EM Waves — Set 5

Correct Answer: A. Decreases

• **Decreases** = In a denser optical medium the speed of light decreases (v = c/n); since frequency is unchanged at the boundary, λ = v/f also decreases by the factor n. • **λ_glass = λ_air/n** — For glass with n = 1.5, wavelength inside the glass is 2/3 of its air value; this compaction of wavelength is what causes the change in propagation direction (refraction). • 💡 Wrong-option analysis: Becomes infinite: an infinite wavelength would mean f = 0 (no oscillation), which contradicts the source continuing to emit at the same frequency; Does not change: wavelength must change when speed changes at constant frequency (λ = v/f); Increases: wavelength increases when light moves to a less dense medium (lower n), not a denser one.

Correct Answer: B. High-speed electrons are suddenly decelerated at a metal target

• **High-speed electrons are suddenly decelerated at a metal target** = When fast electrons are rapidly decelerated at a metal anode, they emit X-rays via bremsstrahlung (braking radiation) and also by displacing inner-shell electrons. • **bremsstrahlung + characteristic X-rays** — The broad bremsstrahlung spectrum is produced by deceleration; sharp characteristic peaks arise when inner-shell vacancies are filled by outer-shell electrons. • 💡 Wrong-option analysis: Atoms vibrate slowly: slow atomic vibrations produce infrared radiation (thermal emission), not X-rays; Cold objects emit heat: room-temperature objects emit far-infrared radiation — producing X-rays requires electron energies of tens of kiloelectron-volts; Sound waves reflect: sound is a mechanical wave and its reflection produces no electromagnetic radiation of any kind.

Correct Answer: A. They are highly energetic and strongly penetrating

• **They are highly energetic and strongly penetrating** = Gamma rays are the highest-energy EM radiation, with photon energies typically from ~100 keV to several MeV, giving them extreme penetrating power. • **photon energy > 100 keV** — Gamma rays can penetrate several centimetres of lead; thick lead or concrete shielding is required for protection in nuclear and medical environments. • 💡 Wrong-option analysis: They have the longest wavelength: gamma rays have the shortest wavelengths (~10^-12 m or less) — radio waves have the longest; They are mechanical waves in air: gamma rays are EM waves, not mechanical; they travel through vacuum at the speed of light; They are always non-penetrating: the opposite is true — gamma rays are the most penetrating of the three common types of nuclear radiation.

Correct Answer: B. Fluorescence

• **Fluorescence** = UV photons excite electrons in certain molecules to higher energy levels; when these electrons relax back they emit visible photons of lower energy (longer wavelength), producing the visible glow. • **UV absorbed → visible emitted** — The emitted visible light has a longer wavelength than the absorbed UV (Stokes' law); fluorescent dyes and 'black-light' posters exploit this mechanism. • 💡 Wrong-option analysis: Magnetic levitation: levitation requires magnetic or electric fields — UV light does not produce the forces needed for macroscopic levitation; Only reflection with no absorption: fluorescence specifically involves absorption of UV followed by re-emission at longer wavelength — pure reflection with no absorption would produce no glow; Superconductivity at room temperature: superconductivity is a quantum low-temperature effect unrelated to UV illumination.

Correct Answer: B. Radio waves

• **Radio waves** = AM broadcasting uses medium-wave frequencies (~535–1605 kHz, λ ~200–560 m) and FM uses VHF (~87.5–108 MHz, λ ~2.8–3.4 m) — both in the radio-wave region. • **AM ~kHz, FM ~100 MHz** — AM waves travel via ground and sky waves over long distances; FM waves have better audio fidelity but shorter range due to line-of-sight propagation. • 💡 Wrong-option analysis: Gamma rays: require nuclear sources and are ionising — completely impractical for broadcasting and hazardous to life; Ultraviolet: UV is scattered and absorbed by the atmosphere — cannot propagate the distances required for broadcast coverage; X-rays: highly absorbed by the atmosphere and require high-energy sources — not usable for audio broadcasting.

Correct Answer: D. Partially horizontally polarized

• **Partially horizontally polarized** = When light reflects from horizontal surfaces (roads, water) near Brewster's angle, it becomes partially or fully polarised with the electric field oscillating horizontally. • **Brewster's angle θ_B = arctan(n)** — For water (n ≈ 1.33), Brewster's angle is ~53°; polarising lenses with a vertical transmission axis block the dominant horizontal component, cutting glare. • 💡 Wrong-option analysis: Circularly polarized: reflected glare is linearly (not circularly) polarised; circular polarisation requires special optical elements not found in natural reflection; Unpolarized always: reflected light from flat surfaces is not fully unpolarised — the partial horizontal polarisation is precisely why polarising lenses are effective; Made of sound waves: light and sound are fundamentally different — glare is an optical (EM) phenomenon, not an acoustic one.

Correct Answer: D. λ/2

• **λ/2** = A half-wave dipole consists of two quarter-wave arms, giving a total physical length of approximately λ/2, which creates resonance at the operating frequency. • **resonance at λ/2** — At resonance the antenna's input impedance is purely resistive (~73 Ω), making it easy to match to a transmission line and maximising radiated power. • 💡 Wrong-option analysis: λ/4: this is the length of a quarter-wave monopole antenna above a ground plane, not a half-wave dipole; λ/10: this is a short antenna that is electrically small and has poor radiation efficiency; 2λ: a full-wavelength dipole creates unwanted side lobes and is not a standard resonant design for basic dipole antennas.

Correct Answer: A. 4.0 × 10^-19 J

• **4.0 × 10^-19 J** = Using E = hc/λ with h = 6.63 × 10^-34 J·s, c = 3 × 10^8 m/s, and λ = 5 × 10^-7 m gives E = (6.63 × 10^-34 × 3 × 10^8)/(5 × 10^-7) ≈ 4.0 × 10^-19 J. • **500 nm = green light ≈ 2.5 eV** — This photon energy (~2.5 eV) is consistent with electronic transitions in atoms; it is in the range relevant to photosynthesis and photovoltaic cells. • 💡 Wrong-option analysis: 4.0 × 10^-29 J: off by 10 orders of magnitude — corresponds to microwave photon energy (~MHz), not visible light; 4.0 × 10^19 J: positive exponent is physically impossible for a single photon — this would be equivalent to ~10^38 eV, far beyond any known particle energy; 4.0 × 10^-9 J: off by 10 orders of magnitude in the other direction — this much energy per photon would correspond to hard gamma rays.

Correct Answer: C. X-rays

• **X-rays** = Wavelengths around 10^-10 m (0.1 nm = 1 Å) fall squarely in the X-ray region, which spans roughly 0.01–10 nm. • **λ ~ 0.1 nm = 1 Å** — This angstrom-scale wavelength is comparable to atomic bond lengths, which is why X-rays are used in crystallography to determine molecular structures. • 💡 Wrong-option analysis: Infrared: IR wavelengths are ~700 nm–1 mm, many thousands of times longer than 10^-10 m; Microwaves: microwave wavelengths are ~1 mm–1 m, even longer than IR and vastly different from 10^-10 m; Radio waves: radio wavelengths range from millimetres to kilometres, the furthest extreme from 10^-10 m.

Correct Answer: B. Gamma rays

• **Gamma rays** = Gamma rays have the shortest wavelengths in the electromagnetic spectrum, typically less than 10^-11 m (shorter than 0.01 nm), corresponding to the highest photon energies. • **λ < 10^-11 m** — Gamma ray photon energies start from ~100 keV and can reach GeV in astrophysical sources; this is why they are the most penetrating and ionising of all EM radiations. • 💡 Wrong-option analysis: Ultraviolet rays: UV wavelengths are ~10–400 nm, far longer than gamma rays; X-rays: X-rays (~0.01–10 nm) overlap with soft gamma rays at the boundary, but the hardest gamma rays are shorter than the shortest X-rays; Microwaves: microwave wavelengths (~1 mm–1 m) are among the longest EM waves, at the opposite end of the spectrum from gamma rays.