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Electromagnetic Waves appeared 31 times across 3 years — 3.6% of Physics. This question is from Electromagnetic Spectrum.

Year 2026 2025 2024 Total
Questions 10 12 9 31

Match the List-I with List-II
List-IList-II
A. Radio-waveI. is produced by Magnetron value
B. Micro-waveII. Due to change in the vibrational modes of atoms
C. Infrared-waveIII. Due to inner shell electrons moving from higher energy level to lower energy level
D. X-rayIV. Due to rapid acceleration of electrons
Choose the correct answer from the options given below:

Solution & Explanation

Core Logic

Mapping production sources to electromagnetic waves: Radio waves: Produced by rapid acceleration and deceleration of electrons in aerials. (A -> IV) Microwaves: Produced by special vacuum tubes like klystrons, magnetrons, and Gunn diodes. (B -> I) Infrared waves: Produced by hot bodies and molecules. They originate due to the change in vibrational modes of atoms/molecules. (C -> II) X-rays: Produced when high-energy electrons are stopped by a metal target, or by transitions of inner shell electrons from a higher to a lower energy level. (D -> III)

Step 1: Final Match

A - IV B - I C - II D - III This corresponds to Option (3).

Pattern Recognition

Memorize fundamental EM wave sources: Radio (accelerating charges), Microwave (magnetrons), IR (vibrational modes/heat), X-ray (inner shell transitions).

Chapter Mix

Class 12 Physics: Electromagnetic Waves

Reference Study Guides

More Electromagnetic Waves Previous-Year Questions — Page 7

Q40 jee_main_2024_31_jan_morning Energy Density Of EM Waves
In a plane EM wave, the electric field oscillates sinusoidally at a frequency of 5 × 10¹⁰ ~Hz and an amplitude of 50 ~Vm⁻¹. The total average energy density of the electromagnetic field of the wave is : [Use ε₀ = 8.85 × 10⁻¹² C² / Nm² ]
  • A. 1.106 × 10⁻⁸ Jm⁻³
  • B. 4.425 × 10⁻⁸ ~Jm⁻³
  • C. 2.212 × 10⁻⁸ ~Jm⁻³
  • D. 2.212 × 10⁻¹⁰ ~Jm⁻³

Solution

Related Formula
Utotal average = (1)/(2)ε₀ E₀²
Core Logic

For an electromagnetic wave, the total average energy density is the sum of the average energy density of the electric field and the magnetic field. They are equal, so:

Uavg = UE + UB = 2UE = 2 ( (1)/(4)ε₀ E₀² ) = (1)/(2)ε₀ E₀²

Where E₀ is the amplitude of the electric field.

Step 2: Substitution

Given: E₀ = 50 V/m ε₀ = 8.85 × 10⁻¹² C²/(N· m²)

Uavg = (1)/(2) × (8.85 × 10⁻¹²) × (50)² Uavg = (1)/(2) × 8.85 × 10⁻¹² × 2500 Uavg = 1.10625 × 10⁻⁸ J/m³
Chapter Mix

Class 12 Physics: Electromagnetic Waves

More Electromagnetic Waves Questions — jee_main_2026_24_january_morning

Practice all Electromagnetic Waves previous-year questions →

Rankbit System
JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)