Polarisation of Light — NEET Physics MCQs with Solutions
Free NEET Physics Polarisation of Light MCQs with step-by-step solutions (42 questions). Part of Wave Theory of Light. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Polarisation of Light · easy · theory
The phenomenon of polarisation of light establishes that light waves are:
A. Both transverse and longitudinal
B. Longitudinal
C. Neither
D. Transverse ✓ Correct
Solution: Only transverse waves can be polarised; the polarisation of light proves that light is a transverse wave.
Q2 — Polarisation of Light · easy · theory
In an unpolarised light beam, the vibrations of the electric field are:
A. Confined to a single plane
B. Along the direction of propagation
C. Absent
D. In all directions perpendicular to the direction of propagation ✓ Correct
Solution: Unpolarised light has electric-field vibrations in all directions in the plane perpendicular to the direction of propagation.
Q3 — Polarisation of Light · easy · theory
Plane (linearly) polarised light is light in which the vibrations are:
A. In all directions
B. Rotating randomly
C. Along the direction of propagation
D. Confined to a single plane containing the direction of propagation ✓ Correct
Solution: In plane-polarised light the electric-field vibrations are restricted to one plane.
Q4 — Polarisation of Light · medium · theory
Which of the following waves CANNOT be polarised?
A. X-rays
B. Light waves
C. Sound waves in air ✓ Correct
D. Radio waves
Solution: Sound waves in air are longitudinal and cannot be polarised; only transverse waves (light, radio, X-rays) can be.
Q5 — Polarisation of Light · easy · theory
A polaroid transmits only those vibrations that are:
A. Of a particular colour
B. Along the direction of propagation
C. Perpendicular to its transmission axis
D. Parallel to its transmission (pass) axis ✓ Correct
Solution: A polaroid selectively transmits the component of vibration parallel to its transmission axis and absorbs the perpendicular component.
Q6 — Polarisation of Light · medium · theory
When an unpolarised beam of intensity I₀ passes through a single ideal polaroid, the transmitted intensity is:
A. $I_0/4$
B. $I_0$
C. Zero
D. $I_0/2$ ✓ Correct
Solution: A polaroid transmits half of the unpolarised light (average of cos²θ over all angles is 1/2), and the emergent light is plane-polarised.
Q7 — Polarisation of Light · medium · theory
A polaroid is rotated in its own plane about the direction of an unpolarised light beam. The transmitted intensity:
A. Remains constant ✓ Correct
B. Doubles
C. Varies between a maximum and zero
D. Becomes zero at one position
Solution: For unpolarised incident light the transmitted intensity is I₀/2 at every orientation, so it does not change as the polaroid is rotated.
Q8 — Polarisation of Light · medium · theory
When a polaroid is rotated in the path of a plane-polarised beam, the transmitted intensity:
A. Stays constant
B. Is always zero
C. Only increases
D. Varies from a maximum to zero (twice in one full rotation) ✓ Correct
Solution: By Malus's law the transmitted intensity varies as cos²θ, reaching a maximum and falling to zero twice per full rotation — this is used to distinguish polarised from unpolarised light.
Q9 — Polarisation of Light · medium · theory
The blue light of the clear sky is partially polarised because of:
A. Diffraction by clouds
B. Dispersion by water droplets
C. Scattering of sunlight by air molecules ✓ Correct
D. Total internal reflection
Solution: Scattering of sunlight by molecules in the atmosphere partially polarises the light, most strongly at 90° to the sun.
Q10 — Polarisation of Light · medium · theory
Which of the following is NOT a method of producing plane-polarised light?
A. Passing light through a converging lens ✓ Correct
B. Selective absorption by a polaroid
C. Reflection at the polarising angle
D. Scattering
Solution: Polarised light is produced by reflection, refraction/double refraction, scattering and selective absorption — but not by simply passing light through a lens.
Q11 — Polarisation of Light · easy · theory
Malus's law for the intensity of light transmitted by an analyser is:
A. $I = I_0 \cos\theta$
B. $I = I_0 \cos^2\theta$ ✓ Correct
C. $I = I_0 \tan^2\theta$
D. $I = I_0 \sin^2\theta$
Solution: Malus's law: the intensity transmitted by an analyser is I = I₀cos²θ, where θ is the angle between the transmission axes of the polariser and analyser.
Q12 — Polarisation of Light · medium · theory
In Malus's law I = I₀cos²θ, the angle θ is measured between:
A. The wavefront and the ray
B. The electric and magnetic fields
C. The transmission axes of the polariser and the analyser ✓ Correct
D. The incident ray and the reflected ray
Solution: θ is the angle between the pass (transmission) axes of the two polaroids.
Q13 — Polarisation of Light · easy · theory
Two polaroids are crossed (their transmission axes at 90°). The intensity of light emerging from the second polaroid is:
A. One quarter of the incident
B. Half of the incident
C. Maximum
D. Zero ✓ Correct
Solution: At θ = 90°, cos²90° = 0, so no light is transmitted — crossed polaroids block the light.
Q14 — Polarisation of Light · medium · theory
Plane-polarised light of intensity I₀ is incident on an analyser whose axis makes 60° with the plane of polarisation. The transmitted intensity is:
A. $3I_0/4$
B. $I_0/2$
C. $I_0/4$ ✓ Correct
D. $I_0$
Solution: I = I₀cos²60° = I₀ × (1/2)² = I₀/4.
Q15 — Polarisation of Light · medium · numerical
For plane-polarised light of intensity I₀ passing through an analyser, at what angle θ is the transmitted intensity equal to I₀/2?
A. 90°
B. 45° ✓ Correct
C. 30°
D. 60°
Solution: I₀cos²θ = I₀/2 ⇒ cos²θ = 1/2 ⇒ cosθ = 1/√2 ⇒ θ = 45°.
Q16 — Polarisation of Light · medium · theory
Plane-polarised light of intensity I₀ passes through an analyser with its axis at 30° to the plane of vibration. The transmitted intensity is:
A. $0.5\,I_0$
B. $0.25\,I_0$
C. $I_0$
D. $0.75\,I_0$ ✓ Correct
Solution: I = I₀cos²30° = I₀ × (√3/2)² = (3/4)I₀ = 0.75 I₀.
Q17 — Polarisation of Light · medium · theory
Unpolarised light of intensity I₀ passes through a polariser and then an analyser whose axis is at 60° to that of the polariser. The final transmitted intensity is:
A. $I_0/4$
B. $3I_0/8$
C. $I_0/8$ ✓ Correct
D. $I_0/2$
Solution: After the polariser: I₀/2. After the analyser: (I₀/2)cos²60° = (I₀/2)(1/4) = I₀/8.
Q18 — Polarisation of Light · medium · theory
Unpolarised light of intensity I₀ falls on two polaroids whose axes are parallel. The intensity of the emergent light is:
A. $I_0/2$ ✓ Correct
B. Zero
C. $I_0/4$
D. $I_0$
Solution: First polaroid: I₀/2. Second (parallel, θ = 0): (I₀/2)cos²0° = I₀/2.
Q19 — Polarisation of Light · medium · theory
Three polaroids are arranged so that the first and third are crossed (90°) and the middle one makes 45° with the first. Unpolarised light of intensity I₀ is incident. The final emergent intensity is:
A. $I_0/2$
B. $I_0/8$ ✓ Correct
C. $I_0/4$
D. Zero
Solution: After P1: I₀/2. After P2 (45°): (I₀/2)cos²45° = I₀/4. After P3 (45° from P2): (I₀/4)cos²45° = I₀/8. (Removing the middle polaroid would give zero.)
Q20 — Polarisation of Light · medium · numerical
Plane-polarised light passes through an analyser. If the intensity falls to one quarter of its maximum value, the angle between the plane of polarisation and the analyser axis is:
A. 75°
B. 45°
C. 30°
D. 60° ✓ Correct
Solution: cos²θ = 1/4 ⇒ cosθ = 1/2 ⇒ θ = 60°.
Q21 — Polarisation of Light · easy · theory
When unpolarised light is reflected at the polarising (Brewster) angle, the reflected light is:
A. Unpolarised
B. Partially polarised only
C. Completely plane-polarised ✓ Correct
D. Circularly polarised
Solution: At the Brewster (polarising) angle, the reflected beam is completely plane-polarised, with vibrations perpendicular to the plane of incidence.
Q22 — Polarisation of Light · easy · theory
Brewster's law relates the polarising angle θ_B to the refractive index n of the medium as:
A. $\sin\theta_B = n$
B. $\cot\theta_B = n$
C. $\tan\theta_B = n$ ✓ Correct
D. $\cos\theta_B = n$
Solution: Brewster's law: the refractive index equals the tangent of the polarising angle, n = tanθ_B.
Q23 — Polarisation of Light · medium · numerical
At the Brewster angle, the angle between the reflected ray and the refracted ray is:
A. 0°
B. 45°
C. 90° ✓ Correct
D. 180°
Solution: At the polarising angle the reflected and refracted rays are perpendicular to each other (θ_B + r = 90°).
Q24 — Polarisation of Light · medium · theory
At the polarising angle θ_B, the angle of refraction r is related to θ_B by:
A. $r = 90^\circ + \theta_B$
B. $r = 2\theta_B$
C. $r = \theta_B$
D. $r = 90^\circ - \theta_B$ ✓ Correct
Solution: Since the reflected and refracted rays are perpendicular, θ_B + r = 90°, so r = 90° − θ_B.
Q25 — Polarisation of Light · medium · theory
The reflected polarised light at the Brewster angle has its vibrations:
A. Perpendicular to the plane of incidence ✓ Correct
B. At 45° to the plane of incidence
C. In the plane of incidence
D. Along the direction of propagation
Solution: The completely polarised reflected beam vibrates perpendicular to the plane of incidence.
Q26 — Polarisation of Light · medium · numerical
The polarising angle for a glass of refractive index 1.5 is approximately:
A. 56.3° ✓ Correct
B. 41.8°
C. 48.6°
D. 33.7°
Solution: θ_B = tan⁻¹(n) = tan⁻¹(1.5) ≈ 56.3°.
Q27 — Polarisation of Light · medium · numerical
For water of refractive index 1.33, the Brewster (polarising) angle is approximately:
A. 37°
B. 60°
C. 53° ✓ Correct
D. 48°
Solution: θ_B = tan⁻¹(1.33) ≈ 53°.
Q28 — Polarisation of Light · medium · numerical
The polarising angle for a certain glass is 60°. Its refractive index is:
A. $\sqrt{3} \approx 1.73$ ✓ Correct
B. 1.5
C. $\sqrt{2} \approx 1.41$
D. 1.33
Solution: n = tanθ_B = tan60° = √3 ≈ 1.73.
Q29 — Polarisation of Light · medium · numerical
Light is incident on a glass plate at its polarising angle of 57°. The angle of refraction in the glass is:
A. 57°
B. 90°
C. 33° ✓ Correct
D. 43°
Solution: r = 90° − θ_B = 90° − 57° = 33°.
Q30 — Polarisation of Light · medium · theory
The Brewster (polarising) angle of a transparent medium depends on the wavelength of light because:
A. The speed of light in vacuum changes
B. The intensity changes with wavelength
C. The frequency of light changes on reflection
D. The refractive index of the medium depends on wavelength ✓ Correct
Solution: Since n = tanθ_B and n varies with wavelength (dispersion), the polarising angle also varies slightly with wavelength.