquely, with edge A touching the mirror at t = 0. Edge B reaches the mirror at point C at time t, the speed of light being v. In Huygens' construction the reflected wavefront is the tangent CE drawn from C to the wavelet of radius AE centred on A. Why does this construction give the angle of reflection equal to the angle of incidence?
  1. The wavelet from A has radius 2vt, as it goes to the mirror and back, which makes the two triangles similar
  2. The wavefronts AB and CE are parallel, so the alternate angles they make with the mirror are equal
  3. Right triangles ABC and CEA share the hypotenuse AC and have BC = AE = vt, so they are congruent
  4. Triangles ABC and CEA have AB = CE = vt, since the wavefront itself moves through a distance vt
Show answer
Answer: (C) Right triangles ABC and CEA share the hypotenuse AC and have BC = AE = vt, so they are congruent
Explanation: Reflected light stays in the same medium, so the wavelet from A grows to radius vt in the time B takes to travel BC = vt. Each triangle is right-angled, at B and at E, because a ray is perpendicular to its wavefront, and AC is common to both. The triangles are congruent, so the angle between AC and each wavefront is the same. These angles equal the angles of incidence and reflection.

Question 3

A plane wavefront travels along the principal axis towards a concave mirror. Which statement correctly describes the reflected wavefront?
  1. It is spherical and diverges from a point behind the mirror, because the middle of the wavefront reaches the mirror first and is reflected ahead of the edges
  2. It is spherical and converges towards the focus, because the middle of the wavefront reaches the mirror first and is reflected ahead of the edges
  3. It is plane, because every point of the wavefront travels the same distance before returning
  4. It is spherical and converges towards the focus, because the edges are reflected first while the middle travels on to the pole and back, and so lags behind
Show answer
Answer: (D) It is spherical and converges towards the focus, because the edges are reflected first while the middle travels on to the pole and back, and so lags behind
Explanation: The edges of a concave mirror stand forward of its pole, so the outer parts of the wavefront are reflected first. The central part travels further, on to the pole and back, and so falls behind the edges. The reflected wavefront is therefore curved with its edges ahead, which is a spherical wavefront converging to the focus.

Question 4

A thin convex lens forms a real image of a point object on its axis. A ray along the axis through the centre of the lens is compared with a ray passing near the edge of the lens. Which statement is correct?
  1. The edge ray has more of its path in air and therefore reaches the image first
  2. The axial ray has the shorter total path and therefore reaches the image first
  3. The axial ray has the longer total path and therefore reaches the image last
  4. The axial ray has the shorter total path but more of it is in slower glass, and both rays take the same time from object to image
Show answer
Answer: (D) The axial ray has the shorter total path but more of it is in slower glass, and both rays take the same time from object to image
Explanation: The total time from a point on the object to the corresponding point on the image is the same along every ray. The central ray covers a shorter distance, but it crosses the thickest part of the lens, where light is slowest. The extra delay in glass exactly balances the longer path of the edge ray. This is why a plane wavefront entering such a lens leaves it as a converging spherical wavefront.

Question 5

Which observation gives no basis for choosing between a model of light as a stream of particles and a model of light as a wave?
  1. Two overlapping beams from one source produce dark bands at some places on a screen.
  2. Light spreads noticeably into the geometrical shadow after passing through a very narrow slit.
  3. Light obeys the laws of reflection at a plane mirror.
  4. Light travels more slowly in water than in air.
Show answer
Answer: (C) Light obeys the laws of reflection at a plane mirror.
Explanation: Reflection follows from both pictures, since elastic particles bouncing off a smooth surface obey the same laws as a reflected wavefront. Light adding to light to give darkness is natural for waves but cannot arise from particles alone. Spreading at a narrow slit is diffraction, a wave effect. The lower speed in water agrees with the wave model and contradicts the corpuscular one.

Question 6

A ray of light passes from air into water and bends towards the normal. Newton's corpuscular model and Huygens' wave model can both account for this bending. How do the two models differ here?
  1. The corpuscular model needs light to be faster in water, the wave model needs it slower, and measurement supports the wave model.
  2. Both models need light to be slower in water, and they differ only in the frequency predicted in water.
  3. The corpuscular model needs light to be faster in water, the wave model needs it slower, and measurement supports the corpuscular model.
  4. The corpuscular model needs light to be slower in water, the wave model needs it faster, and measurement supports the corpuscular model.
Show answer
Answer: (A) The corpuscular model needs light to be faster in water, the wave model needs it slower, and measurement supports the wave model.
Explanation: To bend towards the normal, the corpuscular model requires a force pulling the particles into the water, which raises their speed there. The wave model gives sin i / sin r = v₁/v₂, so bending towards the normal requires a lower speed in water. Foucault's measurement showed that light is slower in water than in air, which favours the wave model.

Question 7

Statement I: If the speed of a wave in a medium is known, Huygens' construction can locate a wavefront at a later time. Statement II: Huygens' construction, by itself, gives the value of the speed of light in vacuum.
  1. Both Statement I and Statement II are true.
  2. Statement I is true, but Statement II is false.
  3. Both Statement I and Statement II are false.
  4. Statement I is false, but Statement II is true.
Show answer
Answer: (B) Statement I is true, but Statement II is false.
Explanation: The wavelets drawn after time t have radius vt, and their forward envelope gives the new wavefront, so the speed is an input to the construction. For the same reason the construction cannot supply the value of that speed. Statement I is thus true and Statement II is false.
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Question 8

A plane wavefront in medium 1 meets a plane boundary at 30°, so the angle of incidence is 30°. One end of the wavefront touches the boundary at point A, and its other end reaches the boundary at point C, where AC = 10 cm. The wave speed in medium 2 is 0.8 times that in medium 1. In the Huygens construction, the radius of the wavelet from A at the moment the wavefront reaches C, and the sine of the angle of refraction, are respectively
  1. 4 cm and 0.4
  2. 4 cm and 0.8
  3. 8 cm and 0.8
  4. 6.25 cm and 0.625
Show answer
Answer: (A) 4 cm and 0.4
Explanation: The far end of the wavefront travels a distance AC sin i = 10 × 0.5 = 5 cm in medium 1 to reach C. In the same time the wavelet from A grows in medium 2 to a radius of 0.8 × 5 = 4 cm. The refracted wavefront is the tangent from C to this wavelet, so sin r = 4/10 = 0.4. This agrees with Snell's law, sin i/sin r = v₁/v₂ = 1.25.

Question 9

A plane wavefront in air meets a flat water surface obliquely, with its edge A reaching the surface first. While the other edge B travels 12 cm in air to reach the surface at C, the secondary wavelet from A travels 9 cm in water. If AC = 20 cm, what are the sine of the angle of refraction and the refractive index of water with respect to air?
  1. sin r = 0.60 and refractive index 4/3
  2. sin r = 0.45 and refractive index 4/3
  3. sin r = 0.45 and refractive index 3/4
  4. sin r = 0.60 and refractive index 3/4
Show answer
Answer: (B) sin r = 0.45 and refractive index 4/3
Explanation: In the Huygens construction the incident wavefront, the refracted wavefront and AC form two right triangles with AC as the common hypotenuse. So sin i = 12/20 = 0.60 and sin r = 9/20 = 0.45. The refractive index is sin i/sin r = 0.60/0.45 = 4/3. This equals the ratio of the speeds, 12 : 9, because both distances are covered in the same time.

Question 10

A plane wavefront in air meets a plane boundary with a medium of refractive index 4/3, the angle of incidence i having sin i = 0.8. End A of the wavefront touches the boundary first, while end B still has 8 cm to travel in air to reach the boundary at point C. When B reaches C, the radius of the Huygens wavelet from A inside the medium and the sine of the angle of refraction are
  1. 6 cm and 0.75
  2. 8 cm and 0.8
  3. 10.7 cm and 0.6
  4. 6 cm and 0.6
Show answer
Answer: (D) 6 cm and 0.6
Explanation: The speed in the medium is 3/4 of that in air, so in the same time the wavelet from A grows to a radius of 8 × 3/4 = 6 cm. The distance along the boundary is AC = BC/sin i = 8/0.8 = 10 cm. The refracted wavefront is the tangent from C to this wavelet, so sin r = 6/10 = 0.6, which agrees with Snell's law.

Question 11

Statement I: A plane wavefront passing through a thin prism emerges as a plane wavefront tilted relative to the original, because the part crossing the thicker region near the base is delayed the most. Statement II: A plane wavefront reflected from a concave mirror becomes a diverging spherical wavefront. Choose the correct option.
  1. Both Statement I and Statement II are false
  2. Statement I is false but Statement II is true
  3. Both Statement I and Statement II are true
  4. Statement I is true but Statement II is false
Show answer
Answer: (D) Statement I is true but Statement II is false
Explanation: Light travels more slowly in glass, so the part of the wavefront crossing more glass near the base lags behind and the emerging wavefront is tilted. A concave mirror delays the edges of the wavefront less than the central part, because the centre travels farther before reflection. The reflected wavefront is therefore a spherical one converging to the focus, not a diverging one.

Question 12

A long, narrow illuminated slit acts as a line source of monochromatic light in a homogeneous, non-absorbing medium. At distances r small compared with its length, the wavefronts and the intensity I are best described by
  1. cylindrical wavefronts, with I ∝ 1/r²
  2. spherical wavefronts, with I ∝ 1/r²
  3. plane wavefronts, with I independent of r
  4. cylindrical wavefronts, with I ∝ 1/r
Show answer
Answer: (D) cylindrical wavefronts, with I ∝ 1/r
Explanation: Points at equal distance from a line source lie on a cylinder coaxial with it, so the wavefronts are cylindrical. The power crossing a length L of such a wavefront spreads over an area 2πrL. Intensity therefore falls as 1/r, which is slower than the 1/r² of a point source.

Question 13

In Huygens' construction for refraction, a plane wavefront meets a plane boundary at an angle of incidence of 30°. In the time one end of the wavefront travels 6.0 cm in medium 1 to reach the boundary, the wavelet from the other end travels 4.0 cm in medium 2. What are the refractive index of medium 2 with respect to medium 1, and the sine of the angle of refraction?
  1. 2/3 and 3/4
  2. 3/2 and 3/4
  3. 2/3 and 1/3
  4. 3/2 and 1/3
Show answer
Answer: (D) 3/2 and 1/3
Explanation: The two distances are covered in the same time, so v₁/v₂ = 6.0/4.0 = 3/2. The construction gives sin i/sin r = v₁/v₂, which is the refractive index of medium 2 with respect to medium 1. Hence the index is 3/2 and sin r = (sin 30°)/(3/2) = 1/3.

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