Electromagnetic Waves Class 12 Physics Numericals 2026-27 | Chapter 8 with Solutions
Core syllabus: displacement-current idea, EM-wave characteristics, qualitative transverse nature, spectrum and elementary uses. Quantitative reinforcement: c = νλ, medium relations, photon-energy and E0/B0 calculations are included because they connect directly with NCERT treatment and recurring exam-preparation practice, but they are not presented as a separate Chapter 8 CBSE mark allocation or a guarantee of numerical questions.
1. Numerical Roadmap
Electric-flux rate, changing electric field, capacitor charge and current continuity.
Frequency, wavelength, speed, time period and propagation in a medium.
Energy–frequency–wavelength calculations and quantitative spectrum comparisons.
E0/B0, direction reasoning and medium-based calculations.
Multi-step calculations combining the core Chapter 8 relationships.
2. Essential Formula Map
| Concept | Formula |
|---|---|
| Displacement current | Id = ε0 dΦE/dt |
| For a parallel-plate capacitor gap | Id = ε0A(dE/dt) |
| Vacuum wave relation | c = νλ |
| Angular frequency | ω = 2πν; T = 1/ν |
| Speed in a medium | v = c/n |
| Wavelength in a medium | λ = λ0/n |
| Photon energy | E = hν = hc/λ |
| Field amplitudes in vacuum | E0 = cB0 |
| Propagation direction | Direction of propagation ∥ E × B |
3. Displacement Current — Calculation Practice
The electric flux between the plates of a charging capacitor is changing at the rate 2.0 × 109 N m2 C−1 s−1. Find the displacement current.
Id = (8.85 × 10−12)(2.0 × 109) = 1.77 × 10−2 A.
A parallel-plate capacitor has plate area 2.0 × 10−2 m2. The electric field between the plates changes at 5.0 × 106 V m−1 s−1. Find the displacement current.
Id = (8.85 × 10−12)(2.0 × 10−2)(5.0 × 106) = 8.85 × 10−7 A.
The charge on a charging capacitor is increasing at 3.5 mC s−1. Find the displacement current between its plates.
Id = 3.5 mC s−1 = 3.5 × 10−3 A.
The charge on a capacitor varies as Q = 2t2 mC, where t is in seconds. Find the displacement current at t = 3 s.
dQ/dt = 4t mC s−1. At t = 3 s, Id = 12 mC s−1 = 0.012 A.
A charging capacitor is connected in a circuit carrying a conduction current of 5.0 mA in the wires. What displacement current exists across the gap between the plates, assuming the ideal charging-capacitor situation?
The same current continuity is maintained through the capacitor gap in the displacement-current description. Therefore Id = 5.0 mA.
The electric field between capacitor plates is constant with time. Find the displacement current through the gap.
For a constant electric field, dE/dt = 0. Hence Id = 0.
Two capacitor gaps have the same plate area. In gap A, dE/dt = 2.0 × 106 V m−1 s−1; in gap B, dE/dt = 5.0 × 106 V m−1 s−1. Find Id,B/Id,A.
With ε0 and A the same, Id is proportional to dE/dt. Thus Id,B/Id,A = (5.0 × 106)/(2.0 × 106) = 2.5.
A charging capacitor has a displacement current of 4.0 mA. Find the rate of change of electric flux.
dΦE/dt = 4.0 × 10−3 /(8.85 × 10−12) ≈ 4.52 × 108 N m2 C−1 s−1.
4. Frequency, Wavelength and Propagation
An electromagnetic wave has frequency 100 MHz. Find its wavelength in vacuum.
λ = c/ν = (3.0 × 108)/(1.0 × 108) = 3.0 m.
An electromagnetic wave has wavelength 600 nm in vacuum. Find its frequency.
λ = 600 nm = 6.0 × 10−7 m. Therefore ν = (3.0 × 108)/(6.0 × 10−7) = 5.0 × 1014 Hz.
An electromagnetic wave has frequency 7.5 × 1014 Hz. Find its wavelength in vacuum.
λ = (3.0 × 108)/(7.5 × 1014) = 4.0 × 10−7 m = 400 nm.
Find the frequency of an electromagnetic wave whose wavelength is 3.0 cm in vacuum.
λ = 3.0 cm = 3.0 × 10−2 m. Hence ν = (3.0 × 108)/(3.0 × 10−2) = 1.0 × 1010 Hz.
A microwave has frequency 2.0 GHz in vacuum. Find its wavelength.
λ = (3.0 × 108)/(2.0 × 109) = 1.5 × 10−1 m.
An electromagnetic wave has frequency 5.0 × 1014 Hz. Find its time period.
T = 1/(5.0 × 1014) = 2.0 × 10−15 s.
An EM wave has wavelength 2.5 m and frequency 1.2 × 108 Hz. Find its speed.
v = (1.2 × 108)(2.5) = 3.0 × 108 m s−1.
An electromagnetic wave of frequency 6.0 × 1014 Hz enters a transparent medium of refractive index 1.5. Find its speed and wavelength in the medium.
v = (3.0 × 108)/1.5 = 2.0 × 108 m s−1. Frequency does not change on entering the medium, so λ = (2.0 × 108)/(6.0 × 1014) = 3.33 × 10−7 m.
A wave has frequency 5.0 × 1014 Hz in vacuum and enters a medium where its speed becomes 2.0 × 108 m s−1. Find its wavelength in the medium.
λ = (2.0 × 108)/(5.0 × 1014) = 4.0 × 10−7 m.
A wave has vacuum wavelength 600 nm and enters a medium of refractive index 2.0. Find its wavelength in the medium.
λ = 600/2 = 300 nm.
5. Photon Energy and Quantitative Spectrum Practice
Find the energy of a photon of wavelength 600 nm. Use h = 6.626 × 10−34 J s.
E = [(6.626 × 10−34)(3.0 × 108)]/(6.0 × 10−7) = 3.31 × 10−19 J.
Find the energy of a photon of wavelength 300 nm.
E = [(6.626 × 10−34)(3.0 × 108)]/(3.0 × 10−7) = 6.626 × 10−19 J.
Find the energy of a photon of wavelength 1.0 μm.
λ = 1.0 × 10−6 m. E = [(6.626 × 10−34)(3.0 × 108)]/(1.0 × 10−6) = 1.99 × 10−19 J.
A photon has wavelength 400 nm. Find its energy in eV. Take 1 eV = 1.602 × 10−19 J.
For 400 nm, E = 4.97 × 10−19 J. Therefore E = (4.97 × 10−19)/(1.602 × 10−19) ≈ 3.10 eV.
A monochromatic 500 nm light pulse carries 1.0 μJ of energy. Estimate the number of photons in the pulse.
Ephoton = [(6.626 × 10−34)(3.0 × 108)]/(5.0 × 10−7) ≈ 3.98 × 10−19 J. Thus N = 1.0 × 10−6 /(3.98 × 10−19) ≈ 2.52 × 1012.
Compare the energy of a 200 nm photon with that of a 600 nm photon.
E200/E600 = λ600/λ200 = 600/200 = 3.
A photon has energy 3.313 × 10−19 J. Find its frequency.
ν = E/h = (3.313 × 10−19)/(6.626 × 10−34) = 5.0 × 1014 Hz.
A photon has energy 6.626 × 10−19 J. Find its wavelength.
λ = [(6.626 × 10−34)(3.0 × 108)]/(6.626 × 10−19) = 3.0 × 10−7 m.
Two equal-energy light pulses contain photons of wavelengths 400 nm and 800 nm. Find the ratio of the number of photons in the 400 nm pulse to that in the 800 nm pulse.
N400/N800 = 400/800 = 1/2.
The frequency of electromagnetic radiation is increased from 4.0 × 1014 Hz to 8.0 × 1014 Hz. By what factor does the energy of each photon change?
Photon energy is directly proportional to frequency. Doubling frequency doubles photon energy.
Estimate the photon energy corresponding to 700 nm red light.
E = [(6.626 × 10−34)(3.0 × 108)]/(7.0 × 10−7) ≈ 2.84 × 10−19 J.
Find the photon energy of radiation with wavelength 200 nm and express it approximately in eV.
E ≈ 9.94 × 10−19 J. In eV, E ≈ (9.94 × 10−19)/(1.602 × 10−19) ≈ 6.20 eV.
6. Field-Amplitude, Direction and Medium-Based Calculations
A plane EM wave in vacuum has magnetic-field amplitude 1.0 μT. Find the electric-field amplitude.
E0 = (3.0 × 108)(1.0 × 10−6) = 3.0 × 102 V m−1.
The electric-field amplitude of an EM wave in vacuum is 60 V m−1. Find the magnetic-field amplitude.
B0 = 60/(3.0 × 108) = 2.0 × 10−7 T.
For an EM wave in vacuum, E0 = 150 V m−1 and B0 = 5.0 × 10−7 T. Find E0/B0.
E0/B0 = 150/(5.0 × 10−7) = 3.0 × 108 m s−1.
An EM wave in vacuum has electric-field amplitude 200 V m−1. If its electric-field amplitude is doubled, what is the new magnetic-field amplitude?
Original B0 = 200/(3.0 × 108) = 6.67 × 10−7 T. Doubling E0 doubles B0, so new B0 = 1.33 × 10−6 T.
At a particular instant, the electric field is along +y and the magnetic field is along +z. Find the direction of propagation.
Using the right-hand rule, +y × +z = +x.
An EM wave propagates along +z and its electric field is along +x. Find the direction of the magnetic field.
We need +x × B = +z. Since +x × +y = +z, B must be along +y.
An EM wave has vacuum wavelength 450 nm and enters a medium of refractive index 1.5. Find its wavelength in the medium.
λ = 450/1.5 = 300 nm.
A wave has frequency 4.0 × 1014 Hz and travels through a medium at 2.0 × 108 m s−1. Find its wavelength and refractive index relative to vacuum.
λ = (2.0 × 108)/(4.0 × 1014) = 5.0 × 10−7 m = 500 nm. n = (3.0 × 108)/(2.0 × 108) = 1.5.
7. Mixed Competency and Multi-Step Numericals
A wave has frequency 5.0 × 1014 Hz and enters glass of refractive index 1.5. Find its speed and wavelength in glass.
v = (3.0 × 108)/1.5 = 2.0 × 108 m s−1. Then λ = (2.0 × 108)/(5.0 × 1014) = 4.0 × 10−7 m.
Radiation has wavelength 500 nm. Find its frequency and photon energy.
ν = (3.0 × 108)/(5.0 × 10−7) = 6.0 × 1014 Hz. E = (6.626 × 10−34)(6.0 × 1014) = 3.98 × 10−19 J.
Radiation A has frequency 3.0 × 109 Hz and radiation B has frequency 3.0 × 1017 Hz. Find their wavelengths in vacuum and identify which has the higher frequency.
For A, λ = (3.0 × 108)/(3.0 × 109) = 0.10 m. For B, λ = (3.0 × 108)/(3.0 × 1017) = 1.0 × 10−9 m. B has the higher frequency and much shorter wavelength.
A UV photon has frequency 1.0 × 1016 Hz and an infrared photon has frequency 1.0 × 1013 Hz. Find the ratio of their photon energies.
EUV/EIR = (1.0 × 1016)/(1.0 × 1013) = 103.
A capacitor has plate area 0.010 m2. The electric field between its plates increases uniformly at 2.0 × 107 V m−1 s−1. Find the displacement current.
Id = (8.85 × 10−12)(0.010)(2.0 × 107) = 1.77 × 10−6 A.
A parallel-plate capacitor has circular plates of radius 10 cm separated by 4.0 cm of air. It is being charged by a constant current of 0.20 A. Calculate (i) its capacitance, (ii) the rate of increase of potential difference between the plates, and (iii) the displacement current across the gap.
r = 0.10 m, d = 0.040 m, so A = π(0.10)² = 3.14 × 10−2 m². Hence C = (8.85 × 10−12)(3.14 × 10−2)/(0.040) ≈ 6.95 × 10−12 F. Next, dV/dt = I/C = 0.20/(6.95 × 10−12) ≈ 2.88 × 1010 V s−1. For an ideal charging capacitor, the displacement current equals the conduction current: Id = 0.20 A.
An EM wave has wavelength 500 nm in vacuum. Calculate (i) frequency, (ii) photon energy, and (iii) electric-field amplitude if its magnetic-field amplitude is 2.0 × 10−7 T.
(i) ν = (3.0 × 108)/(5.0 × 10−7) = 6.0 × 1014 Hz. (ii) Ephoton = (6.626 × 10−34)(6.0 × 1014) ≈ 3.98 × 10−19 J. (iii) E0 = (3.0 × 108)(2.0 × 10−7) = 60 V m−1.
8. Common Numerical Traps
9. A Reliable 5-Step Numerical Method
Write the given quantity and the required quantity.
Select the relation that directly connects the known and unknown values.
Put wavelength, frequency, area and other values into compatible units.
Show substitution and keep the calculation transparent.
Check units and physical direction: higher ν ↔ lower λ; higher ν ↔ higher photon energy.
10. Frequently Asked Numerical Questions
Start with displacement current, c = νλ, wavelength/frequency conversion and simple field-amplitude relations. Then practise medium-based and photon-energy calculations as quantitative reinforcement.
No. Chapter 8 belongs to Unit V, and Unit V is grouped with Optics in the official 2026–27 structure for 18 marks collectively. The syllabus does not provide a separate fixed Chapter 8 mark allocation. CBSE Physics Curriculum 2026–27.
The frequency remains unchanged at the boundary; the speed and wavelength change. Use v = c/n and λ = v/ν when the refractive index is given.
No. They are original Learn Revise Hub practice problems aligned with the current syllabus and recurring concept patterns. They are not claims about exact future CBSE questions.
11. Continue the Chapter 8 Practice Sequence
Read concepts: Electromagnetic Waves Complete Notes
Practise theory: Electromagnetic Waves Important Questions
Practise objective questions: Electromagnetic Waves MCQs
Previous chapter: Chapter 7 — Alternating Current Notes
Foundation link: Chapter 6 — Electromagnetic Induction Notes
Next Chapter 8 resources: Case-Based Questions, Assertion–Reason, PYQs, Formula Sheet + Quick Revision and Chapter Test will be linked after their publication/verification.
12. Final Numerical Readiness Check
✓ I can calculate displacement current from changing electric flux or electric field.
✓ I can convert frequency ↔ wavelength using c = νλ.
✓ I can calculate wave speed and wavelength in a medium when n is given.
✓ I can calculate photon energy from frequency or wavelength.
✓ I can compare photon energies and photon counts quantitatively.
✓ I can use E0 = cB0 for vacuum waves.
✓ I can determine propagation direction from E and B directions.
✓ I show units, substitutions and a clear calculation path.
13. Official and Research References
Official CBSE: Physics Curriculum 2026–27 · Class XII 2026–27 SQP & Marking Scheme.
Official NCERT: Physics Part-I.
Research note: Recent Chapter 8 practice resources and PYQ compilations show recurring student demand around displacement current, EM-wave characteristics, spectrum order/uses, field relationships and short calculations. Those patterns informed the practice mix here, while the official CBSE curriculum remains the syllabus boundary.
Original-content notice: The 45 numerical questions and worked solutions on this page are original Learn Revise Hub practice material. They are not reproduced CBSE board-paper questions.
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