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Electromagnetic Waves Set-3 MCQ Online Practice Test | Class 12 | Quiz Tweek

Electromagnetic Waves Set-3

Practice Important MCQs for Electromagnetic Waves — Physics, Class 12.

Electromagnetic waves unify electric and magnetic phenomena into time‑varying transverse fields described by Maxwell's equations; understanding their propagation speed, energy flow (Poynting vector), polarization and boundary behaviour is essential for solving board-level numerical problems and for competitive exams where multi-step reasoning and physical interpretation are tested. Questions on energy densities, impedance mismatch, skin depth and polarization often combine calculus, vector reasoning and algebra — skills that directly map to JEE/NEET problem styles.

This set focuses on reasoning-rich items: numerical estimates, graph interpretation, and assertion–reason scenarios. Work through the problems to practise converting Maxwellian relations into measurable quantities (intensity, reflectance, skin depth), interpreting time/space graphs of $E$ and $B$ , and avoiding common misconceptions about standing waves, phase relations and polarization.

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Q1. A plane electromagnetic wave propagates in a non-magnetic dielectric medium with relative permittivity $\epsilon_r=4$ and relative permeability $\mu_r=1$ . What is its speed $v$ in the medium?

(A)

Option A $3.00\times10^8\ \mathrm{m/s}$

(B)

Option B $1.50\times10^8\ \mathrm{m/s}$

(C)

Option C $7.50\times10^7\ \mathrm{m/s}$

(D)

Option D $2.12\times10^8\ \mathrm{m/s}$

Q2. A monochromatic plane electromagnetic wave in vacuum has electric field amplitude $E_0=120\ \mathrm{V/m}$ . Calculate the time‑averaged intensity $\langle S\rangle$ (use $\langle S\rangle=\tfrac{1}{2}\epsilon_0 c E_0^2$ if needed and $\epsilon_0=8.85\times10^{-12}\ \mathrm{F/m}$ , $c=3.00\times10^8\ \mathrm{m/s}$ ).

(A)

Option A $9.55\ \mathrm{W/m^2}$

(B)

Option B $38.2\ \mathrm{W/m^2}$

(C)

Option C $19.1\ \mathrm{W/m^2}$

(D)

Option D $1.91\ \mathrm{W/m^2}$

Q3. At a fixed point in space the measured time dependence of fields is $E(t)=E_0\sin(\omega t)$ and $B(t)=B_0\cos(\omega t)$ with $E_0=c\,B_0$ . Based on these traces, which statement about the time‑averaged Poynting vector $\langle\vec{S}\rangle$ is correct?

(A)

Option A The time‑averaged Poynting vector is a nonzero constant in the nominal propagation direction.

(B)

Option B The Poynting vector is zero at all instants (no instantaneous energy flow).

(C)

Option C The instantaneous Poynting vector oscillates and has a nonzero positive time average.

(D)

Option D The instantaneous Poynting vector oscillates but its time average is zero (no net energy transport)

Q4. A plane wave in vacuum is normally incident on a lossless dielectric with relative permittivity $\epsilon_r=2.25$ and $\mu_r=1$ . What fraction of the incident intensity is transmitted into the dielectric (neglect absorption)?

(A)

Option A $96\%$

(B)

Option B $4\%$

(C)

Option C $75\%$

(D)

Option D $60\%$

Q5. Assertion: A time‑varying electric field in vacuum can produce a magnetic field even where there is no conduction current.
Reason: Maxwell added the displacement current density $\epsilon_0\frac{\partial\vec{E}}{\partial t}$ to Ampère's law so that $\nabla\times\vec{B}=\mu_0\vec{J}+\mu_0\epsilon_0\frac{\partial\vec{E}}{\partial t}$ ; this term implies a changing $\vec{E}$ generates $\vec{B}$ .

(A)

Option A Both assertion and reason are true and the reason is a correct explanation of the assertion.

(B)

Option B Both assertion and reason are true but the reason is not a correct explanation of the assertion.

(C)

Option C Assertion is true but reason is false.

(D)

Option D Assertion is false but reason is true.

Q6. A plane electromagnetic wave of frequency $f=1.00\ \mathrm{GHz}$ is incident on copper with conductivity $\sigma=5.8\times10^7\ \mathrm{S/m}$ and $\mu=\mu_0$ . Estimate the skin depth $\delta=\sqrt{\dfrac{2}{\omega\mu\sigma}}$ in copper (give answer in $\mu$ m; use $\mu_0=4\pi\times10^{-7}\ \mathrm{H/m}$ ).

(A)

Option A $0.21\ \mu\mathrm{m}$

(B)

Option B $2.1\ \mu\mathrm{m}$

(C)

Option C $21\ \mu\mathrm{m}$

(D)

Option D $210\ \mu\mathrm{m}$

Q7. Assertion: A $90^\circ$ phase difference between the electric and magnetic fields of a plane electromagnetic wave implies the wave is circularly polarized.
Reason: Circular polarization is produced when two orthogonal components of the electric field have equal amplitude and a $90^\circ$ phase difference.

(A)

Option A Both assertion and reason are true and the reason is a correct explanation of the assertion.

(B)

Option B Both assertion and reason are true but the reason is not a correct explanation of the assertion.

(C)

Option C Assertion is true but reason is false.

(D)

Option D Assertion is false but reason is true.

Q8. Two identical plane waves $E_1=E_0\sin(kx-\omega t)$ and $E_2=E_0\sin(kx+\omega t)$ superpose to give $E(x,t)=2E_0\sin(kx)\cos(\omega t)$ . At $x=\lambda/8$ one finds the instantaneous Poynting vector $S(x,t)\propto\sin(2\omega t)$ . Which statement about energy flow at this point is correct?

(A)

Option A There is no instantaneous energy flow at any time ( $S=0$ for all $t$ ).

(B)

Option B There is a constant nonzero energy flow to the right (positive time average).

(C)

Option C Instantaneous energy flow oscillates in time but its time average is zero.

(D)

Option D There is a constant nonzero energy flow to the left (negative time average).

Q9. Unpolarized light of intensity $I_0$ passes through three ideal linear polarizers. The first is at $0^\circ$ , the second at $45^\circ$ to the first, and the third at $90^\circ$ to the first. What intensity emerges after the third polarizer?

(A)

Option A $I_0/4$

(B)

Option B $I_0/8$

(C)

Option C $I_0/2$

(D)

Option D $0$

Q10. A plane electromagnetic wave in air ( $n_1=1.00$ ) is incident on water ( $n_2=1.33$ ). For non‑magnetic media ( $\mu=\mu_0$ ), the Brewster angle satisfies $\theta_B=\arctan\!\left(\dfrac{n_2}{n_1}\right)$ . What is $\theta_B$ (approximately)?

(A)

Option A $36.9^\circ$

(B)

Option B $23.0^\circ$

(C)

Option C $61.9^\circ$

(D)

Option D $53.1^\circ$

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