Wave Optics – Polarisation: NEET MCQs with Answers

11 free questions on Wave Optics – Polarisation 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: Polarisation: Polarization, plane-polarized light: Brewster's law, uses of plane-polarized light and Polaroid.

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Question 1

Previous-year question
The Brewster angle i(b) for an interface should be [NEET 2020]
  1. 30° < i(b) < 45°
  2. 45° < i(b) < 90°
  3. i(b) = 90°
  4. 0° < i(b) < 30°
Show answer
Answer: (B) 45° < i(b) < 90°
Explanation: Brewster's law gives tan i(b) = µ, and for any medium denser than air µ > 1, so tan i(b) > 1 and the angle lies between 45° and 90°.

Question 2

Previous-year question
Unpolarised light is incident from air on a plane surface of a material of refractive index µ. At a particular angle of incidence i, it is found that the reflected and refracted rays are perpendicular to each other. Which of the following options is correct for this situation? [NEET 2018]
  1. Reflected light is polarised with its electric vector parallel to the plane of incidence
  2. Reflected light is polarised with its electric vector perpendicular to the plane of incidence
  3. i = tan⁻¹(1/µ)
  4. i = sin⁻¹(1/µ)
Show answer
Answer: (B) Reflected light is polarised with its electric vector perpendicular to the plane of incidence
Explanation: At Brewster's angle (i = tan⁻¹µ) the reflected beam is completely plane polarised with its electric vector perpendicular to the plane of incidence.

Question 3

Previous-year question
Two Polaroids P₁ and P₂ are placed with their axes perpendicular to each other. Unpolarised light I₀ is incident on P₁. A third polaroid P₃ is kept in between P₁ and P₂ such that its axis makes an angle of 45° with that of P₁. The intensity of transmitted light through P₂ is [NEET 2017]
  1. I₀/4
  2. I₀/8
  3. I₀/16
  4. I₀/2
Show answer
Answer: (B) I₀/8
Explanation: After P₁ the intensity is I₀/2; after P₃ it is (I₀/2)cos²45° = I₀/4; after P₂, which is at 45° to P₃, it is (I₀/4)cos²45° = I₀/8.

Question 4

Previous-year question
Which of the following phenomena is NOT common to sound and light waves? [AIPMT 1988]
  1. Interference
  2. Diffraction
  3. Coherence
  4. Polarisation
Show answer
Answer: (D) Polarisation
Explanation: Polarisation requires the oscillations to be perpendicular to the direction of propagation, so it occurs only for transverse waves. Light is transverse and can be polarised; sound in a fluid is longitudinal and cannot. Interference, diffraction and coherence are exhibited by both.

Question 5

Previous-year question
Two polaroids P₁ and P₂ are placed with their axes perpendicular to each other. Unpolarised light I₀ is incident on P₁. A third polaroid P₃ is kept in between P₁ and P₂ such that its axis makes an angle 45° with that of P₁. The intensity of transmitted light through P₂ is [NEET 2017]
  1. I₀/2
  2. I₀/4
  3. I₀/8
  4. I₀/16
Show answer
Answer: (C) I₀/8
Explanation: Unpolarised light emerges from P₁ with intensity I₀/2. Passing through P₃ at 45°, Malus's law gives I₂ = (I₀/2)cos²45° = I₀/4. P₂ is at 45° to P₃, so I₃ = (I₀/4)cos²45° = I₀/8.

Question 6

Previous-year question
Unpolarised light is incident from air on a plane surface of a material of refractive index μ. At a particular angle of incidence i, it is found that the reflected and refracted rays are perpendicular to each other. Which of the following options is correct for this situation? [NEET 2018]
  1. i = sin⁻¹(1/μ)
  2. Reflected light is polarised with its electric vector perpendicular to the plane of incidence
  3. Reflected light is polarised with its electric vector parallel to the plane of incidence
  4. i = tan⁻¹(1/μ)
Show answer
Answer: (B) Reflected light is polarised with its electric vector perpendicular to the plane of incidence
Explanation: Perpendicular reflected and refracted rays define Brewster's condition, for which tan i = μ, so i = tan⁻¹(μ). The reflected light is then completely plane polarised with its electric vector perpendicular to the plane of incidence.

Question 7

Previous-year question
The Brewster's angle i_b for an interface should be [NEET (Sep.) 2020]
  1. 30° < i_b < 45°
  2. 45° < i_b < 90°
  3. i_b = 90°
  4. 0° < i_b < 30°
Show answer
Answer: (B) 45° < i_b < 90°
Explanation: Brewster's law states tan i_b = μ. Since the refractive index of any ordinary medium exceeds 1, tan i_b > 1 and hence i_b > 45°; it must also be less than 90° for a reflected ray to exist. Therefore 45° < i_b < 90°.
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Question 8

Previous-year question
A light ray falls on a glass surface of refractive index √3 at an angle 60°. The angle between the refracted and reflected rays would be [NEET 2022]
  1. 60°
  2. 90°
  3. 120°
  4. 30°
Show answer
Answer: (B) 90°
Explanation: Brewster's law gives tan i_p = μ = √3, so i_p = 60°. The incidence angle equals the polarising angle, and at Brewster's angle the reflected and refracted rays are exactly perpendicular. The angle between them is therefore 90°.

Question 9

Previous-year question
If the monochromatic source in Young's double slit experiment is replaced by white light, then [NEET 2024]
  1. the interference pattern will disappear
  2. there will be a central dark fringe surrounded by a few coloured fringes
  3. there will be a central bright white fringe surrounded by a few coloured fringes
  4. all bright fringes will be of equal width
Show answer
Answer: (C) there will be a central bright white fringe surrounded by a few coloured fringes
Explanation: At the centre the path difference is zero for every wavelength, so all colours reinforce and produce a white central fringe. Away from the centre the maxima for different wavelengths fall at different positions (since x ∝ λ), giving a few coloured fringes which soon overlap into general illumination.

Question 10

Statement I: If the plane of vibration of a stretched string is changed randomly in very short intervals of time, the wave on the string is unpolarised, even though the displacement of each point stays perpendicular to the direction of propagation.
Statement II: Interference, diffraction and polarisation are all shown by sound waves in air.
Which of the following is correct?
  1. Statement I is incorrect but Statement II is correct
  2. Both Statement I and Statement II are correct
  3. Statement I is correct but Statement II is incorrect
  4. Both Statement I and Statement II are incorrect
Show answer
Answer: (C) Statement I is correct but Statement II is incorrect
Explanation: An unpolarised wave is transverse, but its plane of vibration changes randomly with time, so Statement I holds. Sound in air is longitudinal, with displacement along the direction of propagation. It shows interference and diffraction, but polarisation is special to transverse waves, so Statement II is wrong.

Question 11

In a polaroid sheet held in the y-z plane, the long chain molecules are aligned along the y-axis. Unpolarised light of intensity I travels along the x-axis and falls on the sheet. Which of the following describes the emergent light?
  1. Intensity I/2, with the electric vector along the z-axis
  2. Intensity I, with the electric vector along the z-axis
  3. Intensity I/2, with the electric vector along the x-axis
  4. Intensity I/2, with the electric vector along the y-axis
Show answer
Answer: (A) Intensity I/2, with the electric vector along the z-axis
Explanation: Electric vectors along the direction of the aligned molecules are absorbed, so the pass-axis is perpendicular to the molecules, here along z. Half the intensity of unpolarised light is removed by this absorption, leaving I/2. The electric vector of light is transverse, so it can never lie along the direction of travel, x.

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