Class 12 Physics | Unit IV
Chapter 8: Electromagnetic Waves
Maxwell's Equations • Displacement Current • EM Spectrum • Properties of EM Waves
1. Displacement Current & Maxwell's Equations
Maxwell's great insight: A changing electric field (dE/dt) produces a magnetic field,
just as a changing magnetic field produces an electric field (Faraday's law). This led Maxwell to
introduce the concept of displacement current.
1.1 Need for Displacement Current
Consider charging a capacitor with current I. Between the plates, no charge flows (gap), yet Ampère's circuital law must still be consistent. Maxwell resolved this by postulating that the changing electric field between plates acts as a current source (displacement current).
Displacement current: Id = ε0 dϕE/dt
where ϕE = electric flux = EA (between parallel plates)
Modified Ampère's law (Ampère-Maxwell law):
∮B·dl = μ0(Ic + Id) = μ0Ic + μ0ε0 dϕE/dt
Id = Ic (at any instant, conduction current outside = displacement current inside the gap)
where ϕE = electric flux = EA (between parallel plates)
Modified Ampère's law (Ampère-Maxwell law):
∮B·dl = μ0(Ic + Id) = μ0Ic + μ0ε0 dϕE/dt
Id = Ic (at any instant, conduction current outside = displacement current inside the gap)
1.2 Maxwell's Four Equations (Summary)
| # | Equation Name | Statement | Physical Meaning |
|---|---|---|---|
| 1 | Gauss's Law (E) | ∮E·dA = Q/ε0 | Electric charges produce E field; flux ∝ enclosed charge |
| 2 | Gauss's Law (B) | ∮B·dA = 0 | No magnetic monopoles; B field lines are always closed loops |
| 3 | Faraday's Law | ∮E·dl = −dϕB/dt | Changing B field produces E field (EMF induction) |
| 4 | Ampère-Maxwell Law | ∮B·dl = μ0I + μ0ε0 dϕE/dt | Current & changing E field both produce B field |
- Equations 3 & 4 together show: changing E → B and changing B → E → self-sustaining electromagnetic waves can propagate through space even in vacuum.
- Maxwell predicted EM waves travel at speed c = 1/√(μ0ε0) = 3×108 m/s — same as light! This proved light is an electromagnetic wave.
⚠️ NEET: Displacement current Id =
ε0dϕE/dt. Between capacitor plates: Id = Iconduction.
Maxwell showed changing E produces B. c = 1/√(μ0ε0). No magnetic
monopoles (∮B·dA=0). Light is an EM wave.
2. Properties of Electromagnetic Waves
EM waves are produced by accelerating charges. They are transverse waves consisting of
oscillating electric and magnetic fields perpendicular to each other and to the direction of propagation.
2.1 Key Characteristics
- E, B, and direction of propagation are mutually perpendicular (E ⊥ B ⊥ c).
- Propagation direction: E × B (cross product gives direction of wave travel).
- EM waves do not require a medium — they can travel through vacuum.
- Speed in vacuum: c = 1/√(μ0ε0) = 3 × 108 m/s.
- Speed in medium: v = c/n where n = refractive index (v < c always).
- E and B oscillate in phase (reach maximum and zero simultaneously).
- Ratio: E0/B0 = c (or E/B = c at all instants).
- EM waves carry energy and momentum (radiation pressure).
Electric field: E = E0 sin(kx − ωt) (say, along y-axis)
Magnetic field: B = B0 sin(kx − ωt) (along z-axis if wave along x-axis)
k = 2π/λ (wave number); ω = 2πf; c = ω/k = fλ
Energy density: u = ½ε0E² + B²/(2μ0) (E and B contribute equally)
Intensity: I = P/A = ½cε0E0² = average power per unit area
Radiation pressure: Prad = I/c (fully absorbed) or 2I/c (fully reflected)
Magnetic field: B = B0 sin(kx − ωt) (along z-axis if wave along x-axis)
k = 2π/λ (wave number); ω = 2πf; c = ω/k = fλ
Energy density: u = ½ε0E² + B²/(2μ0) (E and B contribute equally)
Intensity: I = P/A = ½cε0E0² = average power per unit area
Radiation pressure: Prad = I/c (fully absorbed) or 2I/c (fully reflected)
2.2 Source of EM Waves
- Accelerating charges (not stationary or uniformly moving charges) produce EM waves.
- An oscillating charge (e.g., in an antenna) produces EM waves of the same frequency as its oscillation.
- The electromagnetic nature of light was confirmed by Hertz's experiment (1887), producing and detecting radio waves.
3. Electromagnetic Spectrum
The entire range of electromagnetic radiation, arranged by wavelength or frequency, is the
electromagnetic spectrum. All EM waves travel at the same speed c in vacuum but differ in
wavelength and frequency (c = fλ).
| Type | Wavelength Range | Frequency Range | Source | Key Uses |
|---|---|---|---|---|
| Radio waves | > 0.1 m | < 3 GHz | LC oscillator circuits, antennas | Radio/TV broadcasting, communication |
| Microwaves | 0.1 m to 1 mm | 3 GHz to 300 GHz | Klystron, magnetron, Gunn diode | Microwave ovens, radar, satellite communication |
| Infrared (IR) | 1 mm to 700 nm | 300 GHz to 4×1014 Hz | Hot bodies, sun, LED | Night vision, thermal imaging, TV remote, physiotherapy |
| Visible Light | 700 nm (Red) to 400 nm (Violet) | 4×1014 to 7.5×1014 Hz | Sun, light bulbs, lasers | Vision, photography, optical instruments |
| Ultraviolet (UV) | 400 nm to 1 nm | ~1015 to 1017 Hz | Sun, mercury vapour lamps, very hot bodies | Sterilisation, PUVA therapy, detection of forged documents, vitamin D production |
| X-rays | 1 nm to 10−3 nm | ~1017 to 1019 Hz | High-speed electrons hitting metal target (Coolidge tube) | Medical imaging, CT scan, study of crystal structure, security scanning |
| Gamma rays (γ) | < 10−3 nm | > 1019 Hz | Nuclear reactions, radioactive decay | Cancer treatment, sterilisation, nuclear medicine |
3.1 Mnemonic for EM Spectrum Order (increasing frequency)
R → M → I → V → U → X → G
“Radha Mohan Is Very Ugly X-tra Gross”
(Radio → Microwave → Infrared → Visible → UV → X-ray → Gamma)
Frequency ↑ | Wavelength ↓ | Energy ↑ (from Radio to Gamma)
“Radha Mohan Is Very Ugly X-tra Gross”
(Radio → Microwave → Infrared → Visible → UV → X-ray → Gamma)
Frequency ↑ | Wavelength ↓ | Energy ↑ (from Radio to Gamma)
3.2 Visible Spectrum (VIBGYOR) — Increasing Wavelength
Violet (380 nm) → Indigo → Blue → Green → Yellow → Orange → Red
(700 nm)
Violet: highest frequency, highest energy, most deviated by prism.
Red: lowest frequency, lowest energy, least deviated by prism.
Violet: highest frequency, highest energy, most deviated by prism.
Red: lowest frequency, lowest energy, least deviated by prism.
⚠️ NEET (2014, 2016, 2018, 2020, 2022): EM spectrum order (R M I V U X G).
Sources: Radio = oscillating circuits; Microwave = klystron/magnetron; IR = hot bodies; UV = sun/mercury
lamp; X-ray = Coolidge tube; Gamma = nuclear decay. All EM waves: transverse, speed c in vacuum, don't
need medium. X-rays and γ-rays overlap in spectrum.
4. Applications & Special Properties
4.1 Microwave Oven — How It Works
Microwaves at 2.45 GHz match the rotational frequency of water molecules. Resonance absorption → rapid rotational motion of H2O → temperature rises → food heats. Works best on water-containing food. Metal containers reflect microwaves (should not be used).
4.2 Greenhouse Effect
Sun emits mostly visible and UV light. Earth's surface absorbs and re-emits as infrared radiation. Certain gases (CO2, CH4, H2O vapour) absorb IR and re-radiate back → trap heat → warming. Increasing CO2 levels → enhanced greenhouse effect → global warming.
4.3 Ozone Layer
Ozone (O3) in stratosphere absorbs harmful ultraviolet radiation from sun (especially UV-B and UV-C). Ozone depletion (by CFCs) → more UV reaches surface → skin cancer, cataracts, DNA damage.
4.4 EM Waves and Momentum
EM waves carry momentum: p = U/c (for total energy U absorbed)
Radiation pressure on fully absorbing surface: P = I/c
Radiation pressure on fully reflecting surface: P = 2I/c (momentum change = 2p)
This was experimentally verified by Nichols and Hull.
Radiation pressure on fully absorbing surface: P = I/c
Radiation pressure on fully reflecting surface: P = 2I/c (momentum change = 2p)
This was experimentally verified by Nichols and Hull.
5. Summary Table — Key Facts
| Concept | Formula/Fact | Note |
|---|---|---|
| Speed of EM waves in vacuum | c = 1/√(μ0ε0) = 3×108 m/s | Same for all EM waves |
| E/B ratio | E0/B0 = c | E and B in phase |
| EM wave equation | c = fλ = ω/k | Transverse wave |
| Displacement current | Id = ε0dϕE/dt | Due to changing E field |
| Source of EM waves | Accelerating charges | Not static or uniformly moving charges |
| Intensity | I = ½cε0E0² | Energy per unit area per second |
| Radiation pressure | I/c (absorbed); 2I/c (reflected) | EM waves carry momentum |
| Photon energy | E = hf = hc/λ | h = 6.63×10−34 J·s |
🎓 NEET Previous Year Questions
Q1. [NEET 2022] Which of the following has the highest frequency in EM spectrum?
Answer Gamma rays have the highest
frequency (~1019–1024 Hz), shortest wavelength, and highest photon energy.
Q2. [NEET 2021] The electromagnetic radiation used in RADAR is:
Answer Microwaves. RADAR uses
microwaves because they can be focused into narrow beams and reflected by metallic objects. Wavelength
~1 mm to 30 cm.
Q3. [NEET 2020] Ozone layer absorbs which type of radiation?
Answer Ultraviolet (UV) radiation.
Ozone absorbs UV-B and UV-C, protecting life on Earth from harmful effects of UV.
Q4. [NEET 2019] Speed of electromagnetic waves in vacuum:
Answer c =
1/√(μ0ε0) = 3×108 m/s. It is the
same for all EM waves regardless of their frequency or wavelength.
Q5. [NEET 2018] Which radiation is used in physiotherapy for deep heating?
Answer Infrared (IR) radiation is
used in physiotherapy as it penetrates tissue and provides deep heating through thermal effects.
Q6. [NEET 2016] The source of X-rays is:
Answer High-speed electrons striking a metal
target (Coolidge tube). When accelerated electrons decelerate on hitting the target, they
emit X-ray photons (Bremsstrahlung radiation).
💡 Rapid Revision
- EM waves: transverse | E ⊥ B ⊥ propagation direction | E/B = c
- c = 1/√(μ0ε0) = 3×108 m/s | Same speed for all EM waves in vacuum
- Displacement current: Id = ε0dϕE/dt = Iconduction
- Spectrum (λ ↓, f ↑): Radio → Micro → IR → Vis → UV → X → γ
- Source: accelerating charges | Source of γ: nuclear reactions
- Radiation pressure: I/c (absorbed) | 2I/c (reflected)
- Greenhouse: CO2 traps IR | Ozone: absorbs UV
CLASS 12 PHYSICS | NCERT SOLUTIONS
Chapter 8 — Electromagnetic Waves
20 NCERT Exercise & Exemplar Questions — Step-by-Step Solutions
Key Formulas: c = 1/√(μ0ε0) = 3×108
m/s | c = fλ | E0/B0 = c | Id =
ε0dϕE/dt | I = ½cε0E0² | E = hf =
hc/λ
📝 NCERT Exercise Questions (8.1 – 8.10)
3 MarksQ1 (Ex 8.1). A parallel plate capacitor with
plate area A and plate separation d is charged to potential V. Find displacement current Id
if voltage changes at rate dV/dt.
✓ Solution
E between plates = V/d; Electric flux ϕE = EA = VA/d
E between plates = V/d; Electric flux ϕE = EA = VA/d
Id = ε0 dϕE/dt = ε0
× (A/d) × dV/dt = C × dV/dt
Where C = ε0A/d is the capacitance. So displacement current = C×(rate of change of
voltage). This is exactly equal to the conduction current Ic = C dV/dt flowing in the
external circuit. ✓3 MarksQ2 (Ex 8.2). A parallel plate capacitor of
area 200 cm² and plate separation 1.5 cm is being charged at rate dV/dt = 106 V/s. Find
Id.
✓ Solution
A = 200 cm² = 200×10−4 m² = 0.02 m²; d = 1.5 cm = 0.015 m
C = ε0A/d = 8.85×10−12×0.02/0.015 = 8.85×10−12×1.333 = 11.8×10−12 F
A = 200 cm² = 200×10−4 m² = 0.02 m²; d = 1.5 cm = 0.015 m
C = ε0A/d = 8.85×10−12×0.02/0.015 = 8.85×10−12×1.333 = 11.8×10−12 F
Id = C×dV/dt =
11.8×10−12×106 = 1.18×10−5 A =
11.8 µA
3 MarksQ3 (Ex 8.3). An EM wave has frequency
1010 Hz. Find wavelength. What type of EM wave is it?
✓ Solution
λ = c/f = 3×108 / 1010 =
3×10−2 m = 3 cm
λ = 3 cm falls in the microwave range (0.1 m to 1 mm). Used in radar and microwave
ovens.3 MarksQ4 (Ex 8.4). The magnetic field in an EM wave
is B = 2×10−7 sin(0.5×103x + 1.5×1011t). Find
E0, frequency, wavelength, and speed.
✓ Solution
B0 = 2×10−7 T; k = 0.5×103 /m; ω = 1.5×1011 rad/s
E0 = cB0 = 3×108×2×10−7 = 60 V/m
λ = 2π/k = 2π/(0.5×103) = 1.257×10−2 m ≈ 1.26 cm
f = ω/(2π) = 1.5×1011/(2π) = 2.39×1010 Hz ≈ 24 GHz (microwave)
v = ω/k = 1.5×1011/(0.5×103) = 3×108 m/s = c ✓
B0 = 2×10−7 T; k = 0.5×103 /m; ω = 1.5×1011 rad/s
E0 = cB0 = 3×108×2×10−7 = 60 V/m
λ = 2π/k = 2π/(0.5×103) = 1.257×10−2 m ≈ 1.26 cm
f = ω/(2π) = 1.5×1011/(2π) = 2.39×1010 Hz ≈ 24 GHz (microwave)
v = ω/k = 1.5×1011/(0.5×103) = 3×108 m/s = c ✓
3 MarksQ5 (Ex 8.5). Verify that
E0/B0 = c for an EM wave. What does this ratio signify?
✓ Solution
From Maxwell's equations, E and B are related by E0/B0 = c = 3×108 m/s.
Significance: E and B oscillate in phase but E is always much larger in magnitude than B (since c is large). The ratio is constant and equals speed of light. This means if E0 = 60 V/m, then B0 = 60/(3×108) = 2×10−7 T (very small). Both fields carry equal energy in the wave (uE = uB).
From Maxwell's equations, E and B are related by E0/B0 = c = 3×108 m/s.
Significance: E and B oscillate in phase but E is always much larger in magnitude than B (since c is large). The ratio is constant and equals speed of light. This means if E0 = 60 V/m, then B0 = 60/(3×108) = 2×10−7 T (very small). Both fields carry equal energy in the wave (uE = uB).
3 MarksQ6 (Ex 8.6). An EM wave travelling along
x-axis has E along y-axis. If E0 = 33.4 V/m, find B0 and direction of B.
✓ Solution
B0 = E0/c = 33.4/(3×108) = 1.11×10−7 T
Direction of B: E along y, propagation along x ⇒ B = E × propagation direction requires B along z-axis.
(E × B gives propagation direction: ŷ × ẑ = x̂ ✓)
B0 = E0/c = 33.4/(3×108) = 1.11×10−7 T
Direction of B: E along y, propagation along x ⇒ B = E × propagation direction requires B along z-axis.
(E × B gives propagation direction: ŷ × ẑ = x̂ ✓)
3 MarksQ7 (Ex 8.7). The sun delivers ~1.5 kW/m²
solar energy at Earth's surface. Find E0 and B0 of the EM wave at Earth.
✓ Solution
I = 1500 W/m²; I = ½cε0E0²
E0² = 2I/(cε0) = 2×1500/(3×108×8.85×10−12) = 3000/(2.655×10−3) = 1.13×106
I = 1500 W/m²; I = ½cε0E0²
E0² = 2I/(cε0) = 2×1500/(3×108×8.85×10−12) = 3000/(2.655×10−3) = 1.13×106
E0 = √(1.13×106) = 1063 V/m ≈
1.06 kV/m
B0 = E0/c = 1063/(3×108) = 3.54×10−6
T ≈ 3.54 µT3 MarksQ8 (Ex 8.8). A radio station broadcasts at 600
kHz. Find wavelength. How far is the next crest?
✓ Solution
This is a radio wave (λ > 0.1 m ✓).
λ = c/f = 3×108/(600×103) =
3×108/6×105 = 500 m
Distance between consecutive crests = one wavelength = 500 m.This is a radio wave (λ > 0.1 m ✓).
3 MarksQ9 (Ex 8.9). Name the part of EM spectrum used
in: (a) RADAR, (b) medical imaging of bones, (c) treating cancer, (d) TV remote, (e) sterilisation.
✓ Solution
(a) RADAR: Microwaves
(b) Bone imaging: X-rays
(c) Cancer treatment: Gamma rays (γ)
(d) TV remote: Infrared (IR)
(e) Sterilisation: UV rays (also γ-rays for industrial sterilisation)
(a) RADAR: Microwaves
(b) Bone imaging: X-rays
(c) Cancer treatment: Gamma rays (γ)
(d) TV remote: Infrared (IR)
(e) Sterilisation: UV rays (also γ-rays for industrial sterilisation)
3 MarksQ10 (Ex 8.10). State the reason why microwaves
are used in microwave ovens. Why are metal containers avoided?
✓ Solution
Microwaves (2.45 GHz) match the rotational frequency of water molecules. At this resonant frequency, water molecules absorb microwave energy very efficiently → molecules rotate rapidly → kinetic energy increases → temperature rises → food heats uniformly throughout (not just surface).
Metal containers: Metals reflect microwaves (free electrons in metal oscillate and re-radiate). This means: (a) food won't heat, (b) reflected waves can damage the magnetron source, (c) possible sparking. Hence only microwave-safe glass/ceramic/plastic containers are used.
Microwaves (2.45 GHz) match the rotational frequency of water molecules. At this resonant frequency, water molecules absorb microwave energy very efficiently → molecules rotate rapidly → kinetic energy increases → temperature rises → food heats uniformly throughout (not just surface).
Metal containers: Metals reflect microwaves (free electrons in metal oscillate and re-radiate). This means: (a) food won't heat, (b) reflected waves can damage the magnetron source, (c) possible sparking. Hence only microwave-safe glass/ceramic/plastic containers are used.
🌟 Additional / Exemplar Questions (Q11 – Q20)
3 MarksQ11. A light beam has intensity
3×104 W/m². Find radiation pressure on (a) a perfectly absorbing surface, (b) a
perfectly reflecting surface.
✓ Solution
I = 3×104 W/m²
(a) Absorbing: P = I/c = 3×104/(3×108) = 10−4 Pa
(b) Reflecting: P = 2I/c = 2×10−4 = 2×10−4 Pa
Reflecting surface has double the radiation pressure (momentum changes direction → Δp = 2p).
I = 3×104 W/m²
(a) Absorbing: P = I/c = 3×104/(3×108) = 10−4 Pa
(b) Reflecting: P = 2I/c = 2×10−4 = 2×10−4 Pa
Reflecting surface has double the radiation pressure (momentum changes direction → Δp = 2p).
3 MarksQ12. The amplitude of electric field in an EM
wave is 48 V/m. Find (a) amplitude of magnetic field, (b) average energy density.
✓ Solution
(a) B0 = E0/c = 48/(3×108) = 1.6×10−7 T
(b) Average energy density: uavg = ½ε0E0²×(1/2) × 2 = ½ε0E0²
Wait — total average = ε0Erms² = ε0(E0/√2)² = ½ε0E0²
(a) B0 = E0/c = 48/(3×108) = 1.6×10−7 T
(b) Average energy density: uavg = ½ε0E0²×(1/2) × 2 = ½ε0E0²
Wait — total average = ε0Erms² = ε0(E0/√2)² = ½ε0E0²
u = ½ε0E0² =
½×8.85×10−12×(48)² =
½×8.85×10−12×2304 = 1.02×10−8
J/m³
(Electric and magnetic contributions are equal, so total = 2 × one contribution.)3 MarksQ13. Calculate the photon energy of (a) violet
light (400 nm), (b) red light (700 nm), (c) X-ray (0.1 nm).
✓ Solution
E = hc/λ; hc = 6.63×10−34×3×108 = 1.989×10−25 J·m
(a) Violet: E = 1.989×10−25/400×10−9 = 4.97×10−19 J = 3.1 eV
(b) Red: E = 1.989×10−25/700×10−9 = 2.84×10−19 J = 1.77 eV
(c) X-ray: E = 1.989×10−25/0.1×10−9 = 1.989×10−15 J = 12.4 keV
E = hc/λ; hc = 6.63×10−34×3×108 = 1.989×10−25 J·m
(a) Violet: E = 1.989×10−25/400×10−9 = 4.97×10−19 J = 3.1 eV
(b) Red: E = 1.989×10−25/700×10−9 = 2.84×10−19 J = 1.77 eV
(c) X-ray: E = 1.989×10−25/0.1×10−9 = 1.989×10−15 J = 12.4 keV
3 MarksQ14. The E field of an EM wave is E = 100
sin(2π×108t) V/m. Find frequency, wavelength, B0, and intensity.
✓ Solution
ω = 2π×108 ⇒ f = 108 Hz = 100 MHz (radio/FM band)
λ = c/f = 3×108/108 = 3 m
B0 = E0/c = 100/(3×108) = 3.33×10−7 T
ω = 2π×108 ⇒ f = 108 Hz = 100 MHz (radio/FM band)
λ = c/f = 3×108/108 = 3 m
B0 = E0/c = 100/(3×108) = 3.33×10−7 T
I = ½cε0E0² =
½×3×108×8.85×10−12×104 =
13.3 W/m²
3 MarksQ15. Explain why EM waves are transverse in
nature. What evidence supports this?
✓ Solution
EM waves consist of oscillating E and B fields, both perpendicular to the direction of propagation (and perpendicular to each other). This makes them transverse.
Evidence: (1) Polarisation — light can be polarised (only transverse waves can be polarised; longitudinal waves cannot). A polaroid filter selectively passes E oscillations in one plane. (2) Maxwell's equations themselves predict E ⊥ B ⊥ c. (3) Hertz's experiments confirmed the transverse nature through polarisation observations.
EM waves consist of oscillating E and B fields, both perpendicular to the direction of propagation (and perpendicular to each other). This makes them transverse.
Evidence: (1) Polarisation — light can be polarised (only transverse waves can be polarised; longitudinal waves cannot). A polaroid filter selectively passes E oscillations in one plane. (2) Maxwell's equations themselves predict E ⊥ B ⊥ c. (3) Hertz's experiments confirmed the transverse nature through polarisation observations.
3 MarksQ16. What is the momentum delivered to a
surface per unit time per unit area when a beam of intensity I falls on a perfectly reflecting surface?
✓ Solution
Momentum of EM wave = U/c (for energy U).
For reflection: momentum change = 2U/c (direction reverses).
Force per unit area (pressure) = rate of momentum change per unit area
Momentum of EM wave = U/c (for energy U).
For reflection: momentum change = 2U/c (direction reverses).
Force per unit area (pressure) = rate of momentum change per unit area
P = 2I/c (for perfect reflection)
For perfect absorption: P = I/c. The factor of 2 arises because reflected photons change momentum by 2p
each.3 MarksQ17. Explain why X-rays cannot be produced by
an oscillating charge like radio waves.
✓ Solution
X-rays have very short wavelength (~0.01–10 nm) corresponding to very high frequency (~1017–1019 Hz). To produce EM waves of frequency f, you need a charge oscillating at that frequency.
No practical LC circuit can oscillate at 1018 Hz (would require incredibly small L and C values — physically impossible). Instead, X-rays are produced by decelerating high-speed electrons (Bremsstrahlung) or inner shell electronic transitions in heavy atoms (characteristic X-rays). These atomic-level processes naturally have the required energy scales.
X-rays have very short wavelength (~0.01–10 nm) corresponding to very high frequency (~1017–1019 Hz). To produce EM waves of frequency f, you need a charge oscillating at that frequency.
No practical LC circuit can oscillate at 1018 Hz (would require incredibly small L and C values — physically impossible). Instead, X-rays are produced by decelerating high-speed electrons (Bremsstrahlung) or inner shell electronic transitions in heavy atoms (characteristic X-rays). These atomic-level processes naturally have the required energy scales.
3 MarksQ18. A laser beam of power 10 W is focused on
a spot of area 10−6 m². Find intensity and E0.
✓ Solution
I = P/A = 10/10−6 = 107 W/m²
E0 = √(2I/(cε0)) = √(2×107/(3×108×8.85×10−12))
= √(2×107/2.655×10−3) = √(7.53×109)
I = P/A = 10/10−6 = 107 W/m²
E0 = √(2I/(cε0)) = √(2×107/(3×108×8.85×10−12))
= √(2×107/2.655×10−3) = √(7.53×109)
E0 ≈ 8.68×104 V/m ≈ 86.8
kV/m
3 MarksQ19. Show that c =
1/√(μ0ε0). Calculate the value and verify it equals
3×108 m/s.
✓ Solution
μ0 = 4π×10−7 T·m/A; ε0 = 8.85×10−12 C²/(N·m²)
μ0ε0 = 4π×10−7 × 8.85×10−12 = 4×3.14159×8.85×10−19
= 111.3×10−19 = 1.113×10−17 s²/m²
μ0 = 4π×10−7 T·m/A; ε0 = 8.85×10−12 C²/(N·m²)
μ0ε0 = 4π×10−7 × 8.85×10−12 = 4×3.14159×8.85×10−19
= 111.3×10−19 = 1.113×10−17 s²/m²
c = 1/√(1.113×10−17) =
1/(3.336×10−9) = 2.998×108 ≈
3×108 m/s ✓
This was Maxwell's triumph: speed of EM waves from purely electrical constants equals speed of
light.3 MarksQ20. Arrange in increasing order of
wavelength: γ-rays, X-rays, microwaves, radio waves, IR, UV, visible light.
✓ Solution
Increasing wavelength (decreasing frequency):
Increasing frequency (decreasing wavelength): reverse of above.
Increasing wavelength (decreasing frequency):
γ-rays < X-rays < UV < Visible < IR < Microwaves <
Radio waves
γ-rays: λ < 10−12 m (highest energy) | Radio: λ > 0.1 m (lowest
energy)Increasing frequency (decreasing wavelength): reverse of above.
✍ Score Guide — 20 Questions
NCERT Ex 8.1–8.10: 3 marks × 10 = 30 marks | Additional Q11–Q20: 3 marks × 10 = 30 marks | Total: 60 marks
NCERT Ex 8.1–8.10: 3 marks × 10 = 30 marks | Additional Q11–Q20: 3 marks × 10 = 30 marks | Total: 60 marks
CLASS 12 PHYSICS | FORMULA CAPSULE
Electromagnetic Waves
Chapter 8 — Complete Formula Sheet & EM Spectrum Quick Reference
📅 Section A — Core Formulas
| Quantity | Formula | Key Note |
|---|---|---|
| Speed of EM waves | c = 1/√(μ0ε0) = 3×108 m/s | Same for ALL EM waves in vacuum |
| Wave equation | c = fλ = ω/k | k=2π/λ; ω=2πf |
| E-B ratio | E0/B0 = c | E & B oscillate in phase |
| Displacement current | Id = ε0 dϕE/dt | = C × dV/dt between plates |
| Ampère-Maxwell law | ∮B·dl = μ0(Ic + Id) | Id = Iconduction |
| Intensity | I = ½cε0E0² | Power per unit area |
| Energy density | u = ½ε0E0² = B0²/(2μ0) | E and B carry equal energy |
| Radiation pressure (absorb) | P = I/c | Momentum absorbed |
| Radiation pressure (reflect) | P = 2I/c | Momentum reversed (×2) |
| Photon energy | E = hf = hc/λ | h = 6.63×10−34 J·s |
| Photon momentum | p = h/λ = E/c | Massless particle |
📅 Section B — EM Spectrum Quick Reference
| Type | λ Range | Source | Use |
|---|---|---|---|
| Radio | > 0.1 m | LC oscillators | Broadcasting, communication |
| Microwave | 0.1 m – 1 mm | Klystron, magnetron | Radar, ovens, satellite comm. |
| Infrared | 1 mm – 700 nm | Hot bodies | Night vision, remote, physio |
| Visible | 700–400 nm | Sun, lamps | Vision, photography |
| Ultraviolet | 400 nm – 1 nm | Sun, Hg lamp | Sterilisation, vitamin D |
| X-rays | 1 – 10−3 nm | Coolidge tube | Medical imaging, CT scan |
| Gamma (γ) | < 10−3 nm | Nuclear decay | Cancer treatment, sterilisation |
📅 Section C — Maxwell's 4 Equations (Summary)
| # | Name | Equation | Meaning |
|---|---|---|---|
| 1 | Gauss (E) | ∮E·dA = Q/ε0 | Charges → E field |
| 2 | Gauss (B) | ∮B·dA = 0 | No magnetic monopoles |
| 3 | Faraday | ∮E·dl = −dϕB/dt | Changing B → E |
| 4 | Ampère-Maxwell | ∮B·dl = μ0I + μ0ε0dϕE/dt | Current + changing E → B |
🧠 Memory Tricks
“R M I V U X G”
Radha Mohan Is Very Ugly X-tra Gross. Radio → Microwave → Infrared → Visible →
UV → X-ray → Gamma. Increasing frequency and energy. Decreasing wavelength.
“E/B = c — always!”
The ratio of electric to magnetic field amplitude is always c (3×10⁸ m/s). E is in V/m and B
is in Tesla. Both oscillate in phase and carry equal energy density.
“Accelerating charges = EM waves”
Static charge → E only. Uniform motion → E + B (steady). Accelerating charge → EM
radiation (waves). Only acceleration produces propagating EM waves.
“Reflected → 2× Radiation Pressure”
Absorbed: p = U/c, pressure = I/c. Reflected: momentum reverses, Δp = 2U/c, pressure = 2I/c.
Reflecting mirror has double the radiation pressure of a black absorber.
“Microwave oven = H₂O resonance”
2.45 GHz matches water molecule rotation → resonant absorption → heating. Metal reflects
microwaves → do not use metal containers. Glass/ceramic/plastic are safe.
“Ozone → UV | Greenhouse → IR”
Ozone absorbs UV (protective layer). CO₂ and greenhouse gases trap IR (re-radiated by Earth's
surface). Global warming = enhanced greenhouse effect due to increased CO₂.
🔢 Critical Values
c = 3×10⁸ m/s
μ₀ = 4π×10⁻⁷ T·m/A
ε₀ = 8.85×10⁻¹² C²/N·m²
h = 6.63×10⁻³⁴ J·s
Visible: 400–700 nm
Microwave oven: 2.45 GHz
1 eV = 1.6×10⁻¹⁹ J
hc = 1240 eV·nm
❌ Common Mistakes to Avoid
- All EM waves have same speed in vacuum: c = 3×108 m/s. They differ ONLY in frequency and wavelength. In a medium, different wavelengths travel at different speeds (dispersion).
- EM waves are transverse, not longitudinal: Evidence = polarisation. Sound waves are longitudinal and cannot be polarised.
- Source confusion: γ-rays from nuclear decay (NOT Coolidge tube). X-rays from Coolidge tube (NOT nuclear). Confusing sources is very common in NEET.
- Id is NOT a real current: Displacement current is not a flow of charges. It is the effect of changing E field. But it produces B field just like conduction current does.
- Radiation pressure: Reflection gives 2I/c (NOT I/c). Absorption gives I/c. Don't confuse absorption vs reflection.
- E and B are in phase: In EM waves, E and B reach max and zero together. Don't confuse with AC circuits where V and I can be out of phase.
📱 Practice MCQs for this topic inside our App
📱 Practice MCQs for this topic inside our App
📱 Practice MCQs for this topic inside our App
📱 Practice MCQs for this topic inside our App