Electromagnetic Waves JEE Main previous year questions with solutions

5 solved JEE Main questions on Electromagnetic Waves, free to read — no sign-in needed. The full chapter has 162 questions; sign in to attempt the remaining 157 in the exam simulator.

1. Q1JEE Main 2026 (08 Apr, Shift 2)Electromagnetic Spectrum
   A monochromatic source of light operating at $15$ kW emits $2.5\times 10^{22}$ photons/s. The region of an electromagnetic spectrum to which the emitted electromagnetic radiation belongs to ________. (Take $h=6.6\times 10^{-34}$ J·s and $c=3\times 10^8$ m/s).
   1. A.Microwave
   2. B.Infrared
   3. C.Visible
   4. D.Ultraviolet
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   Answer: (D)

   Power of the source, $P = 15 \text{ kW} = 15 \times 10^3 \text{ J/s}$ Number of photons emitted per second, $n = 2.5 \times 10^{22} \text{ s}^{-1}$ Energy of one photon, $E = \dfrac{P}{n} = \dfrac{15 \times 10^3}{2.5 \times 10^{22}} = 6 \times 10^{-19} \text{ J}$ Using $E = \dfrac{hc}{\lambda}$, we get $\lambda = \dfrac{hc}{E}$ $\lambda = \dfrac{6.6 \times 10^{-34} \times 3 \times 10^8}{6 \times 10^{-19}} = 3.3 \times 10^{-7} \text{ m} = 330 \text{ nm}$ The wavelength $330 \text{ nm}$ lies in the ultraviolet region of the electromagnetic spectrum. Answer: Ultraviolet
2. Q2JEE Main 2025 (07 Apr, Shift 1)Electromagnetic Waves and Maxwell's equations
   Two plane polarized light waves combine at a certain point whose electric field components are $\begin{aligned} & \mathrm{E}_1=\mathrm{E}_0 \sin \omega \mathrm{t} \\ & \mathrm{E}_2=\mathrm{E}_0 \sin \left(\omega \mathrm{t}+\frac{\pi}{3}\right)\end{aligned}$ Find the amplitude of the resultant wave.
   1. A.0.9 E
   2. B.$\mathrm{E}_0$
   3. C.$1.7 \mathrm{E}_0$
   4. D.$3.4 \mathrm{E}_0$
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   Answer: (C)

   $\mathrm{E}=\sqrt{\left(\mathrm{E}_0\right)^2+\left(\mathrm{E}_0\right)^2+2\left(\mathrm{E}_0\right)\left(\mathrm{E}_0\right) \cos \frac{\pi}{3}}$ $\mathrm{E}=\sqrt{2 \mathrm{E}_0^2+\mathrm{E}_0^2}=\sqrt{3} \mathrm{E}_0=1.73 \mathrm{E}_0$
3. Q3JEE Main 2023 (08 Apr, Shift 2)X-Rays
   The waves emitted when a metal target is bombarded with high energy electrons are
   1. A.Microwaves
   2. B.Infrared rays
   3. C.X-rays
   4. D.Radio Waves
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   Answer: (C)

   The bombarding electrons can eject electrons from the inner shells of the atoms of the metal target. Those vacancies will be quickly filled by electrons dropping from higher levels, emitting x-rays with sharply defined frequencies associated with the difference between the atomic energy levels of the target atoms. Hence, X-rays are emitted when a metal target is bombarded with high energy electrons.
4. Q4JEE Main 2026 (24 Jan, Shift 1)Electromagnetic Spectrum
   Match the LIST-I with LIST-II $\begin{array}{|l|l|l|l|} \hline & \text{List-I} & & \text{List-II} \\ \hline \text{A.} & \text{Radio-wave} & \text{I.} & \text{is produced by Magnetron valve} \\ \hline \text{B.} & \text{Micro-wave} & \text{II.} & \text{due to change in the vibrational modes of atoms} \\ \hline \text{C.} & \text{Infrared-wave} & \text{III.} & \text{due to inner shell electrons moving from higher to lower energy level} \\ \hline \text{D.} & \text{X-ray} & \text{IV.} & \text{due to rapid acceleration of electrons} \\ \hline \end{array}$ Choose the correct answer from the options given below:
   1. A.A-IV, B-II, C-I, D-III
   2. B.A-IV, B-I, C-II, D-III
   3. C.A-IV, B-III, C-I, D-II
   4. D.A-II, B-IV, C-III, D-I
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   Answer: (B)

   Matching electromagnetic wave sources: Radio-wave (A) - produced by rapid acceleration of electrons in antennas and transmitters, matches IV. Micro-wave (B) - produced by Magnetron valve, matches I. Infrared-wave (C) - due to change in vibrational modes of atoms and molecules, matches II. X-ray (D) - due to inner shell electrons moving from higher to lower energy levels, matches III. Correct pairing: A-IV, B-I, C-II, D-III
5. Q5JEE Main 2026 (06 Apr, Shift 2)Electromagnetic Waves and Maxwell's equations
   For an electromagnetic wave propagating through vacuum, $\vec{k}$, $\vec{E}$ and $\omega$ represent propagation vector, electric field and angular frequency, respectively. The magnetic field associated with this wave is represented by :
   1. A.$\dfrac{\vec{E} \times \vec{k}}{\omega}$
   2. B.$\dfrac{\vec{k} \times \vec{E}}{\omega}$
   3. C.$\omega(\vec{E} \times \vec{k})$
   4. D.$\omega(\vec{k} \times \vec{E})$
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   Answer: (B)

   For an electromagnetic wave, the direction of wave propagation $\hat{k}$ is along the direction of $\vec{E} \times \vec{B}$. Since $\vec{E}$, $\vec{B}$, and $\vec{k}$ form a mutually perpendicular right-handed system, the direction of the magnetic field is given by $\hat{B} = \hat{k} \times \hat{E}$. The magnitudes of the electric and magnetic fields are related by $B = \dfrac{E}{c}$. Using the relation for the speed of light $c = \dfrac{\omega}{k}$, we get $B = \dfrac{kE}{\omega}$. Combining the magnitude and direction gives: $\vec{B} = B\hat{B} = \dfrac{kE}{\omega} (\hat{k} \times \hat{E})$ Since $\vec{k} = k\hat{k}$ and $\vec{E} = E\hat{E}$, this simplifies to: $\vec{B} = \dfrac{\vec{k} \times \vec{E}}{\omega}$ Answer: $\dfrac{\vec{k} \times \vec{E}}{\omega}$