Polarisation of Light — JEE Main Physics MCQs with Solutions
Free JEE Main 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. Longitudinal
B. Neither
C. Both transverse and longitudinal
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. Along the direction of propagation
B. Absent
C. In all directions perpendicular to the direction of propagation ✓ Correct
D. Confined to a single plane
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. Confined to a single plane containing the direction of propagation ✓ Correct
C. Rotating randomly
D. Along the direction of propagation
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. Light waves
B. X-rays
C. Radio waves
D. Sound waves in air ✓ Correct
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. Perpendicular to its transmission axis
B. Along the direction of propagation
C. Parallel to its transmission (pass) axis ✓ Correct
D. Of a particular colour
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. $I_0/2$ ✓ Correct
B. Zero
C. $I_0$
D. $I_0/4$
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. Becomes zero at one position
D. Varies between a maximum and zero
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. Varies from a maximum to zero (twice in one full rotation) ✓ Correct
B. Stays constant
C. Only increases
D. Is always zero
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. Scattering of sunlight by air molecules ✓ Correct
B. Diffraction by clouds
C. Dispersion by water droplets
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. Scattering
B. Reflection at the polarising angle
C. Passing light through a converging lens ✓ Correct
D. Selective absorption by a polaroid
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. $I = I_0 \tan^2\theta$
B. $I = I_0 \sin^2\theta$
C. $I = I_0 \cos\theta$
D. $I = I_0 \cos^2\theta$ ✓ Correct
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 electric and magnetic fields
B. The wavefront and the ray
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. Half of the incident
B. Maximum
C. One quarter of the incident
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. $I_0/4$ ✓ Correct
B. $3I_0/4$
C. $I_0/2$
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. 30°
B. 60°
C. 45° ✓ Correct
D. 90°
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. $0.75\,I_0$ ✓ Correct
B. $0.5\,I_0$
C. $0.25\,I_0$
D. $I_0$
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. $I_0/8$ ✓ Correct
B. $3I_0/8$
C. $I_0/2$
D. $I_0/4$
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. $I_0$
B. $I_0/2$ ✓ Correct
C. $I_0/4$
D. Zero
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. $I_0/4$
B. $I_0/2$
C. Zero
D. $I_0/8$ ✓ Correct
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. 30°
C. 45°
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. $\tan\theta_B = n$ ✓ Correct
B. $\sin\theta_B = n$
C. $\cot\theta_B = n$
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. 90° ✓ Correct
B. 45°
C. 0°
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. $r = \theta_B$
C. $r = 90^\circ - \theta_B$ ✓ Correct
D. $r = 2\theta_B$
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. Along the direction of propagation
B. At 45° to the plane of incidence
C. In the plane of incidence
D. Perpendicular to the plane of incidence ✓ Correct
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. 48.6°
B. 41.8°
C. 56.3° ✓ Correct
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. 48°
D. 53° ✓ Correct
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. $\sqrt{2} \approx 1.41$
B. 1.5
C. 1.33
D. $\sqrt{3} \approx 1.73$ ✓ Correct
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. 33° ✓ Correct
B. 57°
C. 43°
D. 90°
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 refractive index of the medium depends on wavelength ✓ Correct
B. The frequency of light changes on reflection
C. The intensity changes with wavelength
D. The speed of light in vacuum changes
Solution: Since n = tanθ_B and n varies with wavelength (dispersion), the polarising angle also varies slightly with wavelength.