EM Waves — Set 4

Correct Answer: D. Infrared (near IR)

• **Infrared (near IR)** = Optical fibre communication predominantly uses near-infrared wavelengths because glass fibres have minimum signal attenuation and dispersion in this region. • **~1.3 μm and 1.55 μm** — These two telecom windows in the near-IR offer the lowest fibre loss; 1.55 μm is the standard for long-haul networks with losses as low as ~0.2 dB/km. • 💡 Wrong-option analysis: Ultraviolet: UV is strongly absorbed by glass, making it impractical for fibres over any significant distance; Gamma rays: require nuclear sources, are highly ionising, and would damage the fibre material — completely impractical for communications; X-rays: highly absorbed by glass and biological tissue; also require specialised generation equipment incompatible with communication fibres.

Correct Answer: C. A changing electric field producing a magnetic effect

• **A changing electric field producing a magnetic effect** = Maxwell's displacement current J_D = ε₀ ∂E/∂t arises from a time-varying electric field and produces a magnetic field just like a conduction current. • **J_D = ε₀ ∂E/∂t** — This term completes Ampere's law, ensuring continuity of current in circuits with capacitors and enabling the prediction of self-sustaining EM waves. • 💡 Wrong-option analysis: A constant potential difference only: a constant (DC) potential produces a steady E-field, not a changing one — no displacement current is generated; A steady flow of electrons in a wire: that is conduction current, not displacement current — displacement current occurs in regions (like capacitor gaps) where there is no charge flow; A changing magnetic field producing an electric current only: a changing B induces an EMF (Faraday's law), which is the converse effect — displacement current is about a changing E producing a magnetic effect.

Correct Answer: C. Radiation pressure

• **Radiation pressure** = EM waves carry momentum p = U/c (where U is energy), and when absorbed or reflected by a surface they transfer momentum, exerting a radiation pressure. • **p = U/c** — Solar radiation exerts a pressure of ~9 × 10^-6 Pa on Earth; this is exploited in solar sail spacecraft concepts. • 💡 Wrong-option analysis: Only heat: EM waves can transfer heat (thermal radiation) but the question is specifically about the consequence of carrying momentum, which is pressure; Only electric charge: EM waves carry energy and momentum, not electric charge — they are electrically neutral; Only magnetic poles: EM waves do not carry magnetic monopoles — no isolated magnetic poles have been observed.

Correct Answer: B. Ultraviolet

• **Ultraviolet** = The ozone (O₃) layer in the stratosphere absorbs a large fraction of the Sun's UV radiation, especially UV-B and UV-C, protecting life on Earth from their harmful effects. • **UV-B (280–315 nm) and UV-C (<280 nm)** — UV-C is nearly entirely blocked by ozone; UV-B is partially filtered — ozone depletion allows more UV-B to reach the surface, raising skin-cancer and ecosystem risks. • 💡 Wrong-option analysis: Infrared: IR is partially absorbed by water vapour and CO₂ (greenhouse effect) but ozone is not the main IR absorber; Microwaves: microwaves largely pass through the atmosphere and are not significantly absorbed by the ozone layer; Radio waves: radio waves are not absorbed by ozone — they are reflected by the ionosphere at certain frequencies.

Correct Answer: C. Cause rotation of polar molecules like water

• **Cause rotation of polar molecules like water** = The alternating electric field of microwaves at 2.45 GHz causes polar water molecules to continuously rotate, and molecular friction converts this rotational energy into heat. • **2.45 GHz resonance** — This frequency is close to the rotational absorption band of water; the kinetic energy of rapidly rotating molecules is what we measure as heat. • 💡 Wrong-option analysis: Turn oxygen into ozone: ozone formation requires UV photons with enough energy to dissociate O₂ — microwaves are far too low-energy for this; Stop molecular motion: stopping motion would mean cooling, the opposite of heating; Increase nuclear binding energy: nuclear forces are unaffected by microwave frequencies — these photons are far too low-energy to influence nuclear structure.

Correct Answer: A. 1 m

• **1 m** = Using λ = c/f with c = 3.0 × 10^8 m/s and f = 3.0 × 10^8 Hz gives λ = 1 m exactly. • **f = 300 MHz** — A 300 MHz wave with λ = 1 m falls in the UHF band; quarter-wave antennas for this frequency would be 25 cm long. • 💡 Wrong-option analysis: 10 m: this would require f = 30 MHz (HF band), ten times lower than 3.0 × 10^8 Hz; 0.1 m: this would require f = 3 GHz (S-band microwave), ten times higher; 100 m: this would require f = 3 MHz (MF/HF), a hundred times lower than the given frequency.

Correct Answer: C. 2 m

• **2 m** = Using λ = c/f with c = 3.0 × 10^8 m/s and f = 1.5 × 10^8 Hz gives λ = 3.0/1.5 = 2 m. • **150 MHz = VHF band** — This frequency is in the VHF (Very High Frequency) band; VHF antennas are typically half a wavelength, i.e. about 1 m long. • 💡 Wrong-option analysis: 0.2 m: this corresponds to f = 1.5 GHz (L-band microwave), ten times higher; 200 m: this corresponds to f = 1.5 MHz (MF band), a hundred times lower; 20 m: this corresponds to f = 15 MHz (shortwave HF), ten times lower than 150 MHz.

Correct Answer: C. Infrared

• **Infrared** = All objects above absolute zero emit thermal (blackbody) radiation; at room temperature (~300 K) peak emission is in the mid-infrared (~10 μm), which thermal cameras detect. • **Wien's law: λ_max × T ≈ 2898 μm·K** — A human body at 310 K emits peak IR at ~9.3 μm; thermal cameras capture this to create temperature maps. • 💡 Wrong-option analysis: Visible: visible cameras detect reflected visible light — they cannot image heat differences in dark environments the way thermal cameras can; Radio waves: radio waves are emitted thermally but at far lower intensities than IR at room temperature — not practical for thermal imaging; Ultraviolet: UV emission from room-temperature objects is negligible compared to IR — UV is associated with high-temperature sources like the Sun.

Correct Answer: A. c = 1/√(μ0ε0)

• **c = 1/√(μ₀ε₀)** = Maxwell derived this expression from his equations; μ₀ is the permeability of free space and ε₀ is the permittivity of free space. • **μ₀ = 4π × 10^-7 H/m, ε₀ = 8.85 × 10^-12 F/m** — Substituting these values gives c = 1/√(4π × 10^-7 × 8.85 × 10^-12) ≈ 3 × 10^8 m/s, confirming light as an EM wave. • 💡 Wrong-option analysis: c = √(μ₀ε₀): this gives units of √(H/m · F/m) = √(s²/m²) = s/m, not m/s — the reciprocal is needed; c = μ₀ε₀: units are H·F/m² = s²/m², not m/s; c = μ₀/ε₀: units are H/F = Ω² (impedance squared), not m/s.

Correct Answer: B. Meter

• **Meter (m)** = Wavelength is the spatial distance between two successive points in phase (e.g. crest to crest), so its SI unit is the metre. • **λ = c/f** — For a 100 MHz radio wave, λ = 3 m; for a visible photon at 600 nm, λ = 6 × 10^-7 m — both expressed in metres. • 💡 Wrong-option analysis: Second: the SI unit of time, which is relevant to period (T = 1/f) not wavelength; Hertz: the SI unit of frequency (cycles per second), not the unit of a distance; Volt: the SI unit of electric potential difference, completely unrelated to wavelength.