Electromagnetic Waves — Class 12 Physics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 12 Physics chapter "Electromagnetic Waves" — 8 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 12 Physics chapter "Electromagnetic Waves" — 8 important questions with detailed answers for CBSE board ex…
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NCERT Solutions for Class 12 Physics — Electromagnetic Waves. Step-by-step answers to all 8 textbook questions from this chapter, written for CBSE board preparation and free to use.
Electromagnetic Waves — All 8 Questions Solved
Q1. What is an electromagnetic wave? Write its characteristics.
An electromagnetic wave is a transverse wave consisting of time-varying electric and magnetic fields perpendicular to each other and to the direction of propagation.
Characteristics:
1. Transverse in nature — E and B oscillate perpendicular to direction of propagation
2. E and B are perpendicular to each other
3. Both E and B oscillate in phase
4. Speed in vacuum = c = 3 × 10⁸ m/s
5. Speed in medium = c/n where n is refractive index
6. Obey Maxwell's equations
7. Carry energy and momentum
8. Can be reflected, refracted, diffracted, and polarized
9. Do not require a medium for propagation
10. Relationship: E₀/B₀ = c
Q2. Derive the relationship between c, μ₀, and ε₀ for speed of electromagnetic waves.
From Maxwell's equations:
∇ × E = -dB/dt (Faraday's law)
∇ × B = μ₀ε₀(dE/dt) (Ampere-Maxwell law)
For a plane wave traveling in +x direction:
E = E₀ sin(kx - ωt)
B = B₀ sin(kx - ωt)
From Faraday's law:
-dE/dx = dB/dt
-E₀k cos(kx - ωt) = -B₀ω cos(kx - ωt)
E₀k = B₀ω
E₀/B₀ = ω/k = v
From Ampere-Maxwell law:
dB/dx = μ₀ε₀(dE/dt)
B₀k cos(kx - ωt) = μ₀ε₀ E₀ω cos(kx - ωt)
B₀k = μ₀ε₀ E₀ω
Dividing first by second:
(E₀/B₀)/(B₀k) = ω/(μ₀ε₀ E₀ω)
1/B₀² = μ₀ε₀/ω²
ω/k = 1/√(μ₀ε₀)
Therefore: c = 1/√(μ₀ε₀) ≈ 3 × 10⁸ m/s
Q3. What is meant by polarization of an electromagnetic wave? Describe linear and circular polarization.
Polarization: Restriction of oscillation of electric field vector E to a particular direction perpendicular to propagation.
Linear (plane) polarization:
- E oscillates along a fixed direction (one plane)
- E⃗ = E₀ sin(kx - ωt) ĵ (for oscillation in y-direction)
- Can be obtained using polarizers (Nicol prism, polaroid)
- Malus law: I = I₀ cos²θ (θ = angle between polarizer axes)
Circular polarization:
- E vector rotates in a plane perpendicular to propagation
- E⃗ = E₀[sin(kx - ωt) ĵ + sin(kx - ωt + π/2) k̂]
- Magnitude |E⃗| = E₀ (constant)
- Obtained by superposition of two perpendicular linear waves of equal amplitude with phase difference π/2
- Right-handed: E rotates clockwise when viewed toward source
- Left-handed: E rotates counterclockwise
Q4. Light of intensity 0.50 W/m² is incident on a polarizer. What is the intensity of transmitted light? What if two polarizers are placed with axes at 60° to each other?
Case 1: Single polarizer
Using Malus law: I = I₀ cos²θ
For unpolarized light incident on first polarizer (θ = 0°):
I₁ = I₀/2 = 0.50/2 = 0.25 W/m²
Case 2: Two polarizers at 60° angle
After first polarizer: I₁ = I₀/2 = 0.25 W/m²
After second polarizer at 60° to first:
I₂ = I₁ cos²(60°) = 0.25 × cos²(60°)
cos(60°) = 0.5, so cos²(60°) = 0.25
I₂ = 0.25 × 0.25 = 0.0625 W/m²
Alternatively:
I₂ = (I₀/2) × cos²(60°) = (0.50/2) × 0.25 = 0.0625 W/m²
Final answers:
- Single polarizer: 0.25 W/m²
- Two polarizers at 60°: 0.0625 W/m² or 1/16 W/m²
Q5. State Brewster's law and derive the polarizing angle for reflected light.
Brewster's Law: When unpolarized light is incident at a particular angle (Brewster's angle), the reflected light is completely polarized with E parallel to the surface.
Derivation:
At Brewster's angle θ_B:
- Reflected ray is perpendicular to refracted ray
- θ_B + θ_r = 90° where θ_r is refraction angle
From Snell's law:
sin θ_B = n sin θ_r
Since θ_r = 90° - θ_B:
sin θ_B = n sin(90° - θ_B)
sin θ_B = n cos θ_B
tan θ_B = n
Bruesterer's Angle: θ_B = tan⁻¹(n) or θ_B = tan⁻¹(n₂/n₁)
For air-glass interface (n = 1.5):
θ_B = tan⁻¹(1.5) ≈ 56.3°
At this angle, the reflected light contains only the component perpendicular to plane of incidence, making it completely plane polarized.
Q6. An electromagnetic wave has electric field E = 50 sin(2π × 10⁷ t - 2π × 10⁻² x) V/m in SI units. Find wavelength, frequency, and period.
Given: E = 50 sin(2π × 10⁷ t - 2π × 10⁻² x) V/m
Comparing with standard form: E = E₀ sin(ωt - kx)
E₀ = 50 V/m
ω = 2π × 10⁷ rad/s
k = 2π × 10⁻² rad/m
Frequency:
ω = 2πν
ν = ω/(2π) = (2π × 10⁷)/(2π) = 10⁷ Hz = 10 MHz
Wavelength:
k = 2π/λ
λ = 2π/k = 2π/(2π × 10⁻²) = 1/(10⁻²) = 100 m
Period:
T = 1/ν = 1/(10⁷) = 10⁻⁷ s = 0.1 μs
Verification:
v = νλ = 10⁷ × 100 = 10⁹ m/s
(Note: This is > c, so this may be internal wave or error in problem. If v = c, then ν ≈ 3 × 10⁶ Hz)
Answers:
- Wavelength: 100 m
- Frequency: 10 MHz
- Period: 0.1 μs
Q7. What is the displacement current? Derive its expression in terms of electric field.
Displacement Current: It is the hypothetical current between capacitor plates due to changing electric field, introduced by Maxwell to complete Ampere's law.
Derivation:
For a parallel plate capacitor:
Q = CV = ε₀(A/d) × E × d = ε₀AE
Current I = dQ/dt = ε₀A (dE/dt)
This gives conduction current between plates:
i_d = ε₀(dΦ_E/dt) = ε₀A(dE/dt)
Displacement current density:
J_d = i_d/A = ε₀(dE/dt)
Maxwell's modified Ampere's law:
∮ B⃗ · dl⃗ = μ₀(i_c + i_d) = μ₀[i_c + ε₀(dΦ_E/dt)]
Or in differential form:
∇ × B⃗ = μ₀J_c + μ₀ε₀(dE⃗/dt)
Key points:
- Displacement current has no physical charges moving
- But produces magnetic field like conduction current
- Essential for electromagnetic wave propagation
- Between capacitor plates: i_d = ε₀A(dE/dt)
Q8. A parallel plate capacitor with plate area 100 cm² is charged by a 10 A current. What is the displacement current between the plates?
Given:
Plate area A = 100 cm² = 100 × 10⁻⁴ m² = 0.01 m²
Conduction current i_c = 10 A
For a capacitor, the displacement current equals the conduction current in the circuit:
i_d = i_c
This is because charge flows into one plate at rate i_c, accumulating charge:
Q = ∫i_c dt
This changes the electric field between plates:
E = Q/(ε₀A)
Displacement current:
i_d = ε₀A(dE/dt) = ε₀A × d/dt[Q/(ε₀A)] = dQ/dt = i_c
Therefore: i_d = 10 A
Displacement current density:
J_d = i_d/A = 10/0.01 = 1000 A/m²
Answer:
- Displacement current between plates: 10 A
- Displacement current density: 1000 A/m²
Electromagnetic Waves — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. Which of Maxwell's equations is modified to include the displacement current?
- A. Gauss's Law for Electricity
- B. Gauss's Law for Magnetism
- C. Faraday's Law of Induction
- D. Ampere's Circuital Law
Answer: D. Ampere's Circuital Law — Maxwell modified Ampere's Circuital Law by adding the displacement current term (ε₀ dΦE/dt) to account for the magnetic field produced by a changing electric field, completing the symmetry between changing electric and magnetic fields.
MCQ 2. What is the speed of electromagnetic waves in a vacuum?
- A. Dependent on their frequency
- B. Dependent on their wavelength
- C. A constant value, c = 1/√(μ₀ε₀)
- D. Dependent on the amplitude of the electric field
Answer: C. A constant value, c = 1/√(μ₀ε₀) — The speed of electromagnetic waves in a vacuum is a universal constant, 'c', determined by the fundamental constants permeability of free space (μ₀) and permittivity of free space (ε₀), and is independent of the wave's properties.
MCQ 3. What is the nature of electromagnetic waves?
- A. Longitudinal waves
- B. Transverse waves
- C. Both longitudinal and transverse
- D. Neither longitudinal nor transverse
Answer: B. Transverse waves — Electromagnetic waves are transverse waves, meaning the oscillations of the electric and magnetic fields are perpendicular to each other and also perpendicular to the direction of wave propagation.
MCQ 4. In an electromagnetic wave, the electric field (E) and magnetic field (B) are:
- A. Parallel to each other and in phase
- B. Perpendicular to each other and 90° out of phase
- C. Perpendicular to each other and in phase
- D. Parallel to each other and 180° out of phase
Answer: C. Perpendicular to each other and in phase — The electric and magnetic fields in an electromagnetic wave are perpendicular to each other, and they oscillate in phase, reaching their maximum and minimum values simultaneously.
MCQ 5. Which of the following parts of the electromagnetic spectrum has the longest wavelength?
- A. X-rays
- B. Infrared waves
- C. Radio waves
- D. Ultraviolet waves
Answer: C. Radio waves — Radio waves have the lowest frequencies and consequently the longest wavelengths in the electromagnetic spectrum, ranging from about 1 millimeter to over 100 kilometers.
MCQ 6. Which of the following radiations has the highest penetrating power?
- A. Gamma rays
- B. Microwaves
- C. Visible light
- D. Radio waves
Answer: A. Gamma rays — Gamma rays have the highest frequency and shortest wavelength among electromagnetic waves, which gives them the highest energy and thus the greatest penetrating power.
MCQ 7. The direction of propagation of an electromagnetic wave is given by:
- A. The direction of the electric field vector E
- B. The direction of the magnetic field vector B
- C. The direction of E × B
- D. The direction of E + B
Answer: C. The direction of E × B — The direction of propagation of an electromagnetic wave is given by the Poynting vector, which is proportional to the cross product of the electric field vector (E) and the magnetic field vector (B).
MCQ 8. What is the source of electromagnetic waves?
- A. Stationary charges
- B. Charges moving with constant velocity
- C. Accelerated charges
- D. Neutral atoms
Answer: C. Accelerated charges — Electromagnetic waves are produced by accelerated electric charges. When charges accelerate, they radiate energy in the form of oscillating electric and magnetic fields.
MCQ 9. If the electric field of an electromagnetic wave is given by E = E₀ sin(kx - ωt) ĵ, what is the direction of the magnetic field component?
- A. Along the +x direction
- B. Along the +y direction
- C. Along the +z direction
- D. Along the -z direction
Answer: D. Along the -z direction — The wave propagates along the +x direction (from kx - ωt). Since E is along +y (ĵ), and E, B, and the direction of propagation are mutually perpendicular with E × B in the direction of propagation, B must be along the -z direction (kx - ωt) to satisfy (ĵ × -k̂) = î.
MCQ 10. The energy density of an electromagnetic wave is shared equally between:
- A. Kinetic and potential energy
- B. Electric and magnetic fields
- C. Its wavelength and frequency
- D. Its amplitude and phase
Answer: B. Electric and magnetic fields — In an electromagnetic wave, the total energy density is equally distributed between the electric field energy and the magnetic field energy.
MCQ 11. What is the relation between the peak electric field (E₀) and peak magnetic field (B₀) in an electromagnetic wave in vacuum?
- A. E₀ = B₀
- B. E₀ = cB₀
- C. B₀ = cE₀
- D. E₀ = B₀/c²
Answer: B. E₀ = cB₀ — In an electromagnetic wave propagating in vacuum, the ratio of the magnitudes of the electric field and magnetic field is equal to the speed of light, c, i.e., E₀/B₀ = c.
MCQ 12. Which of the following is NOT a property of electromagnetic waves?
- A. They are transverse waves.
- B. They can travel through a vacuum.
- C. They require a medium for propagation.
- D. They travel with the speed of light in vacuum.
Answer: C. They require a medium for propagation. — Electromagnetic waves are unique because they do not require a material medium for their propagation and can travel through the vacuum of space.
MCQ 13. A capacitor is being charged by an external circuit. The current flowing through the connecting wires is called:
- A. Displacement current
- B. Conduction current
- C. Magnetic current
- D. Polarization current
Answer: B. Conduction current — The current flowing through the conducting wires of the circuit is the conduction current, which is due to the flow of actual charge carriers.
MCQ 14. What is the primary application of microwaves?
- A. Medical imaging (e.g., X-rays)
- B. Sterilization of medical equipment
- C. Cooking and radar systems
- D. Night vision devices
Answer: C. Cooking and radar systems — Microwaves are extensively used in microwave ovens for cooking due to their ability to heat water molecules, and in radar systems for detecting objects and measuring their speed.
MCQ 15. Which region of the electromagnetic spectrum is used for optical fiber communication?
- A. Radio waves
- B. Infrared waves
- C. Ultraviolet waves
- D. X-rays
Answer: B. Infrared waves — Infrared radiation, particularly in specific wavelength windows, is primarily used for optical fiber communication because it experiences low attenuation in glass fibers.
MCQ 16. The intensity of an electromagnetic wave is proportional to:
- A. E₀
- B. B₀
- C. E₀B₀
- D. E₀²/B₀²
Answer: C. E₀B₀ — The intensity of an electromagnetic wave is proportional to the product of the magnitudes of the electric and magnetic fields (E₀B₀), or equivalently, to the square of the electric field amplitude (E₀²) or the magnetic field amplitude (B₀²).
MCQ 17. If the frequency of an electromagnetic wave is doubled in vacuum, what happens to its wavelength?
- A. It doubles
- B. It halves
- C. It remains unchanged
- D. It quadruples
Answer: B. It halves — Since the speed of light (c) in vacuum is constant (c = fλ), if the frequency (f) doubles, the wavelength (λ) must halve to maintain the constant speed.
MCQ 18. What is the physical quantity carried by an electromagnetic wave?
- A. Mass
- B. Charge
- C. Energy and momentum
- D. Temperature
Answer: C. Energy and momentum — Electromagnetic waves transport energy and momentum from one point to another, even through a vacuum, without transferring any mass.
MCQ 19. Which of the following phenomena is NOT explained by the wave nature of light?
- A. Reflection
- B. Refraction
- C. Diffraction
- D. Photoelectric effect
Answer: D. Photoelectric effect — The photoelectric effect, involving the emission of electrons from a metal when light shines on it, can only be explained by the particle (photon) nature of light, not its wave nature.
MCQ 20. The ratio of the average energy density of the electric field to that of the magnetic field in an electromagnetic wave is:
- A. 1:1
- B. 1:2
- C. 2:1
- D. Dependent on frequency
Answer: A. 1:1 — In an electromagnetic wave, the average energy density stored in the electric field is equal to the average energy density stored in the magnetic field, hence their ratio is 1:1.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.