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Polarization, Brewster’s law and Polaroids: NEET MCQs with answers

50 questions on this topic of Optics (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. A beam of linearly polarized light is incident normally on an ideal polaroid. When the polaroid is rotated from an orientation of maximum transmission by an angle θ₁, the transmitted intensity drops to 75% of its initial maximum value. What is the value of cos(θ₁)?

  1. 1/2
  2. 1/√2
  3. √3/2
  4. 3/4
Show answer and explanation

Answer: (C) √3/2

According to Malus's law, transmitted intensity is I = I_max cos²(θ₁). Given I = 0.75 I_max = (3/4) I_max, we have cos²(θ₁) = 3/4. Taking the positive square root gives cos(θ₁) = √3/2 (which corresponds to θ₁ = 30°).

medium

2. Unpolarized light of intensity I₀ is incident on a system of two ideal polaroids P₁ and P₂ having mutually perpendicular pass axes. A third polaroid P₃ is inserted between P₁ and P₂ with its transmission axis oriented at an angle θ with respect to P₁. For what value of θ is the transmitted intensity maximum, and what is this maximum intensity?

  1. θ = 45°, I_max = (1/8) I₀
  2. θ = 30°, I_max = (3/16) I₀
  3. θ = 45°, I_max = (1/4) I₀
  4. θ = 60°, I_max = (1/8) I₀
Show answer and explanation

Answer: (A) θ = 45°, I_max = (1/8) I₀

After passing P₁, intensity is I₁ = I₀ / 2. The angle between P₁ and P₃ is θ, so intensity after P₃ is I₂ = (I₀/2) cos²θ. The angle between P₃ and P₂ is (90° - θ), so the final intensity is I₃ = I₂ cos²(90° - θ) = (I₀/2) cos²θ sin²θ = (I₀/8) sin²(2θ). This reaches a maximum when sin(2θ) = 1, giving 2θ = 90° or θ = 45°, and I_max = I₀ / 8.

hard

3. Four identical ideal polaroids are arranged in a row. The transmission axis of each polaroid is rotated by an angle of 30° clockwise with respect to the preceding one. If unpolarized light of intensity 128 W/m² is incident on the first polaroid, what is the intensity of light emerging from the fourth polaroid?

  1. 40.5 W/m²
  2. 36 W/m²
  3. 27 W/m²
  4. 54 W/m²
Show answer and explanation

Answer: (C) 27 W/m²

Intensity after the first polaroid is I₁ = 128 / 2 = 64 W/m². For each subsequent polaroid, the angle between adjacent axes is Δθ = 30°, giving an attenuation factor of cos²(30°) = 3/4 per stage. There are 3 successive transmissions after the first polaroid: I₄ = I₁ × [cos²(30°)]³ = 64 × (3/4)³ = 64 × (27 / 64) = 27 W/m².

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4. Two polaroids P₁ and P₂ are set up with their pass axes crossed (at 90°). A third polaroid P₃ is introduced between them such that its transmission axis makes an angle of 45° with that of P₁. If an unpolarized beam of light of intensity I₀ is incident on P₁, the intensity of light emerging from P₂ is:

  1. I₀ / 8
  2. I₀ / 4
  3. I₀ / 16
  4. I₀ / 2
Show answer and explanation

Answer: (A) I₀ / 8

After passing through P₁, the intensity is I₁ = I₀ / 2. Passing through P₃ oriented at 45°, intensity becomes I₂ = I₁ cos²(45°) = (I₀ / 2)(1/2) = I₀ / 4. P₂ is at 45° to P₃, so final intensity is I₃ = I₂ cos²(45°) = (I₀ / 4)(1/2) = I₀ / 8.

easy

5. Unpolarised light of intensity I₀ passes through a single ideal polaroid. The transmitted intensity is:

  1. I₀/4
  2. Zero
  3. I₀/2
  4. I₀
Show answer and explanation

Answer: (C) I₀/2

A polaroid transmits only the component along its axis. Averaging cos² θ over all directions gives ½, so the intensity becomes I₀/2.

NCERT Class 12

hard

6. A stack of N identical ideal polaroids is arranged such that the transmission axis of each polaroid is rotated by an angle θ = 90° / (N - 1) relative to the preceding one, so that the first and last polaroids are crossed. An unpolarized beam of intensity I₀ enters the first polaroid. In the limit as N becomes extremely large (N → ∞), the transmitted intensity emerging from the final polaroid approaches:

  1. 0
  2. I₀ / 4
  3. I₀ / 2
  4. I₀ / e
Show answer and explanation

Answer: (C) I₀ / 2

After the 1st polaroid, intensity is I₁ = I₀ / 2. There are (N - 1) successive transmissions, each with angle Δθ = π / [2(N - 1)]. The final intensity is I_final = (I₀ / 2) [cos(π / (2(N - 1)))]^(2(N - 1)). For large N, cos(x) ≈ 1 - x²/2, so [1 - π²/(8(N-1)²)]^(2(N-1)) ≈ 1 - π²/(4(N-1)) → 1 as N → ∞. Therefore, the transmitted intensity approaches I₀ / 2.

easy

7. An unpolarized light beam of intensity 32 W/m² passes through two polaroids whose pass axes are inclined at 60° to each other. The intensity of the transmitted light is:

  1. 16 W/m²
  2. 4 W/m²
  3. 8 W/m²
  4. 2 W/m²
Show answer and explanation

Answer: (B) 4 W/m²

After passing through the first polaroid, the unpolarized light becomes linearly polarized with intensity I₁ = 32 / 2 = 16 W/m². Using Malus's law for the second polaroid, I = I₁ cos²(60°) = 16 × (1/2)² = 4 W/m².

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8. A plate of transparent material of refractive index μ is placed in air. Light strikes the plate at the polarizing angle i_p. The deviation angle suffered by the transmitted (refracted) ray is:

  1. i_p
  2. 180° - 2 i_p
  3. 90° - 2 i_p
  4. 2 i_p - 90°
Show answer and explanation

Answer: (D) 2 i_p - 90°

The deviation of the refracted ray is δ = |i_p - r|. At the polarizing angle, i_p + r = 90°, which means r = 90° - i_p. Substituting r gives δ = i_p - (90° - i_p) = 2 i_p - 90°.

medium

9. A glass plate (μ = 1.5) is immersed in water (μ = 4/3). A light ray traveling in water is incident on the glass plate. The Brewster angle for this water-glass interface is:

  1. sin⁻¹(0.75)
  2. tan⁻¹(0.889)
  3. tan⁻¹(2.00)
  4. tan⁻¹(1.125)
Show answer and explanation

Answer: (D) tan⁻¹(1.125)

Brewster's law states that tan(i_B) = μ_glass / μ_water. Here, μ_glass = 1.5 = 3/2 and μ_water = 4/3. Thus, tan(i_B) = (3/2) / (4/3) = 9/8 = 1.125, giving i_B = tan⁻¹(1.125).

easy

10. When unpolarized light is reflected from a flat glass plate at Brewster's angle, the reflected light is:

  1. Linearly polarized with electric vector parallel to the plane of incidence
  2. Unpolarized
  3. Circularly polarized
  4. Linearly polarized with electric vector perpendicular to the plane of incidence
Show answer and explanation

Answer: (D) Linearly polarized with electric vector perpendicular to the plane of incidence

At Brewster's angle, the reflected beam consists entirely of vibrations perpendicular to the plane of incidence (s-polarized), while vibrations in the plane of incidence are completely refracted into the medium.

easy

11. Which of the following optical phenomena can be exhibited by electromagnetic waves but NOT by sound waves in air?

  1. Polarization
  2. Interference
  3. Refraction
  4. Diffraction
Show answer and explanation

Answer: (A) Polarization

Polarization is unique to transverse waves. Since light is a transverse electromagnetic wave, it can be polarized, whereas sound waves in air are longitudinal and therefore cannot undergo polarization.

easy

12. The phenomenon of polarisation shows that light waves are:

  1. Transverse
  2. Longitudinal
  3. Stationary
  4. Mechanical
Show answer and explanation

Answer: (A) Transverse

Only transverse waves can be polarised, because their vibrations can be confined to one plane perpendicular to the direction of propagation.

NCERT Class 12

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