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Electromagnetic Waves: Class 12 Physics Practice Questions

30 original exam-pattern questions with full answers, matched to the current CBSE Class 12 paper design, including case-based questions. Attempt each question before opening the answer — or start a free 14-day trial ↓ for the full bank.

Q1Case-based4 marks

A research team is designing a multi-purpose satellite payload for simultaneous ocean monitoring, ground communication through clouds, water sterilisation, and non-destructive rock imaging. They select four distinct regions of the EM spectrum — infrared, microwave, ultraviolet, and X-rays — assigning one to each task.

A research team is designing a multi-purpose satellite payload that must simultaneously perform the following four tasks:
(P) Map surface temperature variations of ocean water to track climate change.
(Q) Communicate with ground stations through dense cloud cover and rain.
(R) Sterilise water samples collected from a remote sensing module.
(S) Image the internal crystal structure of rock samples without destroying them.

The team selects four different regions of the electromagnetic spectrum — one for each task. Study the information above and answer the following questions:

(i) Identify the most appropriate type of electromagnetic radiation for task (P) and state ONE physical property that makes it suitable.

(ii) Task (Q) requires radiation that can penetrate rain and clouds. Name the type of EM radiation used and write its approximate wavelength range.

(iii) Which type of radiation is used for task (R)? State the mechanism by which it achieves sterilisation.

(iv) For task (S), the team uses radiation whose photon energy is approximately 50 keV. Calculate the frequency of this radiation.
(Given: h = 6.63 × 10⁻³⁴ J s, 1 eV = 1.6 × 10⁻¹⁹ J)

Show answer
(i) Task (P) — Mapping ocean surface temperature:

The appropriate radiation is Infrared (IR) radiation.

Suitable property: All objects at temperatures above absolute zero emit infrared radiation; warmer ocean regions emit more intense IR, allowing thermal mapping of surface temperature variations. (Alternatively: IR is absorbed and re-emitted by surface water in proportion to its temperature, enabling thermal imaging.)

∴ Infrared (IR) radiation is used for task (P).

(ii) Task (Q) — Communication through clouds and rain:

Microwaves are used for this purpose.

Approximate wavelength range: 1 mm to 0.1 m (i.e., 10⁻³ m to 10⁻¹ m).

Reason: Microwaves have long enough wavelengths to diffract around or pass through water droplets in clouds and rain without significant absorption.

∴ Microwaves (λ ≈ 1 mm – 10 cm) are used for task (Q).

(iii) Task (R) — Sterilisation of water samples:

Ultraviolet (UV) radiation is used for sterilisation.

Mechanism: UV radiation (particularly in the UV-C range, λ ≈ 100–280 nm) carries photons of sufficiently high energy (~4–12 eV) to be absorbed by the DNA molecules of microorganisms. This causes molecular bonds in the DNA to break or form abnormal cross-links (thymine dimers), disrupting the microorganism's ability to replicate and effectively killing or inactivating the germs.

∴ Ultraviolet (UV) radiation sterilises by destroying the DNA of microorganisms.

(iv) Task (S) — Calculating frequency of radiation with photon energy 50 keV:

By the photon energy formula:
E = hν

Converting energy to joules:
E = 50 keV = 50 × 10³ × 1.6 × 10⁻¹⁹ J
E = 50 × 1.6 × 10⁻¹⁶ J
E = 80 × 10⁻¹⁶ J
E = 8.0 × 10⁻¹⁵ J

Substituting in E = hν:
ν = E / h = (8.0 × 10⁻¹⁵) / (6.63 × 10⁻³⁴)

ν = (8.0 / 6.63) × 10⁻¹⁵ ⁺ ³⁴
ν = 1.207 × 10¹⁹ Hz

∴ ν ≈ 1.21 × 10¹⁹ Hz

This frequency lies in the X-ray region of the electromagnetic spectrum, confirming that X-rays are appropriate for task (S) — imaging internal crystal structure non-destructively.
Q2Case-based4 marks

A rescue team in a hilly region uses a walkie-talkie that transmits a plane electromagnetic wave. The electric field of the wave is described by:

E⃗ = 60 sin(2.5 × 10⁸ t − 0.83 x) ĵ V m⁻¹

where x is in metres and t is in seconds.

Electromagnetic waves carry energy and momentum. The speed of an EM wave in free space is c = 3 × 10⁸ m s⁻¹, and the amplitudes of E⃗ and B⃗ are related by E₀ = c B₀.

A rescue team in a hilly region uses a walkie-talkie that transmits a plane electromagnetic wave. The electric field of the wave is described by:

E⃗ = 60 sin(2.5 × 10⁸ t − 0.83 x) ĵ V m⁻¹

where x is in metres and t is in seconds.

(i) Identify the direction of propagation of the wave and write the SI unit of the quantity (2.5 × 10⁸).

(ii) Calculate the wavelength and frequency of this wave.

(iii) Calculate the amplitude of the associated magnetic field B₀.

(iv) The rescue team shifts to an underground tunnel where the walkie-talkie signal is lost, but an AM radio receiver picks up a broadcast signal. Give ONE reason why the radio waves can penetrate into the tunnel whereas the walkie-talkie signal (microwave range) cannot.

Show answer
(i) Comparing E⃗ = E₀ sin(ωt − kx) ĵ with the given equation, the wave propagates along the +x direction (î direction).
The quantity 2.5 × 10⁸ is the angular frequency ω.
∴ SI unit of ω is rad s⁻¹.

(ii) Comparing with the standard form E⃗ = E₀ sin(ωt − kx) ĵ:
Angular frequency: ω = 2.5 × 10⁸ rad s⁻¹
Wave number: k = 0.83 rad m⁻¹

Wavelength:
λ = 2π / k
λ = (2 × 3.14) / 0.83
∴ λ ≈ 7.57 m

Frequency:
ν = ω / 2π
ν = (2.5 × 10⁸) / (2 × 3.14)
∴ ν ≈ 3.98 × 10⁷ Hz ≈ 4 × 10⁷ Hz

(iii) The relationship between electric and magnetic field amplitudes in an EM wave is:
E₀ = c B₀
→ B₀ = E₀ / c
→ B₀ = 60 / (3 × 10⁸)
∴ B₀ = 2 × 10⁻⁷ T

The associated magnetic field oscillates along the z-direction (k̂) since E⃗ is along ĵ and propagation is along î (B⃗ = B₀ sin(2.5 × 10⁸ t − 0.83 x) k̂).

(iv) Radio waves (AM, wavelength ~ hundreds of metres) have much longer wavelengths than microwaves (walkie-talkie, λ ~ cm). Longer wavelength EM waves diffract more easily around and through openings in obstacles such as tunnel walls and mountain terrain. Because the wavelength of radio waves is comparable to (or larger than) the dimensions of gaps and irregularities in the tunnel, they bend (diffract) into the tunnel and reach the receiver. Microwaves, having much shorter wavelengths, do not diffract significantly around such large obstacles and are blocked or absorbed by the concrete/rock walls.
Q3Case-based4 marks

A meteorologist uses a weather radar that emits short pulses of microwave radiation to detect raindrops in the atmosphere. The radar operates at a frequency of 3 GHz. Electromagnetic waves are transverse in nature — the oscillating electric field E⃗ and magnetic field B⃗ are mutually perpendicular and both perpendicular to the direction of propagation. The speed of all electromagnetic waves in vacuum (and approximately in air) is c = 3 × 10⁸ m s⁻¹, and the ratio of electric to magnetic field amplitudes equals c: E₀/B₀ = c.

A meteorologist uses a weather radar that emits short pulses of microwave radiation to detect raindrops in the atmosphere. The radar operates at a frequency of 3 GHz. Study the following statements about this radar system and answer the sub-parts:

(i) Which part of the electromagnetic spectrum does the radar radiation belong to? Name ONE other practical application of this part of the spectrum.

(ii) The electric field amplitude of the microwave pulse at a certain point is E₀ = 60 V m⁻¹. Calculate the corresponding magnetic field amplitude B₀ at that point.

(iii) As the microwave pulse travels through air, which of the following correctly describes the relationship between E⃗ and B⃗?
(a) E⃗ and B⃗ are parallel to each other and to the direction of propagation.
(b) E⃗ and B⃗ are perpendicular to each other but both parallel to the direction of propagation.
(c) E⃗ and B⃗ are perpendicular to each other and both perpendicular to the direction of propagation.
(d) E⃗ is perpendicular to the direction of propagation but B⃗ is parallel to it.

(iv) The meteorologist notices that the radar can detect objects only when they are illuminated by radiation whose wavelength is comparable to or smaller than the size of the object. Raindrops have a diameter of about 2 mm. Show whether 3 GHz microwaves are suitable for detecting raindrops. (c = 3 × 10⁸ m s⁻¹)

Show answer
(i) Microwave radiation belongs to the MICROWAVE region of the electromagnetic spectrum (frequency range ~3×10⁸ Hz to 3×10¹¹ Hz).
One other practical application: Microwave ovens (heating food) / satellite communication / RADAR systems for aircraft navigation.
∴ Region: Microwaves; Application: Microwave oven (or satellite communication). [1 mark]

(ii) For an electromagnetic wave, the relationship between electric field amplitude E₀ and magnetic field amplitude B₀ is:

E₀/B₀ = c

where c = 3 × 10⁸ m s⁻¹.

Substituting the given values:
B₀ = E₀/c = 60 / (3 × 10⁸)
B₀ = 20 × 10⁻⁸ T
∴ B₀ = 2 × 10⁻⁷ T [1 mark]

(iii) Option (c) is correct.
Electromagnetic waves are transverse waves in which E⃗ and B⃗ are perpendicular to each other and both are perpendicular to the direction of propagation of the wave. [1 mark]

(iv) Using the relation c = νλ, the wavelength of 3 GHz microwaves is:

λ = c/ν = (3 × 10⁸) / (3 × 10⁹)
λ = 0.1 m = 100 mm

Diameter of a raindrop ≈ 2 mm.

Since λ = 100 mm >> 2 mm (size of raindrop), the wavelength of 3 GHz microwaves is much larger than the raindrop diameter.
∴ 3 GHz microwaves are NOT suitable for detecting raindrops, because effective detection requires λ ≤ size of object. Higher frequency (shorter wavelength) microwaves, such as those in the range of a few GHz to tens of GHz (λ ~ 1–10 mm), would be more appropriate. [1 mark]
Q4Case-based4 marks

Electromagnetic radiation spans a wide spectrum — from radio waves of very low frequency to gamma rays of very high frequency. Each region of the spectrum has distinct properties and specific applications in science, medicine, and technology. Medical science exploits several regions of this spectrum for diagnosis, therapy, and sterilisation.

A hospital uses different types of electromagnetic radiation for various medical purposes. The radiology department uses radiation that can pass through soft tissue to image bones. The physiotherapy department uses radiation that produces heat in deep muscles for pain relief. The sterilisation unit uses radiation to kill bacteria on surgical instruments without heating them.

(i) Identify the type of electromagnetic radiation used in the radiology department to image bones.
(ii) Identify the type of electromagnetic radiation used in physiotherapy to heat deep muscles.
(iii) A physiotherapy machine produces electromagnetic radiation of frequency 2.4 × 10⁹ Hz. Calculate the wavelength of this radiation. (Given: c = 3 × 10⁸ m/s)
(iv) Identify the type of electromagnetic radiation used to sterilise surgical instruments, and state ONE property that makes it suitable for this purpose.

Show answer
(i) The radiology department uses X-rays to image bones.
X-rays have high penetrating power; they pass through soft tissue but are absorbed by denser bone, producing a shadow image on a photographic plate or detector.

(ii) The physiotherapy department uses infrared (IR) radiation to heat deep muscles.
Infrared radiation is absorbed by body tissues and converted to heat, producing a therapeutic warming effect in deep muscle layers.

(iii) By the wave equation for electromagnetic waves:
c = νλ
∴ λ = c / ν
λ = (3 × 10⁸ m/s) / (2.4 × 10⁹ Hz)
∴ λ = 0.125 m = 12.5 cm
This wavelength falls in the microwave region of the electromagnetic spectrum.

(iv) The sterilisation unit uses ultraviolet (UV) radiation to kill bacteria on surgical instruments.
Property: UV radiation has sufficient photon energy to destroy the DNA of micro-organisms, killing bacteria without raising the temperature of the instruments (non-thermal sterilisation).
Q5Case-based4 marks

A team of scientists is designing a multi-purpose satellite system that requires different parts of the electromagnetic spectrum for different functions — broadcasting, thermal imaging, and medical diagnostics. The theoretical basis of all electromagnetic waves rests on Maxwell's concept of displacement current.

A team of scientists is designing a new multi-purpose satellite system. For different functions, they need to use different parts of the electromagnetic spectrum. Read the following requirements and answer the questions:

(i) The satellite must relay television broadcast signals to remote areas. Name the type of electromagnetic radiation best suited for this purpose and state its frequency range.

(ii) The satellite carries an instrument to study the thermal (heat) profile of Earth's surface and oceans. Name the part of the electromagnetic spectrum used, and state one method of its detection.

(iii) Another instrument onboard is designed to detect cancer cells in the human body by imaging dense tissues. Name the electromagnetic radiation used and give one method of its production.

(iv) The displacement current plays a key role in the theoretical foundation of electromagnetic waves. If the electric flux through a surface is changing at a rate of 2×10¹¹ V m s⁻¹, calculate the displacement current through that surface.
(Given: ε₀ = 8.854×10⁻¹² C² N⁻¹ m⁻²)

Show answer
(i) Electromagnetic radiation used: Microwaves (including radio waves/UHF–SHF band).
Frequency range: ~10⁸ Hz to ~10¹¹ Hz (300 MHz to 300 GHz).
Microwaves can penetrate the ionosphere and are used by communication satellites to relay TV signals to dish antennas in remote areas. [1 mark]

(ii) Part of spectrum: Infrared (IR) radiation (also called heat waves).
Detection method: Infrared radiation is detected using a thermopile (or a bolometer / photographic film sensitive to IR). It produces a measurable voltage/temperature change on absorption. [1 mark]

(iii) Electromagnetic radiation used: X-rays.
Method of production: X-rays are produced when high-energy electrons (accelerated through a high potential difference of the order of 10⁴ V) are suddenly decelerated upon striking a heavy metal target (e.g., tungsten) in an X-ray tube — this is called Bremsstrahlung (braking radiation). [1 mark]

(iv) By Maxwell's hypothesis, displacement current is given by:

I_D = ε₀ × (dΦ_E/dt)

Substituting values:
I_D = (8.854×10⁻¹² C² N⁻¹ m⁻²) × (2×10¹¹ V m s⁻¹)
I_D = 8.854 × 2 × 10⁻¹²⁺¹¹ A
I_D = 17.708 × 10⁻¹ A

∴ I_D ≈ 1.77 A [1 mark]
Q6Case-based4 marks

A scientist at a remote weather station simultaneously operates: (1) a radio-wave transmitter at 10 MHz for satellite communication, (2) a microwave oven at 2.45 GHz for heating food in a plastic container, (3) a UV-sterilisation lamp at ~10¹⁵ Hz for disinfecting equipment, and (4) a thermal infrared sensor to detect radiation from a warm rock surface (~10¹³ Hz). All these are part of the electromagnetic spectrum.

A scientist working at a remote weather station notices that the station's communication system uses radio waves of frequency 10 MHz to send data to a satellite. Simultaneously, the scientist uses a microwave oven (operating at 2.45 GHz) to heat food, and a UV-sterilisation lamp (frequency ~10¹⁵ Hz) to disinfect equipment. Later, she detects infrared radiation from a warm rock surface using a thermal sensor.

(i) Arrange the four radiations used — radio waves (10 MHz), microwaves (2.45 GHz), UV radiation, and infrared radiation — in increasing order of their wavelengths in free space.

(ii) The scientist notices that the microwave oven heats food efficiently but does not affect the plastic container. Give the physical reason for this selective heating.

(iii) Calculate the wavelength of the radio waves used for communication. (Speed of light c = 3 × 10⁸ m/s)

(iv) The UV lamp is suddenly replaced by a source of X-rays of the same power. Will X-rays sterilise more effectively than UV? Justify your answer in one sentence based on a physical property of X-rays.

Show answer
(i) Arranging in increasing order of wavelength (λ = c/ν, so higher frequency → smaller wavelength):

Frequencies: UV (~10¹⁵ Hz) > Microwave (2.45 GHz = 2.45×10⁹ Hz) > Radio (10 MHz = 10⁷ Hz)
Infrared ~10¹³ Hz lies between UV and microwave.

Increasing order of frequency: Radio < Microwave < Infrared < UV
∴ Increasing order of wavelength (λ ∝ 1/ν):

UV < Infrared < Microwave < Radio waves [1 mark]

(ii) Microwaves (at 2.45 GHz) have a frequency that resonates with the natural vibrational frequency of water molecules present in food, causing them to vibrate vigorously and generate heat. Plastic containers contain no free water molecules (or polar molecules that resonate at this frequency), so microwaves pass through them without being absorbed. ∴ Only food is heated selectively, not the plastic container. [1 mark]

(iii) By the wave equation:

λ = c/ν

Given: c = 3×10⁸ m/s, ν = 10 MHz = 10×10⁶ Hz = 10⁷ Hz

λ = (3×10⁸) / (10⁷)

∴ λ = 30 m [1 mark]

(iv) Yes, X-rays will sterilise more effectively than UV because X-rays have much higher photon energy (E = hν, and ν_X > ν_UV), enabling them to penetrate deeper into microbial cells and cause more severe ionisation damage to DNA, thereby destroying micro-organisms more efficiently. [1 mark]
Q7Case-based4 marks

A research team monitoring glacial melting in polar regions uses three electromagnetic wave-based instruments: a microwave radar (frequency 10 GHz) to map ice surface topography, an infrared sensor to measure surface temperature, and an ultraviolet detector to study ozone layer thickness. All three instruments rely on distinct regions of the electromagnetic spectrum, yet share the fundamental property that they are transverse waves travelling at speed c = 3 × 10⁸ m/s in vacuum.

A research team is designing a system to monitor glacial melting in polar regions. They use three different electromagnetic wave instruments: (i) a microwave radar to map ice surface topography, (ii) an infrared sensor to measure surface temperature, and (iii) an ultraviolet detector to study ozone layer thickness above the ice.

(a) Arrange microwaves, infrared radiation, and ultraviolet radiation in ascending order of their wavelengths.

(b) The microwave radar emits waves with frequency 10 GHz (10 × 10⁹ Hz). Calculate the wavelength of these microwaves. (Given: c = 3 × 10⁸ m/s)

(c) The infrared sensor detects radiation because all objects above 0 K emit thermal radiation. Name the physical quantity that remains constant when electromagnetic waves travel from vacuum into a medium, and state why.

(d) The UV detector records a sharp increase in UV intensity reaching the ice surface. Give one reason for this and name one harmful biological effect of increased UV exposure.

Show answer
(a) Ascending order of wavelengths (shortest to longest):

Frequency and wavelength are inversely related: λ = c/ν. Higher frequency → shorter wavelength.

Frequency order: UV > IR > Microwaves
∴ Ascending order of wavelength: Ultraviolet < Infrared < Microwaves

[1 mark]

(b) Wavelength of microwave radar:

Formula: λ = c/ν

Substituting values:
λ = (3 × 10⁸ m/s) / (10 × 10⁹ Hz)
λ = (3 × 10⁸) / (10¹⁰)

∴ λ = 3 × 10⁻² m = 0.03 m (3 cm)

[1 mark]

(c) The physical quantity that remains constant when electromagnetic waves travel from vacuum into a medium is FREQUENCY (ν).

Reason: Because frequency depends solely on the source that produces the wave, not on the medium through which it travels. The speed v = c/n and wavelength λ′ = λ/n both decrease in a medium (where n > 1), but the number of oscillations per second (frequency) is fixed by the source.

∴ Frequency remains unchanged; wavelength and speed change.

[1 mark]

(d) Reason for increased UV intensity reaching the ice surface:
Depletion (thinning) of the ozone layer (O₃) in the stratosphere — ozone strongly absorbs UV radiation, so a thinner ozone layer allows greater UV intensity to penetrate to the Earth's surface.

One harmful biological effect of increased UV exposure:
It causes skin cancer (or damage to DNA / causes sunburn / causes snow blindness / suppresses the immune system — any one accepted).

[1 mark]
Q8Case-based4 marks

A hospital uses different types of electromagnetic waves in its various departments. The radiology department uses waves that can pass through soft tissue to image bones. The physiotherapy department uses waves to provide deep heat treatment to muscles. The sterilisation unit uses waves to kill bacteria and sterilise surgical instruments. The emergency ward uses a device that detects heat emitted by the human body to identify fever-affected patients.

A hospital uses different types of electromagnetic waves in its various departments. The radiology department uses waves that can pass through soft tissue to image bones. The physiotherapy department uses waves to provide deep heat treatment to muscles. The sterilisation unit uses waves to kill bacteria and sterilise surgical instruments. The emergency ward uses a device that detects heat emitted by the human body to identify fever-affected patients.

(i) Identify the electromagnetic wave used in the radiology department and state its frequency range.
(ii) Identify the electromagnetic wave used in the physiotherapy department and state its frequency range.
(iii) Identify the electromagnetic wave used in the sterilisation unit and state its frequency range.
(iv) Identify the electromagnetic wave detected in the emergency ward and state its frequency range.

Show answer
(i) Radiology department — X-rays.
X-rays have high penetrating power through soft tissue but are absorbed by denser bone, making them suitable for bone imaging.
Frequency range: 3×10¹⁷ Hz to 3×10¹⁹ Hz (approximately 10¹⁸ Hz order).
∴ Electromagnetic wave used = X-rays; frequency ≈ 10¹⁷ Hz to 10¹⁹ Hz.
[1 mark]

(ii) Physiotherapy department — Infrared (IR) radiation.
Infrared waves are absorbed by body tissues and converted to heat energy, providing deep heat treatment to muscles and joints.
Frequency range: 3×10¹¹ Hz to 4×10¹⁴ Hz (approximately 10¹¹ Hz to 10¹⁴ Hz).
∴ Electromagnetic wave used = Infrared radiation; frequency ≈ 10¹¹ Hz to 10¹⁴ Hz.
[1 mark]

(iii) Sterilisation unit — Ultraviolet (UV) radiation.
UV radiation has sufficient photon energy to break molecular bonds in bacterial DNA, destroying micro-organisms and sterilising surgical instruments.
Frequency range: 8×10¹⁴ Hz to 3×10¹⁷ Hz (approximately 10¹⁵ Hz to 10¹⁷ Hz).
∴ Electromagnetic wave used = Ultraviolet radiation; frequency ≈ 10¹⁵ Hz to 10¹⁷ Hz.
[1 mark]

(iv) Emergency ward (fever detection) — Infrared (IR) radiation.
All bodies above absolute zero emit thermal radiation. A fever-affected patient emits more infrared radiation than a healthy person; thermal imaging cameras detect this IR emission to map body temperature.
Frequency range: 3×10¹¹ Hz to 4×10¹⁴ Hz (approximately 10¹¹ Hz to 10¹⁴ Hz).
∴ Electromagnetic wave detected = Infrared radiation; frequency ≈ 10¹¹ Hz to 10¹⁴ Hz.
[1 mark]
Q9MCQ1 mark

Which of the following electromagnetic waves has the highest frequency?

Show answer
Option (d) is correct.

Explanation: In the electromagnetic spectrum, frequency increases in the order: radio waves → microwaves → infrared → visible → ultraviolet → X-rays → gamma rays. Gamma rays have the highest frequency (≈ 10¹⁸ Hz to 10²² Hz) and correspondingly the shortest wavelength.
Q10MCQ1 mark

The electromagnetic wave used for cooking food in a microwave oven belongs to which part of the electromagnetic spectrum?

Show answer
Option (B) is correct.

Explanation: Microwaves have wavelengths in the range of 10⁻³ m to 10⁻¹ m. Their frequency (~2.45 GHz) matches the natural resonance frequency of water molecules in food, causing them to vibrate and generate heat — this is the principle used in microwave ovens.
Q11MCQ1 mark

Microwaves are used in RADAR systems. The frequency range of microwaves is:

Show answer
Option (a) is correct.

Explanation: Microwaves lie in the frequency range 10⁸ Hz to 10¹¹ Hz (wavelength range approximately 1 mm to 0.1 m). They are produced by special vacuum tubes (klystrons, magnetrons) and are used in RADAR systems for detecting the position and speed of distant objects. Option (b) corresponds to infrared waves, option (c) to radio waves, and option (d) to ultraviolet waves.
Q12MCQ1 mark

The electromagnetic wave propagating along the z-axis has its electric field oscillating along the x-axis. Along which direction does its magnetic field oscillate?

Show answer
Option (C) is correct.

Explanation: In an electromagnetic wave, E⃗, B⃗, and the direction of propagation are mutually perpendicular. Since the wave propagates along the z-axis and E⃗ oscillates along the x-axis, by the right-hand rule (E⃗ × B⃗ must point along the direction of propagation, i.e., z-axis), B⃗ must oscillate along the y-axis.
Q13Short Answer2 marks

Name the physical quantity that is analogous to conduction current in a capacitor during charging. Write the expression for it in terms of the rate of change of electric flux.

Show answer
The physical quantity analogous to conduction current in a capacitor during charging is the displacement current.

Maxwell showed that a changing electric field between the capacitor plates produces a current equivalent in its magnetic effect to a conduction current. This is called displacement current.

By Maxwell's modification of Ampere's circuital law, the displacement current is given by:

I_D = ε₀ × dΦ_E/dt

where ε₀ = 8.854 × 10⁻¹² C² N⁻¹ m⁻² is the permittivity of free space and dΦ_E/dt is the rate of change of electric flux through the surface.

∴ Displacement current I_D = ε₀ (dΦ_E/dt), measured in ampere (A).
Q14Short Answer2 marks

State two properties of electromagnetic waves. Why are X-rays used in medical diagnostics to detect bone fractures?

Show answer
Two properties of electromagnetic waves: (i) Electromagnetic waves are transverse in nature — the electric field vector E⃗ and magnetic field vector B⃗ oscillate perpendicular to each other and also perpendicular to the direction of propagation. (ii) Electromagnetic waves do not require a material medium for propagation; they can travel through vacuum with speed c = 3×10⁸ m/s. Reason for use of X-rays in medical diagnostics: X-rays have very high frequency (≈ 10¹⁸ Hz) and correspondingly very short wavelength (≈ 10⁻¹⁰ m), giving them high penetrating power. They pass easily through soft body tissues but are absorbed by denser materials such as bone. ∴ X-rays produce shadow images on a photographic plate that clearly reveal bone fractures.
Q15Short Answer2 marks

State two properties of electromagnetic waves. Show that the ratio of the amplitudes of the electric and magnetic fields in an electromagnetic wave equals the speed of light in free space.

Show answer
Properties of electromagnetic waves (any two):
(i) They are transverse in nature — E⃗ and B⃗ are mutually perpendicular and both perpendicular to the direction of propagation.
(ii) They travel through free space with speed c = 3×10⁸ m/s, independent of the frequency of the source.

Ratio E₀/B₀:
For a plane electromagnetic wave travelling along the x-direction, the electric and magnetic fields are:
E⃗ = E₀ sin(kx − ωt) ĵ
B⃗ = B₀ sin(kx − ωt) k̂

From Maxwell's equations applied to a plane EM wave:
∂E/∂x = −∂B/∂t
→ E₀ k cos(kx − ωt) = B₀ ω cos(kx − ωt)
→ E₀/B₀ = ω/k

Since the phase speed of the wave is c = ω/k:
∴ E₀/B₀ = c = 3×10⁸ m s⁻¹
Q16Short Answer2 marks

The oscillating electric field of an electromagnetic wave in vacuum is given by E⃗ = 50 sin(2π × 10¹⁰ t) ĵ V/m. (a) Identify the type of electromagnetic wave. (b) Find the wavelength of this wave.

Show answer
(a) Comparing with the standard form E = E₀ sin(ωt), the angular frequency is:
ω = 2π × 10¹⁰ rad/s

∴ frequency ν = ω/2π = 10¹⁰ Hz

Since ν ~ 10¹⁰ Hz lies in the microwave region of the electromagnetic spectrum, this is a microwave.

(b) The speed of an electromagnetic wave in vacuum is c = 3 × 10⁸ m/s.

Using c = νλ:

λ = c/ν = (3 × 10⁸)/(10¹⁰)

∴ λ = 3 × 10⁻² m = 3 cm
Q17Short Answer3 marks

A scientist working at a remote Antarctic research station uses a satellite communication system. The system transmits signals using electromagnetic waves of frequency 12 GHz. The electric field amplitude of the wave at the receiving antenna is measured to be 0.60 V/m.

(i) Identify the type of electromagnetic wave used. Give ONE practical reason why this type of wave is preferred for satellite communication over radio waves of frequency 10 MHz.

(ii) Calculate the amplitude of the magnetic field (B₀) associated with this wave in free space.

(iii) The wave enters a medium of refractive index n = 1.5. What is the speed of the wave and the new ratio E₀/B₀ in this medium?

(iv) The displacement current concept is essential to justify the propagation of electromagnetic waves through free space (vacuum between satellite and Earth). State the expression for displacement current (I_d) and explain in one sentence how it resolves the inconsistency in Ampere's Circuital Law.

Show answer
(i) The wave of frequency 12 GHz belongs to the microwave region of the electromagnetic spectrum.

Practical reason: Microwaves (12 GHz) are not reflected by the ionosphere and pass through it to reach satellites in orbit, whereas radio waves of 10 MHz are reflected back by the ionosphere and cannot be used for satellite (space) communication.

(1 mark: correct identification + valid reason)

――――――――――――――――――――――――――――――
(ii) In free space, the ratio of electric field amplitude to magnetic field amplitude equals the speed of light:

E₀/B₀ = c

∴ B₀ = E₀/c

Substituting values:

B₀ = 0.60 / (3×10⁸)

∴ B₀ = 2×10⁻⁹ T (= 2 nT)

(1 mark: correct formula + substitution + ∴ answer with unit)

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(iii) The speed of light in a medium of refractive index n is given by:

v = c/n

Substituting:

v = (3×10⁸) / 1.5

∴ v = 2×10⁸ m/s

The ratio E₀/B₀ in any medium equals the speed of the wave in that medium:

E₀/B₀ = v = 2×10⁸ m/s

(Note: Both E₀ and B₀ change on entering the medium, but their ratio remains equal to the wave speed in that medium.)

(1 mark: correct speed + correct new ratio with unit)

――――――――――――――――――――――――――――――
(iv) Maxwell's displacement current is defined as:

I_d = ε₀ · dΦ_E/dt

where Φ_E is the electric flux through the surface considered.

Resolution of inconsistency: The original Ampere's Circuital Law (∮B⃗·dL⃗ = μ₀ I_c) gives different values of the magnetic field depending on which surface is chosen at a capacitor gap (where conduction current I_c = 0 but electric flux is changing); adding the displacement current term I_d = ε₀ dΦ_E/dt makes the total current (I_c + I_d) consistent across ALL surfaces, thereby removing the inconsistency.

(1 mark: correct expression for I_d + correct one-sentence explanation)
Q18Short Answer3 marks

A team of scientists is designing a deep-space communication system. They propose using three different electromagnetic radiations — microwaves (λ ≈ 1 cm), infrared radiation (λ ≈ 10 μm), and ultraviolet radiation (λ ≈ 300 nm) — for different purposes aboard a spacecraft.

(i) The scientists find that microwaves are most suitable for long-range communication with Earth. State the physical property of microwaves that makes them travel through the Earth's ionosphere without being reflected back, unlike AM radio waves of longer wavelengths.

(ii) Infrared sensors are used on the spacecraft to detect the presence of nearby asteroids even in complete darkness. Identify the source of infrared radiation from asteroids and state the principle behind this detection.

(iii) The UV radiation detector on the spacecraft records a UV flux from a distant star. If the photon energy of this UV radiation is 6.63 eV, calculate the frequency and wavelength of this radiation. (Given: h = 6.63 × 10⁻³⁴ J s, c = 3 × 10⁸ m/s, 1 eV = 1.6 × 10⁻¹⁹ J)

(iv) The scientists debate whether to use microwaves or infrared radiation for internal communication within the spacecraft cabin. Give ONE scientific reason why infrared radiation (used in TV remotes) is more suitable than microwaves for short-range indoor communication.

Show answer
(i) The frequency of microwaves (~10 GHz) is higher than the plasma frequency of the ionosphere. As a result, microwaves are not reflected by the ionospheric layer and pass straight through it, enabling direct satellite/deep-space communication. AM radio waves (lower frequency) are reflected by the ionosphere and cannot penetrate it to reach outer space.

(Marking: 1 mark for stating frequency of microwaves > ionospheric plasma frequency / microwaves penetrate the ionosphere, with correct reasoning.)

──────────────────────────────────

(ii) Every body at a temperature above absolute zero emits thermal (heat) radiation. Asteroids, even in darkness, have a finite temperature and therefore emit infrared radiation as thermal emission (blackbody radiation). The infrared sensors detect this thermally radiated infrared energy from the asteroid, allowing its presence to be identified without any visible light source.

(Marking: 1 mark for identifying thermal emission / blackbody radiation at finite temperature as the source, and linking it to infrared detection.)

──────────────────────────────────

(iii) By the photon energy relation:

E = hν

Given E = 6.63 eV

→ E (in joules) = 6.63 × 1.6 × 10⁻¹⁹ J = 10.608 × 10⁻¹⁹ J ≈ 1.06 × 10⁻¹⁸ J

Frequency:
ν = E / h = (1.06 × 10⁻¹⁸) / (6.63 × 10⁻³⁴)

→ ν = (1.06 / 6.63) × 10⁻¹⁸⁺³⁴ Hz

→ ν ≈ 0.1599 × 10¹⁶ Hz

∴ ν ≈ 1.6 × 10¹⁵ Hz

Wavelength:
Using c = νλ → λ = c / ν

λ = (3 × 10⁸) / (1.6 × 10¹⁵)

∴ λ = 1.875 × 10⁻⁷ m ≈ 187.5 nm

(This lies in the UV region of the electromagnetic spectrum, consistent with the given scenario.)

(Marking: ½ mark for correct formula and conversion of eV to J; ½ mark for ∴ ν = 1.6 × 10¹⁵ Hz with unit; ½ mark for λ = c/ν substitution; ½ mark for ∴ λ ≈ 1.875 × 10⁻⁷ m with unit — total 1 mark for complete numerical with both values correct.)

──────────────────────────────────

(iv) Infrared radiation is more suitable for short-range indoor communication because infrared photons are absorbed or blocked by walls and other opaque surfaces and do not penetrate them. This confines the signal within the cabin/room, preventing interference with adjacent systems. Microwaves, having much longer wavelengths, can penetrate walls and other cabin structures, causing unwanted interference with sensitive onboard electronic instruments and communication equipment.

(Marking: 1 mark for the correct scientific reason — infrared is confined/absorbed by walls / does not penetrate cabin structures, avoiding interference with onboard electronics.)
Q19Short Answer3 marks

Displacement current was introduced by Maxwell to remove an inconsistency in Ampere's Circuital Law. (a) State what inconsistency existed in the original Ampere's Circuital Law. (b) Write the expression for displacement current in terms of electric flux. (c) What are the dimensions of ε₀(dΦ_E/dt)?

Show answer
(a) Inconsistency in Ampere's Circuital Law:

The original Ampere's Circuital Law states: ∮ B⃗ · dL⃗ = μ₀ I_enc.

This law gives inconsistent (ambiguous) results when applied to a surface bounded by the same Amperian loop but passing through the gap between the plates of a charging capacitor. For a flat surface cutting the wire, the enclosed current is I, but for a surface passing through the capacitor gap, the enclosed conduction current is zero — yet both surfaces share the same boundary. This violates the requirement that the value of ∮ B⃗ · dL⃗ must be the same for all surfaces sharing a given loop. Thus the original law was incomplete.

(b) Expression for displacement current:

Maxwell introduced displacement current I_D to resolve this inconsistency. The displacement current is defined as:

I_D = ε₀ (dΦ_E / dt)

where Φ_E is the electric flux through the surface, and ε₀ is the permittivity of free space.

The modified (Ampere–Maxwell) law becomes:
∮ B⃗ · dL⃗ = μ₀ (I + I_D) = μ₀ I + μ₀ ε₀ (dΦ_E / dt)

(c) Dimensions of ε₀ (dΦ_E / dt):

Since I_D = ε₀ (dΦ_E / dt) is a current, its dimensions are those of electric current:

∴ [ε₀ (dΦ_E / dt)] = [I_D] = [A] = [M⁰ L⁰ T⁰ A¹]

OR equivalently: SI unit of ε₀ (dΦ_E / dt) is Ampere (A).
Q20Short Answer3 marks

A scientist is designing a remote sensing satellite that uses electromagnetic waves to study Earth's surface through cloud cover. The satellite transmits a wave whose electric field component (in SI units) is given by:

E⃗ = (6.0 V/m) sin(βy − 3.0 × 10¹⁰ t) x̂

where y is in metres and t is in seconds.

(i) Identify the type of electromagnetic wave being used and state ONE practical application of this type, other than remote sensing.
(ii) Determine the value of the propagation constant β (in rad/m).
(iii) Find the amplitude of the associated magnetic field component, and write a complete expression for B⃗.
(iv) Calculate the wavelength of this wave and identify in which part of the electromagnetic spectrum it belongs.

Show answer
(i) Identification and Application

The angular frequency is ω = 3.0 × 10¹⁰ rad/s.

Frequency: ν = ω/2π = (3.0 × 10¹⁰)/(2π) ≈ 4.8 × 10⁹ Hz ≈ 4.8 GHz.

This falls in the microwave region of the electromagnetic spectrum.

Application: Microwave ovens (cooking food); OR RADAR systems; OR satellite communication links.

∴ The wave is a microwave. One application: microwave ovens / RADAR. [1 mark]

(ii) Value of propagation constant β

For an electromagnetic wave travelling in free space:

c = ω/β

∴ β = ω/c = (3.0 × 10¹⁰)/(3 × 10⁸)

∴ β = 100 rad/m [1 mark]

(iii) Amplitude of B⃗ and complete expression

For an EM wave, the relationship between electric and magnetic field amplitudes is:

E₀ = c × B₀

∴ B₀ = E₀/c = 6.0/(3 × 10⁸)

∴ B₀ = 2.0 × 10⁻⁸ T

Direction: Since E⃗ is along x̂ and the wave propagates along +ŷ, by the right-hand rule (ŷ = x̂ × ẑ → E⃗ × B⃗ must point in +ŷ), B⃗ is along ẑ.

∴ B⃗ = (2.0 × 10⁻⁸ T) sin(100y − 3.0 × 10¹⁰ t) ẑ [1 mark]

(iv) Wavelength and spectral region

Using the relation between propagation constant and wavelength:

β = 2π/λ

∴ λ = 2π/β = 2π/100

∴ λ = 6.28 × 10⁻² m ≈ 6.3 cm

Since λ ≈ 6.3 cm lies in the range 1 mm to 1 m, this wave belongs to the microwave region of the electromagnetic spectrum (consistent with part (i)).

∴ λ ≈ 6.3 × 10⁻² m; Spectral region: Microwave. [1 mark]
Q21Short Answer3 marks

A research satellite continuously monitors Earth's surface. Its onboard sensor array detects four different types of electromagnetic radiation simultaneously:
• Radiation P: Used to detect heat signatures of forest fires through cloud cover; wavelength ≈ 10⁻⁵ m
• Radiation Q: Used for long-distance radio communication by reflecting off the ionosphere; wavelength ≈ 10² m
• Radiation R: Emitted by radioactive decay products on Earth's surface; wavelength ≈ 10⁻¹² m
• Radiation S: Used in the satellite's radar imaging system to penetrate clouds; wavelength ≈ 10⁻² m

(i) Identify each radiation P, Q, R and S by name.
(ii) Arrange P, Q, R and S in increasing order of their frequencies.
(iii) The satellite's communication system uses radiation Q. The electric field of this wave at a given instant is directed vertically upward and the wave travels due East. Using the relation between E⃗, B⃗ and the direction of propagation, determine the direction of the magnetic field B⃗ at that instant.
(iv) The onboard radar (radiation S) has its electric field varying as E = E₀ sin(kx − ωt). Write the expression for the corresponding magnetic field B, and calculate the ratio E₀/B₀, given that the wave travels in free space.

Show answer
(i) Identification of each radiation:

By matching wavelengths to the electromagnetic spectrum:

• Radiation P (λ ≈ 10⁻⁵ m): Infrared (IR) waves
• Radiation Q (λ ≈ 10² m): Radio waves
• Radiation R (λ ≈ 10⁻¹² m): Gamma rays
• Radiation S (λ ≈ 10⁻² m): Microwaves

(1 mark for all four correct; partial: ½ mark for any two correct)

(ii) Increasing order of frequency:

Since frequency ν = c/λ, smaller wavelength → higher frequency.

Arranging wavelengths in decreasing order: λ_Q (10² m) > λ_S (10⁻² m) > λ_P (10⁻⁵ m) > λ_R (10⁻¹² m)

∴ Increasing order of frequency: Q < S < P < R

(i.e., Radio waves < Microwaves < Infrared < Gamma rays)

(1 mark)

(iii) Direction of magnetic field B⃗:

In an electromagnetic wave, E⃗, B⃗, and the direction of propagation (k̂) are mutually perpendicular, related by:

k̂ = (E⃗ × B⃗) / |E⃗ × B⃗|

Given:
• Direction of propagation k̂ = East (let this be +x̂)
• E⃗ = vertically upward (+ẑ)

Using k̂ = Ê × B̂:

+x̂ = +ẑ × B̂

→ B̂ = x̂ × (−ẑ) ... rearranging: since ẑ × ŷ = −x̂, we use the right-hand rule:

ẑ × (−ŷ) = +x̂ ✓ ... checking: ẑ × ŷ = −x̂, so ẑ × (−ŷ) = x̂

∴ B̂ = −ŷ = directed toward South (horizontally, in the plane of Earth's surface)

∴ The magnetic field B⃗ is directed toward the South.

(1 mark)

(iv) Expression for B and ratio E₀/B₀:

Since B⃗ is perpendicular to E⃗ and both are transverse to the direction of propagation, and the magnitudes are related by:

E₀/B₀ = c (speed of light in free space)

If E = E₀ sin(kx − ωt), then:

B = B₀ sin(kx − ωt)

where B₀ = E₀/c

Using c = 1/√(μ₀ε₀) = 3 × 10⁸ m/s:

∴ E₀/B₀ = c = 3 × 10⁸ m s⁻¹

(1 mark)
Q22Short Answer3 marks

A hospital uses three different types of electromagnetic radiation in its day-to-day operations: (i) a microwave oven in the staff canteen heats food, (ii) an X-ray machine images bone fractures, and (iii) a remote-controlled physiotherapy device uses infrared radiation to relieve muscle pain.

(a) Arrange these three radiations in increasing order of their frequencies. (1 mark)
(b) The X-ray machine operates by accelerating electrons through a potential difference of 25 kV. Estimate the minimum wavelength of the X-rays produced. (2 marks)
(c) The physiotherapy infrared remote works even when there is no line-of-sight between the emitter and the patient (e.g., radiation bounces off a wall). Identify the property of EM waves that makes this possible, and state ONE other property that is common to ALL electromagnetic waves. (1 mark)

Show answer
(a) Arranging in increasing order of frequency:

Microwaves have the lowest frequency, infrared (IR) is next, and X-rays have the highest frequency.

∴ Increasing order of frequency: Microwaves < Infrared < X-rays

(b) Finding the minimum wavelength of X-rays:

The minimum wavelength corresponds to maximum photon energy, which equals the kinetic energy gained by the electron:

Formula: hc / λ<sub>min</sub> = eV

∴ λ<sub>min</sub> = hc / eV

Substituting values:

h = 6.63 × 10<sup>−34</sup> J s, c = 3 × 10<sup>8</sup> m s<sup>−1</sup>, e = 1.6 × 10<sup>−19</sup> C, V = 25 × 10<sup>3</sup> V

λ<sub>min</sub> = (6.63 × 10<sup>−34</sup> × 3 × 10<sup>8</sup>) / (1.6 × 10<sup>−19</sup> × 25 × 10<sup>3</sup>)

λ<sub>min</sub> = (19.89 × 10<sup>−26</sup>) / (4.0 × 10<sup>−15</sup>)

∴ λ<sub>min</sub> ≈ 0.497 × 10<sup>−10</sup> m ≈ 0.5 × 10<sup>−10</sup> m (≈ 0.5 Å)

(c) Property enabling reflection off walls:

Electromagnetic waves undergo reflection (and also refraction) — they obey the laws of reflection just like light, which allows infrared radiation to bounce off surfaces and reach the patient without a direct line-of-sight.

One property common to ALL electromagnetic waves: They are transverse waves — the oscillating electric field E⃗ and magnetic field B⃗ are perpendicular to each other and to the direction of propagation; they all travel through free space at the speed c = 3 × 10<sup>8</sup> m s<sup>−1</sup>.

(Examiner note: Accept any ONE correct common property, e.g., they do not require a material medium for propagation / they travel at c = 3 × 10<sup>8</sup> m s<sup>−1</sup> in vacuum / they carry energy and momentum.)
Q23Short Answer3 marks

A science journalist is writing an article on how different types of electromagnetic (EM) waves are used in everyday life and technology. She comes across the following four applications:

(P) Microwave ovens used for cooking food.
(Q) X-ray machines used for imaging bones.
(R) TV remote controls using signals to change channels.
(S) Radar systems used to detect aircraft.

Based on your understanding of the electromagnetic spectrum, answer the following:

(i) Identify the specific type of EM wave used in application (P) and state the physical principle by which it heats food.

(ii) Among applications (Q) and (R), which EM wave has a higher frequency? Justify your answer using the relationship between frequency and wavelength.

(iii) Application (S) uses the same type of EM wave as application (P). Give ONE reason why microwaves — and not radio waves — are preferred in radar systems for aircraft detection.

(iv) The journalist states: "All four applications use transverse waves that travel at the same speed in vacuum." Verify this statement by giving the speed of EM waves in vacuum and the mathematical expression that establishes it.

Show answer
(i) Application (P) uses microwaves (frequency range ~ 10⁹ Hz to 10¹¹ Hz).

Physical principle: Microwaves have a frequency that matches the natural resonant frequency of water molecules in food. The water molecules absorb microwave energy and undergo vigorous rotational/vibrational oscillations, generating heat through molecular friction. This heats the food uniformly from within.

(ii) By the relation c = νλ (where c is constant in vacuum), frequency ν = c/λ. A shorter wavelength corresponds to a higher frequency.

X-rays (application Q) have wavelengths in the range 10⁻¹¹ m to 10⁻⁸ m, whereas infrared/IR waves used in TV remote controls (application R) have wavelengths in the range 10⁻³ m to 7×10⁻⁷ m.

Since λ_X-ray ≪ λ_IR,
∴ ν_X-ray ≫ ν_IR.

The EM wave in application (Q) — X-rays — has the higher frequency.

(iii) Microwaves are preferred over radio waves in radar systems because microwaves have a much shorter wavelength (~ a few cm), comparable to the size of aircraft. By the principle of diffraction, waves reflect well (rather than diffract around) objects whose size is comparable to or larger than the wavelength. Radio waves (λ ~ metres to km) would diffract around aircraft and not reflect effectively, making detection unreliable. Microwaves thus give a well-defined reflected signal, enabling accurate detection and ranging.

(iv) The journalist's statement is correct.

All EM waves are transverse in nature — the electric field E⃗ and magnetic field B⃗ oscillate perpendicular to each other and perpendicular to the direction of propagation.

The speed of all EM waves in vacuum is:

c = 1/√(μ₀ε₀)

where μ₀ = 4π × 10⁻⁷ T m A⁻¹ (permeability of free space) and ε₀ = 8.854 × 10⁻¹² C² N⁻¹ m⁻² (permittivity of free space).

Substituting:
c = 1/√(4π × 10⁻⁷ × 8.854 × 10⁻¹²)

∴ c = 3 × 10⁸ m/s

This value is the same for all EM waves — microwaves, X-rays, infrared, and microwaves used in radar — confirming the journalist's statement.
Q24Short Answer3 marks

A research team is designing a wireless communication system for a remote mountain village. They consider three options:
• Option P: Radio waves of frequency 5 × 10⁶ Hz (AM broadcast band)
• Option Q: Microwaves of frequency 1.0 × 10¹⁰ Hz (satellite link)
• Option R: Infrared radiation of frequency 3 × 10¹³ Hz (line-of-sight remote link)

Based on your understanding of the electromagnetic spectrum, answer the following:
(i) Calculate the wavelength in free space of the radiation used in Option Q. (Speed of light c = 3 × 10⁸ m/s)
(ii) The team observes that Option P signals bend around the mountain peaks and reach the village, but Option R signals are completely blocked. Name the wave property responsible for bending in Option P and explain why Option R does not show the same effect.
(iii) Displacement current plays a key role in the propagation of all three radiations. State what displacement current is and write the expression for it.
(iv) The team finally chooses Option Q (microwaves) for long-distance links via a satellite. Give ONE advantage of microwaves over AM radio waves for satellite communication.

Show answer
(i) Wavelength of microwave (Option Q):

By the relation: λ = c / ν

λ = (3 × 10⁸ m/s) / (1.0 × 10¹⁰ Hz)

∴ λ = 3 × 10⁻² m = 3 cm

(ii) The wave property responsible for bending of Option P (radio waves, λ ≈ 60 m) around mountain peaks is DIFFRACTION.

Diffraction is significant only when the wavelength of the wave is comparable to (or larger than) the size of the obstacle or aperture. Mountain peaks have dimensions of the order of tens to hundreds of metres — comparable to the wavelength of AM radio waves (~60 m). Hence, radio waves diffract appreciably around the peaks and reach the village.

Option R uses infrared radiation with a much smaller wavelength (λ = c/ν = 3×10⁸ / 3×10¹³ ≈ 10⁻⁵ m = 10 μm). This wavelength is far too small compared to the mountain dimensions, so diffraction is negligible and the infrared beam is blocked (shadowed) by the peaks.

(iii) Displacement current:

Displacement current is defined as the current that arises due to a time-varying electric field (or a changing electric flux) in a region, even in the absence of any physical charge flow.

Its expression is:

I_D = ε₀ × (dΦ_E / dt)

where ε₀ = 8.854 × 10⁻¹² C² N⁻¹ m⁻² is the permittivity of free space and dΦ_E/dt is the rate of change of electric flux.

(This concept, introduced by Maxwell, completed Ampere's circuital law and led to the prediction of electromagnetic waves.)

(iv) ONE advantage of microwaves (Option Q) over AM radio waves for satellite communication:

Microwaves have a much higher frequency (shorter wavelength) and therefore carry a higher bandwidth, allowing transmission of large volumes of data (e.g., television, internet, telephony) simultaneously. Additionally, microwaves travel in straight lines (line-of-sight) and pass through the ionosphere without reflection, making them suitable for satellite links — unlike AM radio waves, which are reflected by the ionosphere and cannot reach satellites.
Q25Short Answer3 marks

Name the electromagnetic radiations used in each of the following applications:
(i) Treatment of cancer tumours
(ii) Remote sensing of the Earth's surface through clouds
(iii) Sterilisation of surgical instruments
For each, also state the approximate wavelength range.

Show answer
(i) Treatment of cancer tumours — Gamma rays (γ-rays)
Gamma rays are high-energy electromagnetic radiations emitted by radioactive nuclei. They penetrate deep into tissues and destroy malignant cells.
Wavelength range: λ < 10⁻¹² m (i.e., less than 0.001 nm).

(ii) Remote sensing of Earth's surface through clouds — Microwaves
Microwaves can penetrate clouds, fog, and rain, making them ideal for radar-based remote sensing and satellite imaging.
Wavelength range: 10⁻³ m to 0.1 m (i.e., 1 mm to 10 cm).

(iii) Sterilisation of surgical instruments — Ultraviolet (UV) rays
UV radiation has sufficient energy to kill bacteria and other micro-organisms, and is therefore used to sterilise medical equipment and operating theatres.
Wavelength range: 10⁻⁸ m to 4×10⁻⁷ m (i.e., 10 nm to 400 nm).

[Award 1 mark for each correct (name + wavelength range) pair — 3 marks total.]
Q26Short Answer3 marks

A physiotherapist uses a heat-lamp in her clinic to warm deep muscle tissue in patients. Separately, her colleague in the radiology department uses a different device to image a patient's fractured bone. A third colleague uses yet another device that can sterilise surgical instruments by killing bacteria and viruses.

(i) Identify the part of the electromagnetic spectrum used in each of the three devices.
(ii) For the radiation used in the heat-lamp, state ONE method of production and ONE method of detection.
(iii) Which of the three radiations has the highest frequency? Justify your answer using the electromagnetic spectrum.

Diagram for question 26: Electromagnetic Waves
Show answer
(i) Identification of radiations used in each device:

• Heat-lamp (warming deep muscle tissue): Infrared (IR) radiation — also called heat waves.
• Radiology device (imaging fractured bone): X-rays.
• Sterilisation device (killing bacteria/viruses): Ultraviolet (UV) radiation.

[1 mark — all three correctly identified]

(ii) Production and detection of Infrared radiation:

Production: Infrared radiation is produced by hot bodies and molecules undergoing vibrational transitions. All warm objects (including the human body and incandescent bulbs / resistor heating coils) emit IR radiation.

Detection: Infrared radiation is detected by thermopiles, bolometers, or infrared-sensitive photographic film (also: IR photodetectors / snake-venom pit organs — any one valid detector accepted).

[1 mark — one valid method of production AND one valid method of detection stated]

(iii) Highest frequency among the three radiations:

The order of increasing frequency in the electromagnetic spectrum is:
Infrared (IR) < Ultraviolet (UV) < X-rays

∴ X-rays have the highest frequency among the three.

Justification: The electromagnetic spectrum is arranged in order of increasing frequency (or equivalently decreasing wavelength): Radio → Microwave → IR → Visible → UV → X-ray → Gamma. X-rays lie beyond UV toward the high-frequency (short-wavelength) end, so among IR (~10¹³ Hz), UV (~10¹⁵–10¹⁶ Hz), and X-rays (~10¹⁶–10¹⁹ Hz), X-rays possess the highest frequency.

[1 mark — correct identification of X-rays with valid justification referencing the spectrum order]

[Bonus mark / 4th mark note for examiner: Award the 4th mark for any additional correct supporting point, such as: X-rays have shortest wavelength (λ ~ 10⁻¹⁰ m to 10⁻⁸ m) of the three, consistent with highest frequency via c = νλ, or correctly stating the approximate frequency ranges for all three radiations.]

∴ Summary:
(i) Heat-lamp → Infrared; Bone imaging → X-rays; Sterilisation → UV.
(ii) IR produced by hot bodies / vibrational transitions; detected by thermopiles / bolometers.
(iii) X-rays have the highest frequency; their position beyond UV in the EM spectrum toward gamma rays confirms this.
Q27Short Answer3 marks

A hospital uses different types of electromagnetic radiation for various medical purposes. Match the following applications with the correct type of electromagnetic radiation and answer the questions that follow:

(i) A radiologist uses radiation that can pass through soft tissue and create images of bones.
(ii) A physiotherapy unit uses radiation to provide heat treatment to deep muscle tissue.
(iii) The sterilisation unit uses radiation to kill bacteria and sterilise surgical instruments.
(iv) A dermatologist warns patients about radiation from the Sun that causes skin tanning and can lead to skin cancer.

Identify the type of electromagnetic radiation used in each case (i) to (iv) and state ONE distinguishing property of each identified radiation.

Show answer
All electromagnetic waves travel at speed c = 3×10⁸ m/s in vacuum and are transverse in nature; they differ in frequency (wavelength), which determines their penetrating power and biological effects.

(i) Bone imaging in hospitals:
The radiation used is X-rays.
Distinguishing property: X-rays have very short wavelength (0.001 nm to 10 nm) and high penetrating power — they pass through soft tissue but are absorbed by denser bone, forming shadow images on a photographic plate.

(ii) Deep muscle heat treatment (physiotherapy):
The radiation used is Infrared (IR) radiation.
Distinguishing property: Infrared radiation is readily absorbed by body tissues and converted into heat (thermal energy), making it effective for heating deeper muscle layers without visible light.

(iii) Sterilisation of surgical instruments:
The radiation used is Ultraviolet (UV) radiation.
Distinguishing property: UV radiation (wavelength ≈ 10 nm to 400 nm) carries sufficient energy to destroy the DNA of micro-organisms, thus killing bacteria and sterilising equipment.

(iv) Skin tanning / skin cancer from sunlight:
The radiation responsible is Ultraviolet (UV) radiation (from the Sun).
Distinguishing property: UV radiation from the Sun is partially absorbed by the ozone layer; the portion that reaches Earth has enough energy to damage skin cells — causing tanning, premature ageing, and in excessive exposure, skin cancer.

∴ Summary:
• Case (i) → X-rays: high penetration through soft tissue.
• Case (ii) → Infrared: absorbed as heat by tissue.
• Case (iii) → Ultraviolet: destroys microbial DNA (germicidal).
• Case (iv) → Ultraviolet: damages skin cells; causes tanning and cancer.

[Marking: 1 mark for each correct identification with a valid distinguishing property — total 4 marks.]
Q28Short Answer3 marks

A plane electromagnetic wave travels in vacuum along the z-direction.

(i) Write the expressions for its oscillating electric and magnetic fields, clearly indicating their directions.

(ii) State any two properties of electromagnetic waves.

(iii) Identify the part of the electromagnetic spectrum to which a wave of frequency 5 × 10¹⁴ Hz belongs. Name one method of its detection.

Show answer
(i) For a plane electromagnetic wave propagating along the z-direction, the electric field E⃗ oscillates along the x-direction and the magnetic field B⃗ oscillates along the y-direction (since E⃗, B⃗, and the direction of propagation are mutually perpendicular).

The expressions are:

E⃗ = E₀ sin(kz − ωt) î

B⃗ = B₀ sin(kz − ωt) ĵ

where E₀ and B₀ are the peak amplitudes, k = 2π/λ is the wave number, and ω is the angular frequency.

Also, E₀/B₀ = c = 3 × 10⁸ m/s.

(½ mark for correct directions stated; ½ mark for both expressions written correctly)

(ii) Any two of the following properties of electromagnetic waves:

(a) Electromagnetic waves are transverse in nature — the oscillating E⃗ and B⃗ fields are perpendicular to each other and to the direction of propagation.

(b) They travel through vacuum with speed c = 1/√(μ₀ε₀) = 3 × 10⁸ m/s, independent of the frequency or wavelength of the wave.

(c) They carry energy and momentum but do not require any material medium for propagation.

(d) They are not deflected by electric or magnetic fields (being electrically neutral).

(1 mark for any two correct properties — ½ mark each)

(iii) Given: frequency ν = 5 × 10¹⁴ Hz.

The corresponding wavelength is:

λ = c/ν = (3 × 10⁸)/(5 × 10¹⁴)

∴ λ = 6 × 10⁻⁷ m = 600 nm

This wavelength lies in the range 400 nm – 700 nm.

∴ The wave belongs to the visible light region of the electromagnetic spectrum.

One method of detection: Human eye (or photoelectric cell / photographic film).

(½ mark for identifying visible light; ½ mark for a valid detection method)
Q29Short Answer3 marks

A team of engineering students is designing a wireless communication system for a remote hilly area. They shortlist four types of electromagnetic waves for different parts of their system: (i) Radio waves for broadcasting, (ii) Microwaves for point-to-point links, (iii) Infrared waves for short-range remote controls, and (iv) X-rays for security screening at the base station.

(a) Arrange these four electromagnetic waves in ascending order of their wavelengths.
(b) The students note that all four waves travel at the same speed in vacuum. Using the relation c = νλ, calculate the frequency of the microwave signal if its wavelength is 3 cm. (c = 3×10⁸ m/s)
(c) Give ONE reason why microwaves, and NOT radio waves, are preferred for point-to-point satellite communication links.

Show answer
(a) All electromagnetic waves obey the relation c = νλ, so higher frequency corresponds to shorter wavelength.

Ascending order of wavelengths (shortest → longest):

X-rays < Infrared < Microwaves < Radio waves

∴ Ascending order of wavelength: X-rays → Infrared → Microwaves → Radio waves. [1 mark]

(b) By the wave relation for electromagnetic waves in vacuum:

c = νλ

∴ ν = c / λ

Given: c = 3×10⁸ m/s, λ = 3 cm = 3×10⁻² m

ν = (3×10⁸) / (3×10⁻²)

∴ ν = 1×10¹⁰ Hz [1 mark]

(c) Microwaves are preferred over radio waves for point-to-point satellite communication because microwaves have much higher frequencies (and shorter wavelengths), which allows them to pass straight through the ionosphere without being reflected back to Earth. Radio waves (especially of lower frequencies) are reflected by the ionosphere and cannot reach satellites in space. Additionally, microwaves can be focused into a narrow directional beam using dish antennas, reducing signal loss over long distances. [1 mark]

[Examiner note: Award 1 mark for any ONE correct, clearly stated reason — e.g., microwaves penetrate the ionosphere OR can be beamed directionally. Do not penalise if student gives a different valid scientific reason.] [1 mark for part (a) + 1 mark for part (b) + 1 mark for part (c) = 3 marks; the 4th mark is awarded for correct labelling/identification of the wave type in part (a) AND correct formula statement in part (b) — see breakdown below]

─────────────────────────────
MARK BREAKDOWN (Total: 4 marks)
─────────────────────────────
(a) Correct ascending order of wavelengths with all four waves correctly placed → 1 mark
(Penalise ½ mark if one wave is misplaced; penalise full mark only if order is entirely wrong)

(b) Stating formula c = νλ → ½ mark
Correct substitution with units → ½ mark
∴ ν = 1×10¹⁰ Hz (unit mandatory) → 1 mark

(c) Any ONE valid scientific reason (ionosphere penetration OR directional beaming OR higher bandwidth) → 1 mark
Q30Short Answer3 marks

( a ) What is displacement current? Write the expression for displacement current in terms of the rate of change of electric flux. Why was it introduced by Maxwell?
( b ) A parallel plate capacitor has circular plates of radius 8 cm. It is being charged such that the electric field between the plates changes at the rate of 1.5 × 10¹³ V m⁻¹ s⁻¹. Calculate the displacement current between the plates.
( c ) Name the electromagnetic wave having the shortest wavelength and state one of its uses.

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( a ) Displacement current is defined as the current that arises due to the time-varying electric field (or changing electric flux) in a region, even in the absence of any actual charge flow.

Formula: I<sub>d</sub> = ε₀ dΦ<sub>E</sub>/dt

Why introduced: Without displacement current, Ampere's circuital law gave inconsistent (contradictory) results when applied to the region between the plates of a charging capacitor — the magnetic field calculated using a surface passing through the wire gave a non-zero value, while a surface passing between the plates gave zero. Maxwell introduced displacement current to restore the consistency of Ampere's law, making it: ∮ B⃗ · dL⃗ = μ₀ (I<sub>c</sub> + I<sub>d</sub>).

( b ) By definition, the displacement current is:

I<sub>d</sub> = ε₀ dΦ<sub>E</sub>/dt = ε₀ A (dE/dt)

where A = πr² is the area of the circular plates and dE/dt is the rate of change of electric field.

Given: r = 8 cm = 0.08 m, dE/dt = 1.5 × 10¹³ V m⁻¹ s⁻¹, ε₀ = 8.854 × 10⁻¹² C² N⁻¹ m⁻²

Area: A = π × (0.08)² = π × 6.4 × 10⁻³ m²
A = 3.14159 × 6.4 × 10⁻³ = 2.011 × 10⁻² m²

Substituting:
I<sub>d</sub> = 8.854 × 10⁻¹² × 2.011 × 10⁻² × 1.5 × 10¹³
I<sub>d</sub> = 8.854 × 2.011 × 1.5 × 10⁻¹²⁻²⁺¹³
I<sub>d</sub> = 8.854 × 3.0165 × 10⁻¹
I<sub>d</sub> = 26.71 × 10⁻¹

∴ I<sub>d</sub> ≈ 2.67 A

( c ) Gamma rays have the shortest wavelength among all electromagnetic waves.

One use: They are used in radiotherapy to destroy cancer cells / used in food irradiation to kill bacteria and preserve food.

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